Bit Fields example.

**Mr.Lnx** · 06-10-2015

Hello to all. There is a point that I can't understand.

Code:

struct s {
unsigned int a: 4;
unsigned int: 0;
unsigned int b: 8;
}

a 0-BIT FIELD is a signal to the compiler to align the following bit field at the beginning of a storage unit if storage units are 8 bits long the compiler will allocate 4 bits for the a member , skip 4 bits to the next storage unit and then allocate 8 bits for b.If storage units are 16bit long the compiler will allocate 4 bits for a skip 12 bits and then allocate 8 bits for b.

I can't understand the layout. Compiler will allocate 4 bits for the a member in the first case and skip the 4 bits to the next storage but we have 8bits long storage unit the remaining is padding? "skip 4 bits to the next storage unit" from where is these 4 bits? where come from?

In the second case from where we have 12 bits?
I am little confusing about how memory layout is organized when we are dealing with bit fields

**laserlight** · 06-10-2015

Originally Posted by Mr.Lnx

Compiler will allocate 4 bits for the a member in the first case and skip the 4 bits to the next storage but we have 8bits long storage unit the remaining is padding? "skip 4 bits to the next storage unit" from where is these 4 bits? where come from?

Yes, the gap for alignment is padding.

Originally Posted by Mr.Lnx

In the second case from where we have 12 bits?

16-4=12, where 16 is the number of bits of a "storage unit" and 4 is the number of bits for the member named a.

**Mr.Lnx** · 06-10-2015

Code:

#include <stdio.h>
typedef struct s {
unsigned int a: 4;
unsigned int: 0;
unsigned int b: 8;
} my_s;

int main(void) {
	
	printf("%d , %d" , sizeof(unsigned int) , sizeof(my_s));
	
	return 0;
}

I take 4 and 8 size of unsigned int and my_s structure respectively the unnamed field is off? I don't understand how the compiler reduce the size. We have 3 members here 4 + 4 + 4 = 12 bytes I think.

**laserlight** · 06-10-2015

Originally Posted by Mr.Lnx

I take 4 and 8 size of unsigned int and my_s structure respectively the unnamed field is off? I don't understand how the compiler reduce the size. We have 3 members here 4 + 4 + 4 = 12 bytes I think.

sizeof gives the size in bytes, not bits. Assuming that CHAR_BIT == 8, which is almost certainly true, and assuming that an int is the size of a "storage unit", then you have 32-bit storage units, so the compiler will allocate 4 bits for the member named a, skip 32-4=28 bits, then allocate 8 bits for the member named b, hence the 8 bytes in total after padding.

**Mr.Lnx** · 06-11-2015

Yes I meant that each unsigned int occupies 4 bytes in the memory so the result of sizeof(my_s) I expected instead of 12 bytes not 8 that is why I think I have no understand the concept of bit fields. How many unit storage's would compiler need? 2? It is difficult to understand the concept I really need a schema/picture or an output of assembly code although I don't know assembly :/

**laserlight** · 06-11-2015

Originally Posted by Mr.Lnx

Yes I meant that each unsigned int occupies 4 bytes in the memory so the result of sizeof(my_s) I expected instead of 12 bytes not 8

The member named a is in the first 4 bytes; the member named b is in another 4 bytes. Why would there be yet another 4 bytes? It cannot be for the unnamed 0-bit member that is used merely to "signal to the compiler to align the following bit field at the beginning of a storage unit".

**Mr.Lnx** · 06-11-2015

So the unnamed 0-bit member does not occupy any memory ok.

**laserlight** · 06-11-2015

Yes.

**Mr.Lnx** · 06-11-2015

Ok.So we have 8 * 8 = 64bits hence 2 storage units of 32. The layout would be :

Code:

       
 
    |__4 bits for a member___|___28bits padding___|           |___8bits for member b____|__24bits padding__|
 
                1st 32 bit storage unit                                                                    2nd 32bit storage unit

Is that true?

**laserlight** · 06-11-2015

Yes.

**Mr.Lnx** · 06-11-2015

Originally Posted by laserlight

Yes.

Thank you very much for your time laserlight

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