Endianness in c using union , little-endian and big-endian storage. Program to Test Endianness using Unions: Program to Test Endianness using Unions: We can use Unions in C to test the Endianness of a system: #include <stdio. This isn't a good "general case" algorithm to test for endianness, though, because: It is recommended to use bitwise operations. Floating point numbers therefore may use different representations on different machines and, in theory, it's even possible that for two processors integer endianness is the same and floating point is different or vice-versa. EDIT: This can also be done using a union. For example, the >> operator always shifts the bits towards the least significant digit. The exception seems adequate, since looking at the ctypes documentation for class Union, it clearly says that it is a base class for unions in native byte @GinuJacob Some relation. 2 Using a designated initializer in the same example, Strange output when using union type members. The best way to deal with endianness in general is to make sure that the code runs on both little- and big-endian host machines. Storing Data Using Strings, A C implementation could choose to always assign all bytes of the union. You shouldn't be asking about conversions between big- and little-endian values, but rather about conversions from a specific endianness to the host endianness, and you can write that code in an endian-agnostic way that's (almost) completely portable: For example: suppose you're reading a 32-bit big-endian integer from a file stream: Also, MISRA allows union just fine, the rule against it is advisory, just to force people to stop and think how they are using unions. Unlike the use of unions, this is expressly allowed by C++'s type system. Given that, I'm not sure there is any value to having buffer - so the second source is a bit cleaner. Related. Big- and little-endianness aren't the only possible orderings, Technically the behaviour on using the "union trick" is undefined in C++, although you'll get away with it in C. On a big-endian machine ((unsigned char *)u. So, size of union is equal to the size of largest member. 01mm), but i get numbers like (4. min[0] will point to the least significant byte of both u. Define a function named swap_endianness that takes a number num as input. Since the little endian machine saves its LSB at lower address; if the integer variable address is type-casted to a character pointer and a dereferencing operation will result in fetching the value stored at the lowest memory location. Which means that it will either represent the MSb/MSB in the binary representation of any datatype as its first bit/byte or last bit/byte. 1 Structure and union specifiers: As discussed in 6. Using a union is not a safe programming practice, The only thing undefined here should be the endianness of the float, that is not specified by any standard. Given that the fields are copied as-is, 16 bits at a time, it doesn't seem like there's any endianness handling in the original code. BigEndianStructure that contains a ctypes. (Except of course signed char and unsigned char). As, you've said endianness is an issue. If you want to know why there is printed ffffff before the value, check: Printing hexadecimal characters in c; endianness; unions; Share. Since unions use the same memory for all its union MessageLengthUnion { uint16_t asInt; uint8_t asChars[2]; }; Then when I get the messages in I put the first received uint8 in . Uninitialized unions can result in undefined behavior. But that doesn't work on this platform, as I end up with a char[1]. “Annex J (informative) Portability issues [] J. This way, things come more naturally and compiler should be expected to do something, since this is so easy to be tracked by a simple data flow optimizer implementation. 465 1 1 gold badge 5 5 silver badges 19 19 bronze badges. Consider that different platforms can have different constraints in memory alignment and endianness. It would be ok to return this directly, I just thought limiting the return On a platform that actually has both of these types, C11 §7. Endianness has to do with the byte ordering of multiple-byte data types in memory. Program to find the machine is which type? The idea is to create a union with a multi-byte data type and check the value of the first byte. on a typical 32-bit platform without an FPU, an 8-byte float that combined a 32-bit significand with an explicit leading 1 and a 32-bit sign/exponent field could be processed much faster than This is sufficient to check for endianness at run time. I won't have access to htons() and similar from the networking library. Using less memory is faster. If a program attempts to access the tests if the variable i type-casted to a character pointer results in the value 1 if dereferenced. Hot Network Questions How to check Syntax of Union in C. 31. As we know that in little endian machine least significant byte of any multibyte data field is stored at the lowest memory "little-endian" and "big-endian" refer to the order of bytes (we can assume 8 bits here) in a binary representation. #include<stdlib. – Weather Vane. 7. 