Question

I’m maintaining some C code that was written by the person who had my job last. A situation has come up regarding passing by reference. Here’s a shortened, contrived example of what I’m working with:

static int b;

void SetToTen(int *a){
    b = 10;
    /* >>>>>>>  Need to set a equal to b on this line <<<<<<<< */
return;
}

int main{
    int a = 0;
    SetToTen(&a);
    /* Now a should be equal to 10*/
  . 
  .
  .
return 0;
}

In the SetToTen function, I could either write:

*a = b;

OR

a = &b;

I think these two are functionally equivalent (a will be equal to ten with either of them.) But my question is: are there any sneaky subtleties associated with one over the other? Specifically, if I use a = &b does that mean that if I change b in the future, a will change as well? And is this not the case if I use *a = b?

Thoughts/Musings/Comments would be appreciated.

Answer 1

They’re not functionally equivalent at all. This code:

*a = b;

dereferences a and sets its value to that of b. This code:

a = &b;

changes the a pointer to point to b. If you use the second piece of code, nothing will happen because you’re changing the a pointer by value, it won’t actually change a at all.