Friday, 24 February 2017

VARIABLE ARGUMENTS

VARIABLE ARGUMENTS

Sometimes, you may come across a situation, when you want to have a
function, which can take variable number of arguments, i.e., parameters,
instead of predefined number of parameters. The C programming language
provides a solution for this situation and you are allowed to define a function
which can accept variable number of parameters based on your requirement.
The following example shows the definition of such a function.

int func(int, ... )
{
.
.
.
}
int main()
{
func(1, 2, 3);
func(1, 2, 3, 4);
}


It should be noted that the function func() has its last argument as ellipses,
i.e., three dotes (...) and the one just before the ellipses is always an int which
will represent the total number variable arguments passed. To use such
functionality, you need to make use of stdarg.h header file which provides the
functions and macros to implement the functionality of variable arguments and
follow the given steps:

1. Define a function with its last parameter as ellipses and the one just
before the ellipses is always an int which will represent the number of
arguments.

2. Create a va_list type variable in the function definition. This type is
defined in stdarg.h header file.

3. Use int parameter and va_start macro to initialize the va_list variable to
an argument list. The macro va_start is defined in stdarg.h header file.

4. Use va_arg macro and va_list variable to access each item in argument
list.

5. Use a macro va_end to clean up the memory assigned
to va_list variable.

Now let us follow the above steps and write down a simple function which can
take the variable number of parameters and return their average:

#include <stdio.h>
#include <stdarg.h>
double average(int num,...)
{
va_list valist;
double sum = 0.0;
int i;
/* initialize valist for num number of arguments */
va_start(valist, num);
/* access all the arguments assigned to valist */
for (i = 0; i < num; i++)
{
sum += va_arg(valist, int);
}
/* clean memory reserved for valist */
va_end(valist);
return sum/num;
}
int main()
{
printf("Average of 2, 3, 4, 5 = %f\n", average(4, 2,3,4,5));
printf("Average of 5, 10, 15 = %f\n", average(3, 5,10,15));
}


When the above code is compiled and executed, it produces the following result.
It should be noted that the function average() has been called twice and each
time the first argument represents the total number of variable arguments being
passed. Only ellipses will be used to pass variable number of arguments.

Average of 2, 3, 4, 5 = 3.500000
Average of 5, 10, 15 = 10.000000


No comments:

Post a Comment