Puzzle P14

Parameter is a Pointer

What does the following code write to the monitor?

#include <stdio.h>

void newFunction( int *p )  
{
  printf("  *p=%d\n", *p );
  *p = 123;
  printf("  *p=%d\n", *p );
}

void main ( void )
{
  int a = 77 ;
  printf("a=%d\n", a ) ;
  newFunction( &a ) ;
  printf("a=%d\n", a ) ;
  
  system("pause") ;
}

Call by value is used in this example, also. However, the parameter p of newFunction() requires a value that is of type pointer to int:

int *p

main() supplies such a value in the function call newFunction( &a ). The expression &a evaluates to the address of a. (Often this is called a pointer to a.)

When newFunction() is called, the address of a is copied into the parameter p. Now newFunction() can access a by following the pointer. The statement

printf("  *p=%d\n", *p );

follows the pointer in p to get a value (which is then printed). The statement

*p = 123;

follows the pointer in p to the location in which to store the value of the expression on the right of the =. This location is the variable a, which is then changed.