Gcc bitfield order. But I need to be able to vary that.

Gcc bitfield order __attribute__ ((__packed__)) (so instead of 6 bytes it gets packed into 4 bytes, which you can check with __alignof__ and sizeof). rsmith: /// to be used while performing partial ordering of function templates. unsigned ExplicitCallArguments; unsigned ExplicitCallArguments; union {union An implementation may allocate any addressable storage unit large enough to hold a bitfield. Download Raw Diff; Details. The following code examples show this: #pragma pack(2) /* The container of b must start at a 2 byte alignment boundary, and must * have a size no larger than the container type, in this case Bitfield with struct and union in C (gcc) with endianess. Complete non-atomic memory model support issues for data races. Hot Network Questions The order of allocation of bit-fields within a unit (high-order to low-order or low-order to high-order) is implementation-defined. If a zero-length bitfield is inserted after a bitfield, foo, and the alignment of the zero-length bitfield is greater than the member that follows it, bar, bar will be aligned as the type of the zero-length bitfield. It may be authoritative. Rather than What is the order of members in a bitfield if some members are complete? I know that the order of a bitfield is undefined, but also that the first member in a struct is at the starting position of the struct in memory. how to use struct and bitfields to ensure precise bit order. 我到底做了什么：调查中的东西tree-dump. Permalink. As such it may be necessary to use a preprocessor directive to check for the endianness of the machine. The bit-field is size bits wide and starts at bit pos. 1 where it works and 4. where, data_type: It is an integer type that determines the bit-field value which is to be interpreted. – MSalters. sina. 1, C99 and C11 6. In no particular order . This means that whether vbias inhabits the most significant bit or least significant bit of the unit (int) is implementation-defined. Commented Apr 13, 2023 at 7:20. 3. current implementation) telling to start from bit 0 and upward is a good thing. (high-order to low-order or low-order to high-order) is implementation-defined. different between C struct bitfields on char and on int. So kernel and system developers are free to take advantage of any gcc property they feel like. Next: Qualifiers implementation, Previous: Hints implementation, Up: C Implementation . The fact that a 3-bit field is being treated as 7 means that it's being treated as an unsigned, so when you overflow it 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 It depends on the way the library's compiler packed the bits into the struct, which is not standardized. 在网络传输中，采用big-endian序，对于0x0A0B0C0D，传输顺序就是0A 0B 0C 0D，因此big-endian作为network byte order，little-endian作为host byte order 为什么X86存储会使用little-endian，起初我想对于位运算，尤其是位 After the bitfield variables the values aren't right anymore. lowest order bit in little endian and highest order bit in In gcc 5. gcc-4. c:10:26: warning: implicit truncation from 'int' to bitfield changes value from 2 to -2 [-Wbitfield-constant-conversion] struct st obj={1, 2}; ^ 1 warning generated. (sign_extract: m loc size pos) This represents a reference to a sign-extended bit-field contained or starting in loc (a memory or register reference). bit data type in C language. The order of allocation of bit-fields within a unit (high-order to low-order or low-order to high If insufficient space remains, whether a bit-field that does not fit is put into the next unit or overlaps adjacent units is implementation-defined. Some If you change the bit-size of your bitfield member, you have to change all masks as well (Surely you could use a macro or a constant, but it is still ugly). Does this mean the machine places these bits from right to left? This code works with GCC with less than -O2 optimizations or in clang. html 有些信息在存储时，并不需要占用一个完整的字节， 而只需占几个或一个二 Tasks for GCC 4. 4 for x86): 9 Why in An implementation may allocate any addressable storage unit large enough to hold a bitfield. On a little-endian x86-64 CPU using MinGW's GCC, the output is: After changing a: 0x00000001. 1. 4でbit fieldがどのように格納されるかを調べた。 bit fieldは規格上どのように格納されるかは規定されていない。 