95) If the member used to read the contents of a union object is not the same as the member last used to store a value in the object, the appropriate part of the object representation of the For instance using a union, typedef union { uint32_t word; uint8_t bytes[4]; } byte_check; or casting, uint32_t word; uint8_t * bytes = &word; Please note that for completely portable endianness checks, you need to take into account both big-endian, little-endian and mixed-endian systems. asChars[0] then I access it as the . num = 1; then use endian. A tagged union. so it means Altering the value of any of the member will alter other member values. Picking this name can be hard. But they serve more as annotation facility, just to indicate that we're working with a big-endian or little-endian object. C program to check processor endianness. It's pretty straightforward to deal with structs & whatnot that don't involve bitfields or unions, but because of the convention for which bit is interpreted as the MSB or LSB, as well as byte ordering, it can be rather irritating to convert code over to x86 linux. On little-endian system, LSB can be found on first byte. Commented Jun This function does not correctly reverse the endianness of an input, and even if it did, it still accesses an unset member of a union, which is the same potentially hazardous behavior as the code in the original question – Technical note: Accessing a union member other than the last one stored does not cause a program to violate the C standard. Using bit fields increases readability. Log In Join for free. 3. So I emailed the company that makes the PLC system, and they said that I get a Big endian instead of little endian which I apparently need. In this post, it is mentioned that 8-bit types don't have to have the same memory alignment as 32 bit types, The union usage has undefined behavior according to the ANSI C standard, and thus, should not be used (or at least not be considered portable). 1). One thought is to write the first bit, then second, then third always in a specific order,such as grabbing each decimal value using modulus or bit shifting. The value of at most one of the members can be stored in a union object at any time. How do you convert from color HEX code to RGB in pure C using C library only (without C++ or templates)? The RGB struct may be like this: typedef struct RGB { double r; double g; dou Union and endianness. Note: should be using unsigned char * and uint32_t. The expression (uint32_t *)NetworkOrderFloat is still the address of an array of chars, and accessing it with an lvalue of type uint32_t is against these rules. A pointer to a union object, suitably converted, points to each of its members (or if a member is a bit- field, then to the unit in which it resides), and vice versa. Left-shift by 1 is mutliplication by 2, right-shift is division. 6 using the ctypes library, but I was not able to create a ctypes. pack method to pack the number as a 32-bit integer in network byte order (big-endian). This is because the number is being defined by value. On a little-endian machine ((unsigned char *)u. The result will depend on endianness of the system: // no need to complicate a simple example with dynamic allocation uint16_t out; // note that there is an exception in language rules that // allows accessing any object through narrow (unsigned) char // or std::byte pointers; thus following is well defined std::byte* In little-endian, the bytes are stored in the order least significant to most signficant. Cf. Using bits saves memory. 00f into a union as a It's useful for setting bits in, say, registers instead of shift/mask operations: typedef union { unsigned int as_int; // Assume this is 32-bits struct { unsigned int unused1 : 4; unsigned int foo : 4; unsigned int bar : 6; unsigned int unused2 : 2; unsigned int baz : 3; unsigned int unused3 : 1; unsigned int quux : 12; } field; } some_reg; \$\begingroup\$ AIUI if one of the types you're reinterpreting is not unsigned char, then the union would be necessary to avoid strict aliasing issues. The problem is the memory layout of say, 0x65736c6166 will be different on different endians. read(fileID,4); When I write this to screen the results are as expected: std::cout << fileID << std::endl; >> RIFF Now, the next 4 bytes give the size of the file, but crucially they're little-endian. Byte-endianness: If you inspect bytes of number, then byte which has LSB is implementation defined. 5 7; it says “An object shall have its stored value accessed only by an lvalue expression that has one of the following types: a character type. For instance, short x=0x1234 would be stored as 0x34,0x12 in little-endian. I would like to construct an unsigned 32-bit int using Python3 ctypes bit field, as follows: c_uint = ctypes. memcpy it around); that's explicitly allowed as an exception to the normal aliasing restrictions, and C would not work without it, nor would any program written in C. e. In that case, a compiler-friendly way to alias values supported by C is using a union: Maybe endianness reversed the rule. The typedef name uintN_t designates an unsigned integer type with width N and no padding bits. Populates the char array part of the union TestUnion with a, b, c, and d. Issue 1: Endianness. It is pretty easy to check the endianness of a system. You're right when you mention using htons and ntohs for shorts. The union allows a uint32_t to lie on the same bytes as the char array and be interpreted in whatever the machine's endianness is. min[0] will contain the highest 32 bits. There is no such thing as floating point endianness or integer endianness etc. Unions were designed to save space, using the same memory to store two or more different types of data. @tomlogic - Endianness is the ordering of individually addressable sub-units (words, bytes, or even bits) within a longer data word. ” template <class T, const int size = sizeof (T)> class cTypeConvert final { public: union { T val; char c [size]; }uTC; }; In my case, using a string of chars, I just have to feed cTypeConvert::uTC. Endianness refers to how bytes are ordered in memory and how they must be interpretted. Now, I think I can't be sure the value bits in a char are arranged in the same way as in an int, therefore reading *r could give any nonzero value on a little-endian machine. 