char * gives garbage value when pointing to a variable of a function in C

Question

When i pass a pointer to a function and then change the pointer to a another location of the memory, I get SIGSEV or garbage values. Where is a code to demonstrate that:

#include <stdio.h>

void fn(char *n) {
  /*
   * imagine this variable 'b' is
   * a part of a structure of
   * some library
   */
  char *b = "hello, world!";

  n = b;
}

int main() {
  char *a = NULL;
  fn(a);

  /* it throws garbage or SIGSEV */
  printf("%s\n", a);
}

I know what's the reason for this problem. The function fn, when called creates a variable b but when fn ends, b is deleted. And therefore, a is pointing to a memory that the program doesn't own. What do I do to fix this program. I know, I can solve the issue just by using C++ with it's std::string. But I have to do it in C. Thanks.

Answer 1

It means that you need to pass the pointer a defined in main to the function fn by reference. In C passing by reference means passing an object indirectly through a pointer to it. Thus dereferencing the passed pointer the function will have a direct access to the object pointed to by the pointer and can change it.

void fn(char **n) {
  /*
   * imagine this variable 'b' is
   * a part of a structure of
   * some library
   */
  char *b = "hello, world!";

  *n = b;
}

And in main you need to write

fn( &a );

Pay attention to that the pointer a will be valid after the function call because it points to a string literal used in fn and string literals have static storage duration.