2. Pointers and Function Arguments
Since C passes arguments to functions by value, there is no direct way for the called function to alter    a variable in the calling function. For instance, a sorting routine might exchange two out-of-order    arguments with a function called swap.
Call by Value
int a = 5, b = 7;

swap(a, b);

void swap(int x, int y) /* WRONG */
     int temp;
     temp = x;
     x = y;
     y = temp;
– Because of call by value, swap can't affect the arguments a and b in the routine that called it. The    function above swaps copies of a and b.
Call by Reference
int a = 5, b = 7;

swap(&a, &b);

/* interchange *px and *py */
void swap(int *px, int *py)
     int temp;
     temp = *px;
     *px = *py;
     *py = temp;
Variables passing out of scope
Pointer invalid after variable passes out of scope.
What is wrong with this code?
#include <stdio.h>
char *get_message ( )
    char msg[] = “Aren't pointers fun?”;

    return msg;
int main ( void )
    char *string = get_message();
    puts(string );

    return 0;
– warning: returning address of local variable or temporary
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