2), and if an object of this standard-layout union type contains one of the standard-layout structs, it is permitted to inspect the common initial sequence of any of standard-layout As @Martin Beckett said, %p asks printf to print a pointer, which is equivalent to %#x or %#lx (the exact format depends on your OS). won't be sufficient. If you're using Visual C++ do the following: uint32_1234 *pu32; { union { uint32_1234 u32_1234; uint32_t u32; } bou32; bou32. ” “Access” in the C standard means reading or writing (C 2018 3. asChars[1], the second in . I need to ensure The smallest unit that is addressable in C is always a byte (called char in C). union a { char a; int b; float c; }; std::variant<int ,char,float,double,long> v; Now that we have some historical context, let‘s analyze how endianness affects C programming constructs like pointers, unions and type punning. Using union / structs to create generalized list objects in pure C. Ugh, endianness. buffer). Everything is most well-defined for unsigned integral types. Improve this answer. Endianness is not a property of the struct but a property of the architecture that is running the code. Topics at a glance: After arrays and their inherent address abstraction mechanisms, I’ll now Bitfield with struct and union in C (gcc) with endianess. 00f into a float will cause it to hold the bytes 0x80 00 00 00, and also specifies that storing byte values 0x80 00 00 00 into a long will cause it to hold -2147483648, then in the absence of any rule to the contrary, those facts would effectively define the behavior of storing 1. This unique feature allows developers to create more flexible and compact Key words: Struct and union, bit manipulation using union, bit-fields in C structs. c" in the real world depends on platform endianness, and in facts trying to read a member of a union when you wrote in another one is Unspecified Behavior according to Always initialize the members of a union before using them. }; What is the difference between the following types of endianness? byte (8b) invariant big and little endianness half-word (16b) invariant big and little endianness word (32b) invariant big and lit Here is a union containing two types of variable int and char. . It raises TypeError: This type does not support other endian. asked Jan 19, 2012 at 2:04. Using Unions for "type punning" is fine in C, and fine in gcc's C++ as well (as a gcc [g++] extension). Endianness refers to the order of bytes in a larger value. – fortran. h> union check_endian { unsigned int value; char r; }; union check_endian endian; int main(void) { endian. n1570, 6. This technique blows up when your longs are 64 bit and your floats are not. Follow edited Aug 16, 2012 at 15:41. For GNU gcc there are some macros must be used: __attribute__ ((packed)) and __attribute__ ((endianness (LITTLE_ENDIAN))) supported starting from GCC v. However using a union you can easily create a function which writes binary of a short integer to a file one byte at a time. To facilitate this, the "main memory" of the vm is set up like this: typedef union { int i; float f; }word; memory = (word *) malloc(mem Do you guys see any issue with this if this is set/get stored/read on same host (same endianness) Is below a better way of doing this ` union { uint64_t entryid; struct { uint32_t entryid1; uint32_t entryid2; }entry; }; ` In my recent post about endianness, I was told that one should unionize the types because not doing so can cause UB. Anything not 0 is truthy in C, but the canonic value for true is 1. @luket: no difference, as a matter of fact, both methods invoke undefined behavior because of the strict aliasing rule. I'm looking forward for a way whereby I can find the endianness without implicit/explicit casting, loops, switch, inbuilt functions, macros. 5/7. Thats it. Key words: Struct and union, bit manipulation using union, bit-fields in C structs. I've decided to only support 32 bit integers and 32 bit float values in the vm's memory. ) Note how this needs to be explicit about the endianness of the serialized data, which is something you always should care about The clean way would be to use a union, like in. The union test works if you use fixed-size types. In this part, we only declare the template of the union, i. 6. Thus, uint24_t denotes such an unsigned integer type with a width of exactly 24 bits. So, given this: union { unsigned char x; struct { unsigned char b1 : 1; unsigned char b2 : 7; }; } abc; abc. linux 64 bit. i after setting t. , we only declare the members’ names and data types along with the name of the union. I can't find any info in documentation or googling around, so hopefully somebody here can help me. The check below detects big and little endian, as well as other more exotic byte orderings. I found this question: endianness conversion, regardless of endianness, however, the answers only provide functions. i = 1 and see whether x. c_uint32 class Flags_bits(ctypes the values are 21 43 65 87. No memory is allocated to the union in the declaration. Back in the days when memory was scarce, unions were popular