Sometimes, you may come across a situation, when you want to have a function that can accept a variable number of arguments (parameters) instead of a predefined number of arguments. The C programming language provides a solution for this situation.
Read this chapter to learn how you can define a function that can accept a 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 get such a functionality, you need to use the stdarg.h header file which provides the functions and macros to implement the functionality of variable arguments.
Follow the steps given below −
- 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.
- Create a va_list type variable in the function definition. This type is defined in stdarg.h header file.
- 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.
- Use va_arg macro and va_list variable to access each item in argument list.
- Use a macro va_end to clean up the memory assigned to va_list variable.
Example
Let us now 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 = %fn", average(4, 2,3,4,5)); printf("Average of 5, 10, 15 = %fn", average(3, 5,10,15)); }
Output
When the above code is compiled and executed, it produces the following output −
Average of 2, 3, 4, 5 = 3.500000 Average of 5, 10, 15 = 10.000000
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 are used to pass variable number of arguments.