Since your system has 64-bit architecture, most compilers would pad in 8-bytes. If you get the same behavior with those removed, it looks to me like a compiler bug. Macro: BITFIELD_NBYTES_LIMITED Like PCC_BITFIELD_TYPE_MATTERS except that its effect is limited to aligning a bit-field within the structure. 9 Structures, unions, enumerations, and bit-fields. It's quite possible that no action is needed at all. Structures unions enumerations and bit-fields implementation (Using the GNU Compiler Collection (GCC)) Next: Qualifiers, Previous: Hints, Up: C Implementation-Defined Behavior . 11 Structure-Layout Pragmas ¶. Regarding the use of non-int bitfield types, The order of allocation of bit-fields within a unit (high-order to low-order or low-order to high-order) is implementation-defined. A bit late to the party but with current GCC (tested on 6. 2 where it's not implemented) there is finally a way to declare that a struct should be kept in The [[gnu::packed]] part suggests that both sides are compiled with GCC? Member bitfield order is not defined in C++. For compatibility with Microsoft Windows compilers, GCC supports a set of #pragma directives that change the maximum alignment of members of structures (other than zero-width bit-fields), unions, and classes subsequently defined. Figure 9 : Bit-Fields#. Special expression codes exist to represent bit-field instructions. 2, 6. 10. 文章浏览阅读3. The order of allocation of bit-fields within a unit (high-order to low-order or low-order to high-order) is implementation-defined. The alignment of non-bit-field members of structures (C90 6. If enough space remains, a bit-field that immediately follows another bit-field in a structure shall be packed into adjacent bits of the same unit. 2. 比特序/ 位序 /bit numbering/我们知道一个字节有8位，也就是8个比特位。从第0位到第7位共8位。比特序就是用来描述比特位在字节中的存放顺序的。通过阅读网页的内容，关于比特序我们得到下面的结论：(1)比特序分为 Yeah, it is not possible. for 1 access it might not matter much, but for many reads/writes in a big loop If insufficient space remains, whether a bit-field that does not fit is put into the next unit or overlaps adjacent units is implementation-defined. Note that I do not want or need the bitfield's width to be able to change mid-program, which is what I'm assuming GCC's trying to prevent me from doing. 然后我决定从bit_field_ref参数 0（对结构的引用）类型中获取名称record_type，它存储所有字段的大小和偏移量。. The compilation option BITS_BIG_ENDIAN says which end of the memory unit pos counts Working in C11, the following struct: struct S { unsigned a : 4; _Bool b : 1; }; Gets layed out by GCC as an unsigned (4 bytes) of which 4 bits are used, followed by a _Bool (4 bytes) of which 1 bit is used, for a total size of 8 bytes. Structures, unions, enumerations, and bit-fields [Whether a "plain" int bit-field is treated as a signed int bit-field or as an unsigned int bit-field (6. Anyway, GCC and AFAIK every other GNUish compiler has a __builtin_offsetof used to directly hook into the compiler’s logic. 4. c main. Obviously there is no need for system-level code to conform to any standard so long as it works correctly. com. I found a good article about bitfields jkz. Can somebody give an example where that would matter. h there's a comment saying: // GCC packs bit fields in a different order on big endian machines The gccのbit field使用上の注意. The alignment of the addressable storage unit is unspecified. 1 #15) (probably related to #83784) The setting : STM32L476 / Cortex-M4 GCC 8. It's the same order you declared the members in. 11 Bit-Fields. But this bitfield datatype size "optimization" or this 'volatile' "trick" is 4. Even though the number of bits in the object Programs to communicate with low-level hardware interfaces need to define bit fields laid out to match the hardware data. The other possibility is the order in which gcc evaluates a expression is backwards from what you are thinking. Share. There is no reliable and portable way to determine the order. (VC++ kept making my 16-bit bitfield into a 4-byte structure. However if you don't plan to mix it you better take raw out and you will be safe. /* On GCC this correctly results in a 6 byte structure, but in MSVC the structure comes out 8 bytes. Also, structure padding is totally compiler dependent. Figure 8 : Bit – ordering in small Endianess architecture. width_of_bit-field: The number of bits in the bit-field. The ordering of bits in memory is architecture dependent and memory padding rules vary from compiler to compiler (see wikipedia), so exact position of each bitfield may differs, in other words you never can be sure which bit of raw each bitfield corresponds to. c; To run: . 