after the fuction, same struct variacles.

[nullifyed]

Another question:

the reason that the variables does not change is because:
the whole box (made for the function) is deleted when going back in main. What i mean is that the variables CHANGE but they do not return to the main function so they can printed. Am i correct?

#include <stdio.h>
#include <stdlib.h>

struct Foo{
	int x;
	int y;
};

void f1(struct Foo foo){
	foo.x=12;
	foo.y=13;
}

int main(void){
	struct Foo foo;


	foo.x=10;
	foo.y=11;
	
	printf("x: %d\n", foo.x);
	printf("y: %d\n", foo.y);

		f1(foo);

	printf("x: %d\n", foo.x);
	printf("y: %d\n", foo.y);

	return 0;
}

[cas]

Yes. You're simply passing a copy of the struct to f1(), so any changes are made to a copy. It's like if you make a photocopy of, say, a contract, and give that to somebody. If he burns his copy, you still have yours.

Contrast this to passing a pointer to a struct: if you do that, you can modify the struct and it will be reflected in the caller. This is like telling somebody where you keep the contract; now if it is burned, it's yours that gets destroyed.

[nullifyed]

i am not sure if this is what you are talking about, but there is a problem (red markers)...what is going wrong?

#include <stdio.h>
#include <stdlib.h>

struct Foo{
	int x;
	int y;
}group1,group2;

void f1(struct Foo *foo){
	foo.x=12;
	foo.y=13;
}

int main(void){
	struct Foo foo;

	foo.x=10;
	foo.y=11;
	
	printf("x: %d\n", foo.x);
	printf("y: %d\n", foo.y);

		f1(&foo);

	printf("x: %d\n", foo.x);
	printf("y: %d\n", foo.y);

	return 0;
}

[Adak]

It's a bit confusing because you named the pointer the same as the struct instance-- please don't do that! If nothing else, put a p in the front of the pointer's name, like *pfoo.

If you are using a pointer to a struct, then use the pointer to struct access operator ->, not the struct member access operator (the dot).

void f1(struct Foo *foo){
	foo->x=12;
	foo->y=13;
}

[nullifyed]

ok i' ll try to put "p".
insteed of the code you wrote can we use:

	*(foo.x)=12;
	*(foo.y)=13;

?

[laserlight]

No, because then you are dereferencing foo.x, not dereferencing foo to access x.

[nullifyed]

No, because then you are dereferencing foo.x, not dereferencing foo to access x.

apart from the the "->" i have seen somewhere something like this that i wrote....is there another way to do it?

[jimtuv]

I think you are thinking of

(*foo).x=12

see the dot . is evaluated first before *. So to force the dereference to occur first you add the parenthesis. This is exactly the same as saying

foo->x=12

[nullifyed]

exactly!! thanks....