Structures to Equal

**Crosility** · 02-22-2011

Code:

#include <stdio.h>

	struct bb {
		unsigned char a : 3;
		unsigned char b : 2;
		unsigned char c : 2;
		unsigned char d : 1;
	} b;
	
	struct aa {
		unsigned char a : 3;
		unsigned char b : 2;
		unsigned char c : 2;
		unsigned char d : 1;
	} a;

int main (int argc, char * const argv[]) {
	b.a = 3;
	b.b = 2;
	b.c = 1;
	b.d = 0;

	//a = b;
	
	printf("%d\n", b);
	
	printf("a : %d\nb : %d\nc : %d\nd : %d\nsize of bb : %d\n", b.a, b.b, b.c, b.d, (int)sizeof(b));

	printf("a : %d\nb : %d\nc : %d\nd : %d\nsize of aa : %d\n", a.a, a.b, a.c, a.d, (int)sizeof(a));

getchar(); <-- simple way to keep open
}

I'm currently using a bitfield to set data. (This is just for practice, not actual use).
I'm wondering if there is any possible way to take one structure, have another one equal it. Or should I just quit using bitfield, and manipulate a 1 bit char?

**itsme86** · 02-22-2011

You could always use memcpy():

Code:

memcpy(&a, &b, sizeof(b));

**GReaper** · 02-22-2011

.....
2) If i understand correctly, you want to have only a bit of data. That's impossible because bits don't have addresses, bytes do. A byte is the smallest addressable block of information in a computer.

**Clairvoyant1332** · 02-22-2011

Generally speaking, you shouldn't wholesale compare two structs of differing types. In this particular case, they're laid out the same, so it should work. What you'd have to do is get the address of each one and run memcmp() against them.

EDIT: I misread. For an assignment, memcpy() would work, but again not generally a good idea to do this with differing types.

**Crosility** · 02-22-2011

Originally Posted by itsme86

You could always use memcpy():

Code:

memcpy(&a, &b, sizeof(b));

Never needed to use memcpy before.
Okay, thank you. I'll try that.

----

Splendid.

Output:

51
a : 3
b : 2
c : 1
d : 0
size of bb : 1
a : 3
b : 2
c : 1
d : 0
size of aa : 1

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