Question

Can somebody help me explaining the following code:

Why the char *s doesn’t receive the point location of memory allocated at foo()?

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

char *foo()
{
    char *s = (char *)malloc(20);
    s = "Hello Heap.";
    return s;
}

void bar(char *s)
{
    s = foo();
    printf("bar: %s\n", s); // Works fine just as expected.
}

int main()
{
    char *s;
    bar(s);
    printf("%s\n", s); // Output some undefined content like `H?}?H??`, other than `Hello Heap.`
}

Answer 1

The code fixed

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

char *foo()
{
    char *s = (char *)malloc(20);
    strcpy(s,"Hello Heap.");
    return s;
}

void bar(char **s)
{
    *s = foo();
    printf("bar: %s\n", *s); // Works fine just as expected.
}

int main()
{
    char *s;
    bar(&s);
    printf("%s\n", s); // Output some undefined content like `H?}?H??`, other than `Hello Heap.`
}

Explanation:

1) Your code contains:

char *s = (char *)malloc(20);
 s = "Hello Heap.";

This not good. In this way you are not copying the "Hello Heap." message into the allocated memory. In fact you have pointed the s pointer to allocated memory and then you have pointed your pointer to a constant string address

2) Your code contains

void bar(char *s)
{
    s = foo();
    printf("bar: %s\n", s); // Works fine just as expected.
}

in this function s is getting the pointer (pointer to allocated memory) from the foo() function. but you are not communicating the s pointer address to the function higher leve (main). to do you have to return the address of s at the end of the function or you can pass via address of address of pointer by using input parameter cha ** s