0): 8 clang (8. Target Hook: bool TARGET_ALIGN_ANON_BITFIELD (void) When And that GCC likely avoids this because accessing a bit-field that cross a boundary would require two load instructions (depending on target processor) and additional bit-manipulation instructions to combine the loaded values. For example: 文章浏览阅读1k次。Endianess是个比较tricky的问题，特别是当数据在不同类型间转换时。 先看看在一台32位的little-endian机器上，以下代码的行为： #include // memcpy#include // printftypedef struct { char a; char b; char c; char d;} st;void main() {_stm32 gcc bitfield order To compile: gcc bit-field1. However, a CPU will normally do the write atomically (simply because it's one instruction). In order to read and write the data to a binary stream, it would need to be in the correct order or the fields will get reversed depending on the compiler and platform. Closed Public. The best way to know, is to check your compiler Figure 8 helps us understand how the bits ordering takes place. 问题是要理解它bit_field_ref不引用位本身：它使用 like bit_field_ref & integer_cst this project could be run on many boards and in the context we send the handler and the read and write functions to write and read to the specific bus depending on the board chosen. 9. On c99 standard (§6. Authored by tianqing on Nov 4 2019, 2:00 AM. R. 1. Considering that the bitfield's "natural" type is that of the addressable storage unit it is stored in when it comes out of memory, and that C99's usual arithmetic conversions philosophy only require it to be promoted up to int to be operated on, GCC's behavior seems more normal after reading your answer. After changing b: 0x00000002. from gcc (GCC) 4. Because struct packing order varies between compilers and architectures, the best option is to use a helper function to pack/unpack the binary data instead. struct bitfield { int i; unsigned int u1:8; In this article, the author tells embedded C programmers why writing code involving bit fields needs to be done carefully, failing which the results may not be as expected. However, compiling The order of allocation of bit-fields within a unit (high-order to low-order or low-order to high-order) is implementation-defined. 23. Determined by ABI. The width must be less than or Bitfield with struct and union in C (gcc) with endianess. 28106. Bit 7 6 5 4 3 2 1 0 f5 f4 f4 f3 f3 f2 f2 f1 GCC puts bit fields starting from "first bit" i. Whether a “plain” int bit-field is treated as a signed int bit-field or as an unsigned int bit-field (6. 5. 1 Writing reg. struct tmp 13. 3 it works as long as I don't use the attribute to pack the struct, i. gcc bitfield order control Jon Grant 2011-07-11 14:19:25 UTC. AVR-GCC does my wanted alignment by default: uint16_t 0x1234. 1). So if we assume the compiler packs the members of your bitfield in the order they are declared, starting at the first bit, then the layout should look like this. ; Allowable bit-field types other than _Bool, signed int, and unsigned int (6. With this case in mind, using a uint8_t value with #defined constants and the usual bit manipulations would appear to be more Bitfield Order in struct [duplicate] Running the below code give me this result: 01010101. But if you always use 32bit platforms with the same endianess, bit order should not change. [Allowable bit-field types other than _Bool, signed int, and unsigned int (6. It is safer to use bit shift operations instead of making any assumptions on bit field ordering or alignment when trying to write portable code The order of allocation of bit-fields within a unit (high-order to low-order or low-order to high-order) is implementation-defined. Improve this answer. 0 下，位域的声明顺序对其占用空间也是有影响的，而gcc-TDM 下则无影响，说明两者在处理“存储单元”时的逻辑是略有不同的，而 unix 下所有编译器都执行了存储单元的合并优化实现了紧凑布局，与声明位域所使用的类型 The size of t1 would be 8 bytes with the zero-length bitfield. c我最终了解到它们已知的位字段的名称来自结构的 die 并且难以跟踪。. I am using GCC struct bit fields in an attempt interpret 8 byte CAN message data. I've found that bitfields compiled on GCC2. Using bit with int in structure. Gcc in its present version still pads 4 bytes. . 3). If you run the same code in big Endian architecture, you will get the output given in Figure 9. 0, the code works as expected. 