多次分配后C指针丢失

Question

I am new to C programming and having some issues with the pointers in the following code. I am using the C89 standard.

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

char * b(char *p) {

    char *z;

    z = p;

    printf("b(): %s\n", z);

    return z;

}

void a(char *p, char *q) {

     q = b(p);
     printf("a(): %s\n", q);
}

void main(void) {

    char *p = (char *) malloc(15);
    char *q;

    strcpy(p, "This is a test");

    a(p,q);

    printf("main(): %s\n", q);

    free(p);
}

For some reason, my printf function is printing nothing in the main() function. The printf calls in both of the functions a() and b() are working. I am not sure why my pointer "q" is lost in the main function. I even tried allocation dynamic memory to "q" in main() function but still no results.

Answer 1

q in main() is uninitialized. You probably assume that q is changed after a(p,q) , but as that pointer is passed by value (as everything in C), after returning from a , any changes done to q are discarded (they were local to a )

To solve this, while preserving most of your function signatures, you would have to add another level of pointers

In following code, p and q in a are local to a , but you are not directly changing any of them, insted, you are modifying value, that is pointed at by q (that value is of type *char , so another ponter, easy to get lost)

void a(char *p, char **q) {

     *q = b(p);
     printf("a(): %s\n", q);
}

void main(void) {
    ...
    a(p,&q);

    printf("main(): %s\n", *q);

    ....
}

Or simply return that pointer, and assign to q in main (as you have done in b )

To better understand why this is happening, consider this simpler example:

void foo(int a) {
    print("value is %d\n", a);
    a = 42; // modifies only a local copy of passed parameter a
    print("modified value is %d\n", a);
}

int main() {
    int a = 0
    printf("initial value is %d\n", a);
    foo(a);
    printf("value after returning is (still) %d\n, a");
}