C Incrementing pointer address passed to a function operator-CodePudding

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.

CodePudding user response：

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)  ;

CodePudding user response：

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.

CodePudding user response：

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.
}