1: The warning is not shown anymore -> Looks like it really was a compiler bug – Just to note: what GCC wants me to do is this: struct { unsigned int a : 8; } test; And it works without any problem. 0. Long ago it was decided that to compile the Linux kernel, you needed gcc. You access the bitfields by their names; how the compiler has organised them does not matter in that case. (sign_extract:m loc size pos) This represents a reference to a sign-extended bit-field contained or starting in loc (a memory or register reference). An example of this follows: 我想知道是否有GCC C编译器指令，允许我确定结构包装的位顺序？对像：#pragma bit_order left这种需要的理由是，我有以下结构：struct { union { unsigned char BYTE; struct { unsigned char B0: 1; The result you are looking for can be obtained by making the structure union instead, with two bit-fields overlapping. The C11 standard (and probably C99 too) also states under §6. The order of allocation of bit-fields within a unit The size of a bitfield container is the least multiple of the alignment that fully covers the bitfield, but no larger than the size of the container-type. " No, GCC (in this case including g++, which inherits C implementation-defined I know that the order of a bitfield is undefined, but also that the first member in a struct is at the starting position of the struct in memory. constexpr uint16_t KEY_BITS = 70; constexpr unsigned __int128 KEY_BITS_MASK = (((unsigned __int128) 1) << KEY_BITS Some other options might be -faapcs-bitfield-width such an undefined bit order, byte order, alignment, padding, "storage unit boundaries", integer types actually supported, signed vs unsigned and so on. Re-mapping the bits in bit fields Some C compilers use the left to right order, while other C compiler use the right to left order Example: struct MyMap { unsigned int b31:1; unsigned int b30:1; unsigned int b29:1; unsigned int b28:1; unsigned int b27:1; unsigned int b26:1; unsigned int b25:1 Using the GNU Compiler Collection (GCC): Structures unions enumerations and bit-fields implementation. 7. Bitfields and alignment. Will the order of declared fields be respected versus any optimization or possible implementation of gcc/g++? The order of the members in the memory layout will not be tampered with by the compiler. One reply: GoForSmoke: Bit fields are nice but the compiler may not put the bits in the order you want. They are not stored with the size and alignment appropriate for the specified type, so they cannot be addressed through The order of bit-fields within an allocation unit (on some platforms, bit-fields are packed left-to-right, on others right-to-left). Idem. child1? In this particular case, yes. 1 20070105 (Red Hat 4. Bitfield load/store issues never got finished ; Tests for store data races and optimizer changes ; May require coordination witrh richi over a bitfield re-implementation. The Module Information shows that the data size is 2 bytes (16 bits). (§6. An implementation may allocate any addressable storage unit large enough to hold a bitfield. Benchmarking the above on gcc x86 for example yields a single instruction: mov DWORD PTR [rsp+12], -553648128 转自：https://blog. The order of bit-fields within an allocation unit (on some platforms, bit-fields are packed left-to-right, on others right-to-left). Transgender visa holders and If this prints 2 and 5, then the compiler’s behavior is what you would get from PCC_BITFIELD_TYPE_MATTERS. 3. 6. Reviewers . 1-52) GCC conversion warning when assigning to a bitfield. arm-none-eabi-gcc (Sourcery G++ Lite 2010. However, the compiler may organize the fields in bytes or in, for example, 32-bit words, and it may order the Forum: ARM programming with GCC/GNU tools About using bitfield with arm-elf-gcc. I have updated the union struct and bitfield in C fixing both the endianess of the bit-field and the warning message if the -Wpedantic is used with gcc (bit-field is a GCC extension). [Whether a bit-field can straddle a storage-unit boundary (6. In a thread, I proposed the use of bitfields. Updated Feb 9, 2012 . /a. A member of a union object is accessed using a member of a different type (C90 6. You need to be aware of endianness when defining bit-fields in this way. If you want to get technical, the minute you used __attribute__ (an identifier containing two consecutive underscores) your code has/had undefined behavior. The order of allocation of bit-fields within a unit (high-order to low-order or low-order to high-order) is implementation-defined. That means that they start at the low order bit for little-endian, and the high order bit for big-endian. The type may be int, signed int, or unsigned int. accessing bit fields in structure. Cast a uint32 variable to a bit field - undefined behavior? 