your compiler has arranged all of the members of test3 on integer size boundaries. Once a block has been used for a given type (integer bit-field, or boolean bit-field), the compiler does not allocate any further bit fields of a different type until the next boundary. I doubt it is a bug. It probably has … 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
It’s a bitfield member. Your code means dumpable occupies exactly 1 bit in the structure. Bitfields are used when you want to pack members in bit-level. This can greatly reduce the size of memory used when there are a lot of flags in the structure. For example, if we define a struct having 4 members … 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
Bit-fields are typically only used when there’s a need to map structure fields to specific bit slices, where some hardware will be interpreting the raw bits. An example might be assembling an IP packet header. I can’t see a compelling reason for an emulator to model a register using bit-fields, as it’s never going to … 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
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
All of your bitfield members are signed 1-bit integers. On a two’s complement system, that means they can represent only either 0 or -1. Use uint8_t if you want 0 and 1: struct Bits { uint8_t b0:1, b1:1, b2:1, b3:1, b4:1, b5:1, b6:1, b7:1; };
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
Bit-fields are incredibly poorly defined by the standard. Given this code struct mystruct {int enabled:1;};, then we don’t know: How much space this occupies – if there are padding bits/bytes and where they are located in memory. Where the bit is located in memory. Not defined and also depends on endianess. Whether an int:n bitfield … Read more