Published with kind permission of DevCentral
Copyright © 1998 Interface Technologies, Inc. All Rights Reserved.

 program samp;  
 var  
     a , b : integer; 
  
 procedure swap(var i, j : integer);    
 var t:integer;    
 begin      
     t := i;      
     i := j;      
     j := t;    
 end; 
  
 begin    
     a := 5;    
     b := 10;    
     writeln(a,b);    
     swap(a,b);    
     writeln(a,b);  
 end. 

Because this code uses variable parameters, it swaps the values a and b correctly.

C has no formal mechanism for passing variable parameters: It passes everything by value. Enter and execute the following code and see what happens:

 #include <stdio.h>   
  
 void swap(int i, int j)    
 {    
     int t; 
     
     t = i;    
     i = j;    
     j = t;  
 } 
 
 void main()  
 {    
     int a,b; 
     
     a = 5;  
     b = 10;    
     printf("%d %d\n",a,b);    
     swap(a,b);    
     printf("%d %d\n",a,b);  
 } 

No swapping takes place. The values of a and b are passed to swap, but no values are returned.

To make this function work correctly, you must use pointers, as shown below:

 #include <stdio.h>   
 
 void swap(int *i, int *j)    
 {    
     int t; 
     t = *i;    
     *i = *j;    
     *j = t;  
 } 
 
 void main()  
 {    
     int a,b; 
     a = 5;    
     b = 10;    
     printf("%d %d\n",a,b);    
     swap(&a,&b);    
     printf("%d %d\n",a,b);  
 } 

To get an idea of what this code does, print it out, draw the two integers a and b, and enter 5 and 10 in them. Now draw the two pointers i and j, along with the integer t. When swap is called, it is passed the addresses of a and b. Thus, i points to a (draw an arrow from i to a) and j points to b (draw another arrow from b to j). Because the pointers have been established, *i is another name for a, and *j is another name for b. Now run the code in swap. When the code uses *i and *j, it really means a and b. When the function completes, a and b have been swapped.

Suppose you accidentally forget the & when the swap function is called, and that the swap line accidentally looks like this: swap(a,b);. This causes a segmentation fault. When you leave out the &, the value of a is passed instead of its address. Therefore, i points to an invalid location in memory and the system crashes when *i is used.