variable argument list program behaving strangely - cstdarg

Question

This program gives the correct result just once. I have been trying to understand how the macros in cstadarg can be used to create and call functions that have a variable number of arugments.

#include <iostream>
#include <cstdarg>
using std::cout;
using std::endl;

int sum(int count, ...)
{

    if(count <= 0)
    return 0;

    va_list myarg_ptr; //create pointer to argument list
    va_start(myarg_ptr, count); // initate pointer to argument list

    int sum(0);

    for(int i=0; i<count; i++)
        sum += va_arg(myarg_ptr, int); // use and increment pointer to arugment list

    va_end(myarg_ptr); // set argument list pointer to NULL 
    return sum;
}


int main(int argc, char* argv[])
{
    cout << sum(9, 11, 22, 33, 44, 55, 66, 77, 6) << endl;
    cout << sum(6, 2, 4, 6, 8, 10, 5) << endl;
    cout << sum(9, 1, 2) << endl;

    std::system("pause");
    return 0;
}

The output I get is:

273156986
35
-173256537
Press any key to continue...

The 2nd output is correct, the other 2 make no sense at all. — quantum231
– quantum231, Commented Jul 26, 2013 at 14:57

P0W · Accepted Answer · 2013-07-26 15:03:49Z

1

Fix your sum() call , number of elements for summing should the first argument for sum()

so,

cout << sum(8, 11, 22, 33, 44, 55, 66, 77, 6) << endl; //8 arguments
cout << sum(6, 2, 4, 6, 8, 10, 5) << endl; // 6 arguments
cout << sum(2, 1, 2) << endl; //2 arguments

answered Jul 26, 2013 at 15:03

P0W

48.1k9 gold badges79 silver badges124 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

quantum231 Over a year ago

OH MY GOD!!! I did not realize that. I copied code from the book and than wrote some of my own. I expected that count will automatically have the correct number of arguments and that I won't have the pass on the number of arguments myself. Hmm, if we also have to pass on the number of arguments than this is not as flexible as I had thought it would be. What a blunder.

trojanfoe · Accepted Answer · 2013-07-26 15:03:39Z

1

The first argument to sum() is the number of following (variable) arguments. You aren't calling the function with the correct value the first and third time.

You want:

cout << sum(8, 11, 22, 33, 44, 55, 66, 77, 6) << endl;
cout << sum(6, 2, 4, 6, 8, 10, 5) << endl;
cout << sum(2, 1, 2) << endl;

answered Jul 26, 2013 at 15:03

trojanfoe

123k23 gold badges219 silver badges249 bronze badges

Comments

Dietmar Kühl · Accepted Answer · 2013-07-26 15:07:20Z

The first argument you explicitly mention in the portotype is the part of the arguments! If you want to get the count passed to you function you need to explicitly pass it yourself, e.g., the last call should look like this:

std::cout << sum(3, 9, 1, 2) << '\n';

(you should also stop excessive use of std::endl).

A possibly preferrable C++2011 approach could be the use of a proper variadic argument list:

template <typename T>
int sum(T value) { return value; }
template <typename T, typename... S>
int sum(T value, S... values) { return value + sum(values...); }

Variadic argument lists can correctly detect the number of arguments. When using variable argument lists you need to get the list of detected arguments exactly right which is sometimes nont entirely trivial.

Collectives™ on Stack Overflow

variable argument list program behaving strangely - cstdarg

3 Answers 3

1 Comment

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

1 Comment

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related