Modify struct via a reference taking function

Question

I want to assign the struct a to the struct b. Printing the address and the values of the array before and after function call it shows that during the function call the assignment works and both pointers point to the same struct address. However, after returning from the function the changes are reversed. Why?

typedef struct arrayA {
    int a[3]; 
}arrayA; 

void f(arrayA *a, arrayA *b){
    a = b; 
    printf("address of a: %p\n", a); 
    printf("address of b: %p\n", b); 
}

int main(int argc, char *argv[]) {
    printf("------------ Array assignment test -----------------\n"); 
    arrayA a = { { 0, 0, 0} }; 
    arrayA b = { { 1, 1, 1} }; 
    printf("address of a: %p\n", &a); 
    printf("address of b: %p\n", &b); 
    printf("a[0] : %d a[1] : %d\n", a.a[0], a.a[1]); 
    f(&a, &b); 
    printf("a[0] : %d a[1] : %d\n", a.a[0], a.a[1]); 
    printf("address of a: %p\n", &a); 
    printf("address of b: %p\n", &b); 
    printf("----------------------------------------------------\n"); 

    return 0;
}

Prints

 ------------ Array assignment test -----------------
address of a: 0x7ffd3fc17b80
address of b: 0x7ffd3fc17b90
a[0] : 0 a[1] : 0
address of a: 0x7ffd3fc17b90
address of b: 0x7ffd3fc17b90
a[0] : 0 a[1] : 0
address of a: 0x7ffd3fc17b80
address of b: 0x7ffd3fc17b90
----------------------------------------------------

a = b;. Standard beginner issue. Function parameters are passed by value in C. Modifications to parameters inside a function have no effect on the caller's variable. — kaylum
– kaylum, Commented Jan 16, 2017 at 22:11
What exactly do you want to achieve? If you want to copy the contents use memcpy(a, b, sizeof(*b)) or *a=*b. — kaylum
– kaylum, Commented Jan 16, 2017 at 22:14
Exactly. I want to copy the content. Calling a=b in the main function works perfectly fine. — clickMe
– clickMe, Commented Jan 16, 2017 at 22:15
Note that a=b in main is equivalent to *a=*b in the function. — kaylum
– kaylum, Commented Jan 16, 2017 at 22:18

user1952500 · Accepted Answer · 2017-01-16 22:16:50Z

2

You are passing the pointers by value and expecting them to get modified. They will be modified within the function as there is a copy of the pointer (address) that is made local to the function. When the function returns the originals remain unchanged. (You could try to print the address of the local variables within the function to understand better.)

If you want to change the structs, you'll want to dereference the pointers:

void f(arrayA *a, arrayA *b){
    *a = *b; 
    printf("address of a: %p\n", a); 
    printf("address of b: %p\n", b); 
}

If you want to change the pointers themselves, you'll need an extra indirection:

void f(arrayA **a, arrayA **b){ // Note the ** here
    *a = *b; 
    printf("address of a: %p\n", a); 
    printf("address of b: %p\n", b); 
}

answered Jan 16, 2017 at 22:16

user1952500

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tdao · Accepted Answer · 2017-01-16 22:15:46Z

1

However, after returning from the function the changes are reversed. Why?

Your f() function really just change the pointer inside it, and those pointer changes are not reserved after the function.

You can copy struct by pass-by-pointer:

void f(arrayA *a, arrayA *b){
    *a = *b; 
}

That will ensure you can copy struct in main() via

f(&a, &b);

As all you need is copy struct, there is no need to print out the addresses at all. In case you do need to debug the address, you should convert to (void *) to avoid all the warnings with printf("%p")

answered Jan 16, 2017 at 22:15

tdao

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