Variable number of parameters in function in C++

Question

How I can have variable number of parameters in my function in C++.

Analog in C#:

public void Foo(params int[] a) {
    for (int i = 0; i < a.Leng         


        

Answer 1

If you don't care about portability, you could port this C99 code to C++ using gcc's statement expressions:

#include <cstdio>

int _sum(size_t count, int values[])
{
    int s = 0;
    while(count--) s += values[count];
    return s;
}

#define sum(...) ({ \
    int _sum_args[] = { __VA_ARGS__ }; \
    _sum(sizeof _sum_args / sizeof *_sum_args, _sum_args); \
})

int main(void)
{
    std::printf("%i", sum(1, 2, 3));
}

You could do something similar with C++0x' lambda expressions, but the gcc version I'm using (4.4.0) doesn't support them.

Answer 2

These are called Variadic functions. Wikipedia lists example code for C++.

To portably implement variadic functions in the C programming language, the standard stdarg.h header file should be used. The older varargs.h header has been deprecated in favor of stdarg.h. In C++, the header file cstdarg should be used.

To create a variadic function, an ellipsis (...) must be placed at the end of a parameter list. Inside the body of the function, a variable of type va_list must be defined. Then the macros va_start(va_list, last fixed param), va_arg(va_list, cast type), va_end(va_list) can be used. For example:

#include <stdarg.h>

double average(int count, ...)
{
    va_list ap;
    int j;
    double tot = 0;
    va_start(ap, count); //Requires the last fixed parameter (to get the address)
    for(j=0; j<count; j++)
        tot+=va_arg(ap, double); //Requires the type to cast to. Increments ap to the next argument.
    va_end(ap);
    return tot/count;
}