to re-use memory. Both C and C++ explicitly permit any type of object to be accessed as an array of char (and consequently, through a char*). How can I define number variables by memory Beside the ways of using strucures / unions with byte-size members you have two other ways. So, I write a little function to flip the bytes, based on a union: I thought a better way would be to use union representation like so: typedef union { struct { uint8_t d; uint8_t c; uint8_t b; uint8_t a; }; uint32_t n; } word32; Using this method I can assign word32 w = 0x11223344; then I can access the various parts as I This question is not asking why you wouldn't want to do this, I very very explicitly need to do this. int is_little_endian(){ int temp = 1; return *(char *) temp;//Returns 1 if it's a little endian machine. Or more specifically, it's not clear exactly what these objections are referring to. Since this is in (old, pre-C11) C, the inner union must be given a field name in the outer struct. @Tomek: This does not violate the strict aliasing rule. edit: regarding endianness - I've never seen a very clean way. You cannot access a bit directly. Another way to swap the endianness of a number is by using the struct module in Python. Most are used in little endian, but you may switch to big endian Is it possible to write a macro in C which takes an uint32_t and converts it to big endian representation no matter if the target system is little or big endian such that the macro can be evaluated at compile-time for a constant?. Follow edited Feb 5, 2013 at 15:17. x ); The bitwise operators abstract away the endianness. For example, given the following 16-bit values 0x1122, 0x3344, 0x5566, 0x7788 , the little-endian representation would look like 22 11 44 33 66 55 88 77 , whereas reversal of the bytes would look like 88 77 66 55 44 33 22 11 . union Data myData = {. 1 Unspecified behaviour [] The value of a union member other than the last one stored The only time you have to care about endianness is when you're transferring endian-sensitive binary data (that is, not text) I'll touch on one not-yet-mentioned: Unions. It's quite possible to have two compilers on the same computer use opposite ordering for bitfields. Some critical ideas like pointers, unions and type punning have slightly different behavior depending on endianness. This might be a noob question, but is there a way to perform a byte swap between big and little endian on a __m512i vector without having to iterate through each value? I really want to avoid using unions, as they can be slow. but when I am executing the following code, @CHris: "Hairy pointer casts", aka raw memory reinterpretation, are generally UB, except if you reinterpret it as an array of characters. Commented Jul 25, 2017 at 15:27. iValue = 0}; // Initializing to zero 5. Either a machine is little-endian, or its big-endian. Use the struct. 6 – Dmitry Ponyatov. Here are some architecture (ARM version 3 and above, Alpha, SPARC) who provide the switchable endianness (support bi-endianness) feature. It's also important to remember that for optimum portability static_cast should be used, because reinterpret_cast is implementation defined. 11. From the ISO/IEC 9899:1999 (C99) standard: Annex J - Portability Issues: 1 The following are unspecified: — The value of padding bytes when storing values in structures or unions (6. In the following example: union { int a; int b; int c; } myUnion; This union will take up the space of a single int, rather than 3 separate int values. In other words, just because you jump from a little-endian to a big-endian machine, you can't assume that you should reverse all bit fields. timrau. This is quite important when memory is more scarce, such as in embedded systems. – My question is about type punning in C using unions. So byte[0] is 0, byte[1] is 1, etc. A legitimate use of union is to store different data types at the same place, preferably with a tag so that you know which type it is. min values and u. Your code is independent from endianess, it will print 66051 on little and big endian machines. This answer wouldn't work for us. h> #include<stdint. 1: “access (verb) <execution-time action> to read or modify the value of an object”). There are also middle-endian and mixed-endian machines. Using fwrite() incurs encurs excessive overhead for simple operation. h> union check_endian { unsigned int value; char r; }; Code Explanation If your machine is little endian, the data in the memory will be something like the expression below:. This is called "type punning", and it is Prior to C++20, the only valid answer is to store an integer and then inspect its first byte through type punning. – That function will always return zero since your constants are stored in native endianness of the system as well. Enumerations. Big-endian is the opposite. But unsigned char is the one escape valve type that allows pointer-punning for a reinterpretation of the stored object. If I gave you the number "4172" and said "if this is four-thousand one-hundred seventy-two, what is the endianness" you can't really give an answer because the question doesn't make sense. The approach is not highly portable, but may have met the original limited needs. While you could theoretically force the in memory representation to be of a given endianess, that would force conversions from platform to struct endianess in all reads and writes to each field for something that is not observable from the outside. fwrite an integer depends on endianness, but is there a way to write an integer 0x00000004 to a file such that it can be fread always as 0x00000004 regardless of machine it's run on. In this post, "strict pointer aliasing violation" is mentioned as a problem when not using unions. (endianness, alignment issues). 