Recall the following postscript ( click here )
 

  • Although C only has the pass-by-value parameter passing mechanism, C can achieve the effect of the pass-by-reference mechanism !!!

  • C has a very powerful reference data type that no other (non-C related) programming language provides....

  • This reference data type and its operators provide the C programmer the tools to achieve the effect of the pass-by-reference mechanism

You have learned everything you need to do this in C

Example of "Do it yourself" pass-by-reference in C

Consider the following example: 

#include <stdio.h>        // DEMO this program   

void f(int x)  
{
   x = x + 1;
}

int main(int argc, char *argv[])
{
   int a = 4;

   printf("Before calling f(a), a = %d\n", a); 
   f(a);
   printf("After  calling f(a), a = %d\n", a); 
}

C will pass the value of the variable a to function f( )

Example of "Do it yourself" pass-by-reference in C

We can pass the address (value) of the variable a using &a: 

#include <stdio.h>

void f(int x)  
{
   x = x + 1;
}

int main(int argc, char *argv[])
{
   int a = 4;

   printf("Before calling f(a), a = %d\n", a); 
   f(&a); // Pass the value of the address of var a 
   printf("After  calling f(a), a = %d\n", a); 
}

However:   the data type of the parameter int x is now incorrect.

Example of "Do it yourself" pass-by-reference in C

The correct type is int *x because x stores an address of an int typed variable: 

#include <stdio.h>

void f(int *x)  
{
   x = x + 1;
}

int main(int argc, char *argv[])
{
   int a = 4;

   printf("Before calling f(a), a = %d\n", a); 
   f(&a);
   printf("After  calling f(a), a = %d\n", a); 
}

However:   the statement x = x + 1 says: use the variable x in an addition
x is not an variable that you can use for addition !! (x contains an address !!)
The statement x = x + 1 must be changed to:   use the variable at the address given by x !!!

Example of "Do it yourself" pass-by-reference in C

The correct statement is *x = *x + 1 because *x is an alias of the int typed variable a: 

#include <stdio.h>        // DEMO this program   

void f(int *x)  
{
   *x = *x + 1;
}

int main(int argc, char *argv[])
{
   int a = 4;

   printf("Before calling f(a), a = %d\n", a); 
   f(&a);
   printf("After  calling f(a), a = %d\n", a); 
}

And this is how C achieves the effect of the pass-by-reference mechanism (DIY) !!!

Comment:   the reference parameter is just another example of aliasing

When we pass the address &a as parameter: 

#include <stdio.h>    

void f(int *x) // x will contain &a !! 
{  // It's as if the program has executed: x = &a;
   *x = *x + 1;
}

int main(int argc, char *argv[])
{
   int a = 4;

   printf("Before calling f(a), a = %d\n", a); 
   f(&a);
   printf("After  calling f(a), a = %d\n", a); 
}

The parameter variable x will contain &a.
Therefore:   *x will be an alias for a   and *x = *x + 1 will update the variable a

Functions with reference typed parameters

  • Recall that:

      • C only has the pass-by-value parameter passing mechanism

  • Reference typed parameters in functions will enable the (C) programmer to:

      • achieve the effect of the pass-by-reference mechanism using the * operator

    Example: (we have just seen this !) 

       void f(int *x)  
 {
    *x = *x + 1;            
 }

Declaring a function with reference typed parameters

Situation where you need to use function declaration: 

#include <stdio.h>



int main(int argc, char *argv[])
{
   int a = 4;

   printf("Before calling f(a), a = %d\n", a); 
   f(&a);  // Call function before it's defined
   printf("After  calling f(a), a = %d\n", a); 
}  

void f(int *x)     
{
   *x = *x + 1;
}

Declaring a function with reference typed parameters

Situation where you need to use function declaration: 

#include <stdio.h>

void f(int *x);  // Declare function

int main(int argc, char *argv[])
{
   int a = 4;

   printf("Before calling f(a), a = %d\n", a); 
   f(&a);  // No error now
   printf("After  calling f(a), a = %d\n", a); 
}  

void f(int *x)    
{
   *x = *x + 1;
}