No, you can’t. There is an open issue about supporting bitfields, which doesn’t seem to be active. In the issue, @retep998 explains how bitfields are handled in winapi. That might be helpful if you need to handle bitfields in C interface. OP seems to aim at C interoperation, but if you just need bitfields functionality, … Read more
C standard allows compiler to put bit-fields in any order. There is no reliable and portable way to determine the order. If you need to know the exact bit positions, it is better use plain unsigned variable and bit masking. Here’s one possible alternative to using bit-fields: #include <stdio.h> #define MASK_A 0x00FF #define MASK_B 0x3F00 … Read more
You use a zero-length bitfield as a hacky way to get your compiler to lay out a structure to match some external requirement, be it another compiler’s or architecture’s notion of the layout (cross-platform data structures, such as in a binary file format) or a bit-level standard’s requirements (network packets or instruction opcodes). A real-world … Read more
If you mainly want to be able to name your bit fields and easily manipulate them, e.g. to work with flags represented as single bits in a communications protocol, then you can use the standard Structure and Union features of ctypes, as described at How Do I Properly Declare a ctype Structure + Union in … Read more
It is the order that matters. The following code will give Output: 8 #include<stdio.h> typedef struct size { unsigned int b:32; unsigned int a:1; unsigned int c:1; }mystruct; int main(int argc, char const *argv[]) { mystruct a; printf(“\n %lu \n”,sizeof(a)); return 0; } Unsigned int is a 32 bit integer, occupying 4 bytes. Memory is … Read more
I’d probably knock together something using attributes, then a conversion class to convert suitably attributed structures to the bitfield primitives. Something like… using System; namespace BitfieldTest { [global::System.AttributeUsage(AttributeTargets.Field, AllowMultiple = false)] sealed class BitfieldLengthAttribute : Attribute { uint length; public BitfieldLengthAttribute(uint length) { this.length = length; } public uint Length { get { return length; … Read more
I recommend you to do it exactly as you have shown, since it is the most straight forward one. Initialize to -1 which will work always, independent of the actual sign representation, while ~ will sometimes have surprising behavior because you will have to have the right operand type. Only then you will get the … Read more
This construct specifies the length in bits for each field. The advantage of this is that you can control the sizeof(op), if you’re careful. the size of the structure will be the sum of the sizes of the fields inside. In your case, size of op is 32 bits (that is, sizeof(op) is 4). The … Read more
The 1 and the 7 are bit sizes to limit the range of the values. They’re typically found in structures and unions. For example, on some systems (depends on char width and packing rules, etc), the code: typedef struct { unsigned char a : 1; unsigned char b : 7; } tOneAndSevenBits; creates an 8-bit … Read more
By the C standard, the compiler is free to store the bit field pretty much in any random way it wants. You can never make any assumptions of where the bits are allocated. Here are just a few bit-field related things that are not specified by the C standard: Unspecified behavior The alignment of the … Read more