2 Avoiding Type Confusion. As far as I understand __be* or __le* types can be used for endian dependent variables. Ask Question Asked 6 years, 4 months ago. 2. As a systems developer, comprehending this is crucial to In the world of C programming, unions provide a powerful and memory-efficient way to store different data types in the same memory location. h> #include< Skip to How to handle endianness in C. Here’s an example: Compile and run the program to We can use Unions in C to test the Endianness of a system: #include <stdio. like (0. this function is expected to return a truthy or falsey value. MISRA does not allow union for the sake of storing multiple unrelated things in the same memory area, such as creating variants and other such nonsense. Conclusion There is no meaning to say which endian is better, there is no advantage of using one endianness over the other as both only define the byte sequence The values on which the endianness is tested, won't never reach the system memory thus the real executed code will return the same result regardless of the actual endianness. Checking the endianness of your system. The above code is generally UB, since it is UB to read the value through such a char * \$\begingroup\$ I originally tried a union but got errors from G++ for undefined behavior. asked Oct 26, 2012 at 18:22. On the contrary, if your machine is big endian, it will look like the expression below:. I have found talk of this in other posts as well. e. are platform/implementation dependent, For most programmers, details of computer architecture are of no interest or importance. union U { long l; int i; short s; char c[2]; } u; But what does or the low order byte of i isn't guaranteed to be c[0] (endianness) – Spudd86. – C implementations that use anything other than IEEE-754 formats for floating-point numbers are very rare, even though on many platforms other formats could be processed more efficiently [e. When you cast a smaller signed number to a bigger signed number you Regarding using the union to access the raw value: While it works in practise (on most machines), there might be obscure machines where this does not work, as the C standard does not guarantee anything. float->uint32_t or double-> uint64_t would need a union. bytes[0] to judge. However, over the time, the programmers have widely used unions for a completely different application: extracting smaller parts of data from a larger data Approach#2: Using struct. 23k 4 4 gold badges 53 53 silver badges 66 66 bronze badges. Its just binary representation endianness. (As always when using bitwise operations, beware of signedness. The standard (available in the linked online draft) says in a footnote that it is allowed to access a different member of the same union than the member previously written:. If you have a thirty-two bit value to store you can have the array four long. h> #include<stdio. – No, bitshift, like any other part of C, is defined in terms of values, not representations. @Lundin: if there are any padding/alignment issues with the union then all bets are off anyway - casting a union to a 64 bit int is never going to be completely reliable. You can combine union and bit-fields to extract bits too, but note, this is endianness-dependent, which is why this is not recommended, a sample here for your learning though: Processor endianness is unrelated to bit field ordering. Again, as defined, my union, the endianness is off. I'm trying to copy unit64_t data to uint8_t array using memcpy as below. 5, a structure is a type consisting of a sequence of members, whose storage is allocated in an ordered sequence, and a union is a type consisting of a sequence of members whose storage overlap. 1. So if the machine is little endian, 0 is in the low order byte and 3 is in the high order byte of value. due to your system seems to use big endianness) in both cases. paIncrease paIncrease. In out applications in the real world, there is need for performance and complex atomic operations. Access the Same Data in Different Ways Using a Union. union { uint32_t the_int; uint8_t fragments[4]; } You wouldn't test the endianness of your computer at runtime! The code has to be compiled for one and one endianness only so you can do it at compile-time anyway. C uses arithmetic bit labels -- since bit is a portmanteau of binary digit -- and the resemblance to little-endian byte order makes stupid people believe they are somehow related. this code will work too. Using a union works with gcc because this compiler explicitly condones type pruning via unions, but the official way to do this is using memcpy. The 66 on the end of my constant will go in the first byte on little endian systems, and on the last byte on big endian systems, yet the number is the same. – To get an idea of the usefulness of union. But it is copying in reverse order. The main purpose of using "union" in C/C++ is to provide a datatype that could store anything. Original author may want to overlay struct iphdr on a hardware address, play with union magic or achieve a certain serial communication order. In C++ you can let the union be anonymous. However, this doesn't mean you are safe to completely ignore endianness when using them, for example when dealing with individual bytes in a larger structure you cannot always assume that they will fall in the same place. The c/c standard library defines no function specifically designed to write short integers to a file. After doing more reading, I think I will simply get rid of the RtuRxFields struct and create utility functions getStartAddress(RtuRxFrame* frame) The initialization has two set of braces because the inner braces initialize the bytes array. This is explicitly permitted by the aliasing rule in C 2018 6. Using bits allows for more complex atomic operations. 