Pointer parameter was set but never used warning

Question

I have the following function:

void foo(char *ptr_1)
{
    char *ptr_2;

    bar(ptr_2);

    ptr_1 = ptr_2;
}

And get this warning:

parameter "ptr_1" was set but never used

I understand that the warning is technically true, but is irrelevant at the same time. I could suppress it with:

(void)(ptr_1)

But is there a better way?

Answer 1

It is not an irrelevant warning because the assignment has no effect. You could completely remove ptr_1 from the code without changing its behaviour. Your code is equivalent to this:

void foo(char *)
{
  char *ptr_2;
  bar(ptr_2);
}

In other words, the function parameter isn't used for anything. If your intention was to change the pointer at the caller side, you need to either pass a pointer to a pointer and de-reference it, or return the new value and let the caller use its value:

void foo(char **ptr_1)
{
    char *ptr_2;
    bar(ptr_2);
    *ptr_1 = ptr_2;
}

or

char* foo()
{
    char *ptr_2;
    bar(ptr_2);
    return ptr_2;
}