2. gcc (9. Here's one possible alternative to using bit-fields: return (all & mask) >> shift; return (all & ~mask) | ((value << shift) & mask); The order of allocation of bit-fields within a unit (C90 6. So right now for me, the hex output is in reverse order of the font_pack's declaration order. Within a class, bitfields are "packed into some addressable allocation unit". (bitfield layouts aren't portable) but in general, endian-ness only changes the order of bytes in primitive types (int, short 值得注意的是，在我测试的 Win10 mingw-w64 gcc-8. Thus, C programming bitfield in structure. The type of a bit-field can only be integral (including bool) or (possibly cv-qualified) enumeration type, an unnamed bit-field cannot be declared with a cv-qualified type. 1 Binary is no different. 4. out. The alignment of the addressable storage unit is unspecified. 8. Hello I have a build with a lot of structures in a big-endian style layout. The compilation option BITS_BIG_ENDIAN says which end of the memory unit pos counts from. 9 Structures The order of allocation of bit-fields within a unit (C90 6. Explanation. This is because the structure was declared with int (or unsigned int). This behavior of GCC (endiannesss controlling the bit order for bitfields) is arbitrary and doesn't reveal anything about the hardware. Within a structure object, the non-bit-field members and the units in which bit-fields reside have addresses that increase in the order in which they 14. I am perplexed. Can I be confident that &father == &father. Just tested GCC >= 5. The bits "used" by the first bitfield will be marked as "reserved" in the second one: Explanation. If you need to know the exact bit positions, it is better use plain unsigned variable and bit masking. This is the "right" way to do things. See this comment in the GCC mailing list: Bit-fields are always assigned to the first available bit, possibly constrained by other factors, such as alignment. [The order of allocation of bit-fields within a unit (6. cn/s/blog_6240b5980100tcba. 95 (linux) in little-endian order the bitfields such that the first bit defined in the bitfield is the least significant. But the compiler will not (!) allow you full control of bit fields. The alignment of the addressable storage unit is unspecified I'm getting different result between gcc\clang and Visual C++ (Microsoft CL), when I compare the assembly code in Compiler Explorer and I got the following: If you want to go the other way, and force Microsoft's compiler to use GCC's bitfield layout, I don't think there's any attribute or option to do that. The relevant bytes of the representation of the object are treated as an object of the type used for the access. I've just spent a few hours figuring out why jbig2enc compiled for Android/arm produces output that cannot be decoded In jbig2structs. Bitfield endianness in gcc. If the actual type specifier used is int or a typedef-name defined as int, then it is implementation-defined whether the bit-field is signed or unsigned. For example, with gcc 4. An int is 16 bit by default in avr-gcc, while the word size the processor can read and store atomically is 8 bit. ; Whether a bit-field can straddle a storage-unit boundary (6. It's just convenient, because it means that if you reinterpret the adjacent bitfields as a single scalar (of their underlying type), each bitfield ends up occupying contiguous bits, and the order of bitfield declarations matches the their order in the Given those, while a compiler warning would clearly be in order, it seems that throwing away upper-order bits is guaranteed by the standard. Hot Network Questions Syntax of C Bit Fields struct { data_type member_name: width_of_bit-field;};. 1 #4) A bit-field shall have a type that is a qualified or unqualified version of _Bool, signed int, unsigned int, or some other implementation-defined type. 5k次，点赞2次，收藏9次。本文深入探讨了C语言中结构体(struct)、联合体(union)和位域(BitField)的概念，特别是结构体的内存占用和字节对齐原则。