Pointer is "passed by value"?

August 22, 2022

After calling f() on ptr, I expect it to point to a single byte with value of A.
But instead ptr is copied by value and it is only available in the f function(?)
What am I doing wrong?

void f(char* ptr) {
    ptr = (char*) malloc(1);
    *ptr = 'A';
}

int main() {
    char* ptr;
    f(ptr);
    printf("%c\n", *ptr); // Segmentation fault, But it should be 'A'
    // free(ptr);
}

Thanks!

>Solution :

Yes, it’s passed by value. If you want the changes you make to the pointer to be visible at the call site, you need to pass a pointer to the pointer.

Example:

#include <stdlib.h>
#include <stdio.h>

void f(char **ptr) {          // pointer to the pointer
    *ptr = malloc(1);
    if(*ptr)                  // precaution if malloc should fail
        **ptr = 'A';
}

int main(void) {
    char *ptr;
    f(&ptr);                  // take the address of `ptr`
    if(ptr)                   // precaution again
        printf("%c\n", *ptr); // now fine
    free(ptr);                // without this, you have a memory leak
}