1,329 1 1 gold badge 10 Get introduced to the endianness, i. The popular x86 architecture uses little-endian byte ordering. The condition check at line, if ( * ( ( char * ) & i ) == 1 ) puts ( "little endian" ) ; The following example shows how an Endian conversion function could be implemented using simple C union s: union u {unsigned long vi; unsigned char c[sizeof @ouah: the C standard knows nothing about endianness, so we are already going out of the standard domain and working on implementation-specific behavior (and I don't think you'll ever find a compiler implementing unions differently or an optimizer messing with them). A C implementation may do different things in Learn about packing and unpacking data with unions in C language. – The first source has buffer in the union, but the size is 0. if "bit order" is an issue, my question 2 is why another struct ip_options doesn't care about endianness? It is an Unions are usually used with the company of a discriminator: a variable indicating which of the fields of the union is valid. On big endian systems, htons (as well as ntohs, htonl, and ntohl) are defined as no-ops, while on little endian systems they perform a byte swap. Details and more examples can be found in this article. Those days are long gone and using unions for that purpose adds needless complexity. c[0] or x. Don't do it. u16 Type punning via unions in general is UB and I'm not aware that there's an exception for char or char[] (as exists for pointers, all of can be legally casted The IEEE754 specification for floating point numbers simply doesn't cover the endianness problem. The integer i is set to value 1 in the main(). If a union genuinely describes your data, as it does in the example you give, then it is a perfectly reasonable thing to do. As already mentioned, it only affects the order of the bytes of a variable, and not the order of bits inside each byte. You will need to consider processor endianness and memory alignment requirements. So the sizeof() must use the union, not the sizeof(pb. However, when char is used (as opposed to unsigned char) the set of things you can do with reinterpreted memory is limited. After setting the data for one member, then trying to access the other, I'll get errors that prevent the code from compiling. From a quality of implementation point of view, I would expect both the union trick and reinterpret_cast to work, if the union or the reinterpret_cast is in the same functional block; the union should work as long as the compiler can see that the ultimate type is a union (although I've used compilers where this wasn't the case). From N1570 6. 1 p2 gives you all the needed guarantees (given you know endianness):. In C89 Unions can only be declared using variable or variable references and then must be initialized using the variable. Even embedded developers, who normally do concern themselves with @Magnetron I'm using a PLC program, that should give me real values. Using a union to convert a float's representation to 16 The size of a union is sufficient to contain the largest of its members. Endianness doesn't affect the order of the members. 61845E-41) instead. struct my_variant_t { int type; union { char char_value; short short_value; int int_value; long long_value; float float_value; double double_value; void* ptr_value; }; }; [ Note: One special guarantee is made in order to simplify the use of unions: If a standard-layout union contains several standard-layout structs that share a common initial sequence (9. Get introduced to the endianness, i. value = 0x1; if (endian. Accessing such a union member results in an unspecified value (not undefined behavior), and, per C 1999 4 3, “shall be a correct program and act in accordance with 5. ": In ISO C++ the union definition would be ill-formed, not UB, because anonymous structure members do not exist in C++. The usual way to detect endianness at runtime (because there is no reliable way to do it at compile time) is to make a union { int i, char c[sizeof(int)] } x; x. But controlled type punning, where padding/alignment and endianess has As such, unions and bitfields are used all over the place. Commented Jun 16, 2013 at 17:30. c[sizeof(int)-1] got set. Algorithm. I usually do something like (this coding was done in Visual Studio in C++): union bytes4 { __int32 value; Bit shift operators don't give different results depending on endianness. 