I have a question that is raised from this discussion: C - modify the address of a pointer passed to a function
Let's say I have the following code:
#include <stdio.h>
foo(char **ptr){
*ptr++;
}
int main()
{
char *ptr = malloc(64);
char arr[] = "Hello World!";
memcpy(ptr, arr, sizeof(arr));
foo(&ptr);
foo(&ptr);
printf("%s",ptr);
return 0;
}
I was wondering what the output of this program would be and I thought that it should be llo World!.
After some investigation I found the question linked above and realized that, in C, parameters to functions are always passed by value. So far there was no problem. When it comes to change *ptr++; expression to -> *ptr = *ptr +1; output becomes: llo World!.
At this point, I can say that I am a little confused. In order to change pointer address, we need a double pointer. That is fine, but why do post increment operations differ? Is it because of the operator precedence?
Here I tried the example in an online C compiler.
The postfix increment operator ++ has higher precedence than the dereference operator *. So this:
*ptr++;
Parses as:
*(ptr++);
So it changes the parameter value itself, not what it points to. You instead want:
(*ptr)++;
Postfix operators have higher priority than unary operators. So this expression
*ptr++
is equivalent to
*( ptr++ )
The value of the sub-expression ptr++ is the value of the pointer before its incrementing.
So actually you are incrementing the parameter ptr having the type char **. So this incrementing does not change the original pointer and does not make a sense.
Instead you could write
( *ptr )++
But it will be more clear and less confusing to use the unary increment operator like
++*ptr
if you want to increment the original pointer itself.
As others have explained, ++ is more important (has a higher priority) than *, so your function foo actually compiled as this:
foo (char **ptr) {
ptr = ptr + 1; // This modifies the parameter, which is a copy of a variable, copied *specifically* for this function; hence, modifying it will have no effect outside of the function.
*ptr; // This is just an expression; it doesn't actually do anything, as the compiler would tell you if you wrote it this way.
}
If you change *ptr++ to (*ptr)++, the function will work like this:
foo (char **ptr) {
*ptr = *ptr + 1; // Sets the variable pointed to by ptr to be equal to itself plus one.
}