通过实例解析了不同数据类型在结构体中的内存布局，以及如何通过合理设计结构体成员顺序来优化内存 The point is that the compiler will almost certainly lay out the fields in the order you write them, so all you are really worried about is big-endian integers. The order of allocation of In the following example listing, you can see that the bits are arranged from the highest order byte to the lowest order byte and within each byte, from the LSB to the MSB. Bit fields are special in that you cannot take their address with ‘ & ’. The compiler should be smart enough to figure it out on it's own. ビットフィールド構造体のレイアウトについて The order of allocation of bit-fields within a unit (high-order to low-order or low-order to high-order) is implementation-defined. e. @kiner_shah In GCC using the switch -mms-bitfields it behaves as MS compiler. 0. Note that C99 and C11 specifically permit _Bool as a bit-field member. Consecutive bit fields are "gcc lays out bitfields such that the first listed field corresponds to least significant bit of a byte. Follow Initialize a constant array of struct with bitfield. REG32 = 1, results in BIT31 containing 1. wtf/bit-field-packing-in-gcc-and-clang – Handling race conditions must be done by operating system level calls (which will indeed use read-modify-writes), GCC won't do that. Can you tell me why is this so? Thanks for your help The only thing you can do is change the order of the bitfields in the declaration. If the zero-length bitfield were removed, t1's size would be 4 bytes. or register layout perhaps, (but be aware that not solve byte ordering issues), but using bitfields is not conducive to this purpose since the packing and ordering is also compiler dependent. But I need to be able to vary that. So unless you are sure you already know the behavior on your system and that you 5. 9 Structures, Unions, Enumerations, and Bit-Fields The order of allocation of bit-fields within a unit (C90 6. This section explains how to do that. I have, by the way, never seen anything else than The order of allocation of bit-fields within a unit (high-order to low-order or low-order to high-order) is implementation-defined. Bitfield concept in structure. struct bit field in C. 67. Even though the number of bits in the object representation of _Bool is at least CHAR_BIT, the width of it might depend on patforms, in riscv it might even slower when not word-aligned, as each access might result into 2 accesses + extra work to shift. GCC for 8bit AVR microcontrollers) I would expect the bitfield to be stored in an int. , and no GCC does not generate read-modify-write instructions for volatile. I wrote a small program as an example of one possible message layout. There are no bit-field prvalues: lvalue-to-rvalue conversion always produces an object of the underlying type of the bit-field. 1): 8 cl (19. A bit-field cannot be a static data member. 1 from ARM That the "standard" never really ruled about bitfield order is already baffling and common sense (ie. Bit order from low to high / Mapping structs Using the GNU Compiler Collection (GCC) 4. Georgioss-MacBook-Pro:~ gsamaras$ gcc -Wall main. /* 位域是在结构体内部定义的，通过冒号和数字来指定每个成员所占用的位数。// flag1占用1位// count占用7位// flag2占用1位// 保留位，通常初始化为0在这个例子中，flag1countflag2和reserved。它们分别占用1位、7位、1位和13位。注意，位域的类型通常是无符号整数类型（如），因为位域主要用于存储非负整数。 The size of t1 would be 8 bytes with the zero-length bitfield. But I must still worry about other gcc promises all sorts of things above and beyond the standard. 09-51) 4. Commented Oct 2, 2013 at 15:41. unaligned access causes trap (irq with mcause=4 value) and then it needs to be handled in IRQ handler, extra jumps, possibly saving registers to stack. The n value below always is required to be a small power of two and specifies the new alignment in C言語にはbitfieldという機能がありますが、この機能にはさまざまな落とし穴があり、注意深く利用しても予期せぬバグや移植性の問題を引き起こす可能性があるとよく言われています。 実際、jpcertを確認すると複数の勧告が出ています。 EXP11-C. member_name: The member name is the name of the bit field. The anonymous nested struct declares a new class object, so any bitfields within it will be independent of any bitfields in the outer class. This causes 1-byte to be wasted [NFC] Supress GCC "Bitfield too small to hold all values of enum" warning. Add C1x support. Transgender visa holders and Executive Order 14201: ‘Keeping Men Out of Women’s Sports’ With avr-gcc (i. now I know why – nirbruner. 2. yqexeev rawtl jumefzck jjkr vutl obcomy hizpl zzvzu btqw gbdmmq vfzblnv kewx ybby vzvnszo vlxafa