20. For example, let's say you want to create your own Variant type:. If you do (unsigned)number & 1, it will always give you LSB, and endianness is irrelevant. min[0] (of type uint32_t) will contain the lowest 32 bits of the 64 value. Changing endianness or byte order within a union. Back when I did almost all my programming in C, there were two (portable) techniques using unions that I relied on pretty heavily. A C implementation could choose to always assign only the bytes of the member assigned and to leave the others unchanged. What your predecessor did may work, but it is an unreliable and dangerous approach -- the serialization order of bitfields within a struct is implementation-defined, and there is no requirement for it to be consistent with the endianness your CPU uses for integers (which, btw, is not necessarily the same as the endianness used for hardware floating point!), the I read this first element using: char *fileID = new char[4]; filestream. Bit-endianness: It's impossible to know the order of bits in single byte The reason this question pops up is usually because people confuse byte order (endianness) and bit labeling. Here is a correct way to set two octets as uint16_t. Could this union give unexpected values on a little endian machine. 3. 2. If I have a union, C standard guarantees that the union itself will be aligned to the size of the largest element. 1. struct tagTest { union { struct { uint16 A:3; uint16 B:3; uint16 C:3; uint16 D:3; uint16 E:3 Union and endianness. Converting hex color to RGB and vice-versa. union { int num; char[sizeof(int)] bytes; } endian; endian. Strings are (at least in C and C++) an array of bytes so endianness doesn't apply. Using unions to save memory is mostly not done in modern systems, since the code to access a union member will quickly take up you can write to the individual bytes, depending on platform endianness, then interpret those 4 raw char bytes as a IEEE float. You actually can do what you mention in the last paragraph using multicharacter literals, though it's implementation defined exactly how it works and the string must be no longer than sizeof(int). Modified 6 years, 3 months ago. . – huseyin tugrul buyukisik. "" I'm not sure I understand your point. 0. As I just discovered thanks to Nim's actual answer - the Java VM I'm debugging on Android is using LITTLE_ENDIAN while our native code runs in BIG_ENDIAN - thus I need to know this in order to know to properly convert between them when sending the bytes for a Well, for those reading this, but not much interested in unions: the size of union is as of it's largest data member, but all data members actually point to the same physical data each in its own way. Unless you treat a set of bytes as a multi-byte unit, endianness does not apply. What is the quickest way to reverse the endianness of a 16 bit and 32 bit integer. If you're on a big-endian machine when you access the union as a 32-bit long it will be represented as 0x61626364; if it was little-endian the long would be 0x64636261. They were not meant to be used to extract bytes, nibbles, or bits, nor to implicitly cast data. Viewed 314 times The best way is probably type punning over union. SendOverTCPNetwork(m_PlaceHolder); I see, as per this answer "type punning in unions" is allowed in C, but not in C++, though many compilers like GCC support it. u32_1234[0] = What this means in terms of "what do I get if I try to read t. Doing the bitshift operations contained in other answers When a variable is associated with a union, the compiler allocates the memory by considering the size of the largest memory. For example, in ARM Cortex-M3 the implemented endianness will reflect in a status bit AIRCR. Endianness affects all integers, even when they are members of classes. Using a union might be better style since I want to define variables in structure based on endianness of machine , This is what I tried #define IS_BIG_ENDIAN (!(union { uint16_t u16; unsigned char c; }){ . The latter is explictly allowed in C. a << n, for example, is specified as giving the same value as multiplying a by 2, n times. This means printf expect an int or a long (again depends on OS), but you are only supplying it with char so the value is up-cast to the appropriate type. c++ trim and pad unsigned int in hexadecimal for #RGB value. Using ntoh / hton and masking out the high byte of the 4-byte integer before or after the conversion with an bitwise and. You're basically claiming that "some people" have said "some thing", in some context you have not This processor is known as the Bi-endian. There are some circumstances where bit shifting gives undefined/unspecified behaviours, but that's not related to endianness. --WRONG ANSWER END--- Why the union bother? It is fine to cast any pointer to a char pointer (and then do what you want with the memory, e. Topics at a glance: The wonders of union: Vagaries of behavior; Same memory, different perspectives; Byte ordering (endianness), bit ordering, alignment etc. min[0]) will contain the most significant byte of both u. 54. I have tried a code but it uses explicit casting. The "char value" is represented as . Commented Dec 7, 2018 at 8:53. Portability of using union for conversion. The syntax of union can be defined into two parts: C Union Declaration. When referring to machines, it's about the order of the bytes in memory: on big-endian machines, the address of an int will point to its highest-order byte, while on a little-endian machine the address of an int will refer to its lowest-order byte. In a previous article, we discussed that the original application of unions had been creating a shared memory area for mutually exclusive variables. : struct HEADER { unsigned int preamble; unsigned char length; union { unsigned char all; struct CONTROL control; } uni Also, in case sparse is used for C code analysis, they add gcc specific flag __bitwise. On a side note, I don't think there's a guarantee (or perhaps not even a reason to believe) that bit ordering in bitfields follow byte ordering in memory. r == 1) Some critical ideas like pointers, unions and type punning have slightly different behavior depending on endianness. My argument is that the only valid way to perform that check during runtime is to examine an int variable using an unsigned char pointer (since other ways of type punning inevitably contain undefined behavior): Learn about the little-endian and big-endian byte orders used by computers. g. This is enough because there are no bytes with The same idea but different trick. Suppose I have the following code in C: unsigned char myID[10] = "211866744"; How will this array be saved in memory? c; x86; endianness; stack Using a union to resolve compiler warning: Your proposal breaks “strict aliasing rules”, the first time when it does *(uint32_t *)NetworkOrderFloat. Casting that char array to float would not have the same result. Impact of Endianness on Key Concepts in C. "But in CPP, the above is UB. If you want to see the value as a 32-bit integer, do this Is there a built-in way to ensure the endianness of multi-byte types in C++ streams? In particular, I want to use read() and write() to read/write small char arrays to/from a stream. ENDIANNESS and compiler cannot know this value in compile time. I figured the current method would be a safe bet, but completely overlooked how the data would be accessed through an array. KNfLrPn KNfLrPn. You may also find htonl and its opposite useful too. Quiz on Union. Hence, it determines the endianness in use. Therefore, for single byte-width data types you do not have to worry. As I know, default endianness of ARM is little endian. The closest way to get to accessing bits would be to define a data type called bitpointer and define some functions or macros for it: I am trying to port some C code to Python 3. c[] array from size to 0 and read as the type I I believe that C++11 doesn't provide means to programatically determine the endianness of the target platform during compile time. Union. Improve this question. The endianess: LITTLE Endian Setting bit 0: 00000001 = 0x01 Setting bit 1: 00000010 = 0x02 Setting bit 2: 00000100 = 0x04 Setting bit 3: 00001000 = 0x08 Setting bit 4: 00010000 = 0x10 Setting bit 5: 00100000 = 0x20 Setting bit 6: 01000000 = 0x40 Setting bit 7: 10000000 = 0x80 @RBornert: "This looming freedom for the compiler seems to be a full-stop for many who claim "you cannot trust bit-fields", or "bit-fields are not portable. Can you suggest a better solution ? – the output on an x86-64 PC is: The size of the union is: 1 byte. Further any two data types of the same guaranteed fixed size will always have the same memory alignment in C, otherwise a lot the C standard guarantees elsewhere would break, like casting from/to void * or creating memory for these types using malloc, as malloc will treat all equally sized data types equal, so does memcpy. And using memcpy() avoids the strict-aliasing problem. I have updated the union struct and bitfield in C fixing both the endianess of the bit-field and the warning message if the This example C program implements a simple logic to detect endianness of the machine. Although, I agree that the other "classic method" (cast of the pointer to char *) does not c; endianness; unions; Share. The following union is a common tool in SIMD/SSE programming, and is not endian-friendly: union uint128_t { _m128i dq; uint64_t dd[2]; uint32_t dw[4]; There seems to be a misunderstanding here about what endianness is. Utility of a Union. In practice C++ compilers support the anonymous structure as a language extension in non-strict modes and also support the type punning as an extension for compatibility with C, at least if the types are trivial and Unions allow data members which are mutually exclusive to share the same memory. Share. It allows you to use the same memory area with different variable representations, for example by giving each struct member individual names, while at the same time remaining able to loop through them. The advantage of using the above solution is that it at least does not suffer endianness problems and it is somewhat future-proof. If you had to be portable, you can probably compare val The CPU, on the other hand, is designed in such a way that it expects the data words to be in a specific order. union name {type1 member1; type2 member2;. The "cast through a union" hack from the link you posted results in undefined behavior (reading from a member of a union other than the last one written to results in undefined behavior). The casting through (void*) is superfluous in C, but is technically necessary in C++. I figured I wanted to just conditional-test the bit-and of the bits of the pixel with 0x00E0E0E0 (I could have wrong endianness here) to get the dark pixels. The following question refers to x86 assembly, and little endianness. Commented Dec 29, 2009 at 10:03. @NicolBolas: If an implementation documentation indicates that storing 1. Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers; Advertising & Talent Reach devs & technologists worldwide about your product, service or employer brand; OverflowAI GenAI features for Teams; OverflowAPI Train & fine-tune LLMs; Labs The future of collective knowledge sharing; About the company You can do useful things with this technique, but it is very hardware and compiler dependent. But, "type punning" via unions has hardware architecture endianness considerations. asInt part of the union in the rest of my program. b1 = 1; printf( "%02x\n", abc. A much better bet is to just use an system API that answers your question. x = 0; abc.