C variable parameters

Question

I have this homework to do: "Determine the minimum of 10 precision double numbers from a string (implicit values or from the KB) using a function with a variable number of parameters. The first 7 values will be considered initially, next the last 3 and at the end these 2 values." Well I made it all but I don't know why it gives me some strange results. Here's the code:

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <stdlib.h>
#include <stdarg.h>
#include <conio.h>
double min(double,...);

void main(){
    double a,b,c;
    printf("Introduceti numerele: ");
    scanf("%lf%lf%lf",&a,&b,&c);
    printf("\nMinimul este %lf",min(10,1.34,4.34,7,5.23,6.23,2,8.232,a,b,c));
    _getch();
}

double min(double x,...){
    int i;
    double y;
    va_list ap;
    va_start(ap,x);
    y=va_arg(ap,double);
    for(i=0;i<x;i++){
        if(y>va_arg(ap,double))
            y=va_arg(ap,double);
    }
    va_end(ap);
    return y;
}

Also i don't know why the compiler knows about what argument is next cause i is not used in va_arg(ap,double).

for(i=0;i<x;i++){
    if(y>va_arg(ap,double))
        y=va_arg(ap,double);

What is a strange result (apart from when Donald Trump leaves the hairdressers)? — Ed Heal
– Ed Heal, Commented Mar 6, 2016 at 15:14
Try int min = va_arg(ap,double), y; for(i = 1; i < x; i++){ y = va_arg(ap, double); if(min > y) min = y; } — Spikatrix
– Spikatrix, Commented Mar 6, 2016 at 15:23
since the header file: conio.h is not portable, suggest elimination of that statement and instead of calling _getch() to call getchar() — user3629249
– user3629249, Commented Mar 7, 2016 at 21:34

Jonathan Leffler · Accepted Answer · 2016-03-06 20:25:45Z

3

The first parameter in the call to your function min is the number of arguments, and it has the type int:

#include <stdarg.h>

double min( int numberOfArgs, ... )
         // ^^^
{
    va_list argptr;
    va_start( argptr, numberOfArgs );           // initialize argument pointer

    double minData = va_arg( argptr, double );  // initialize the minimum with the first argument
                                                //   and increment argument pointer
    for ( int i = 1; i < numberOfArgs; i ++ )   // for all of the following arguments
    {
        double data = va_arg( argptr, double ); // get argument and increment argument pointer
        if ( data < minData )                   // test if argument is less than mnimum
            minData = data;
    }
    va_end( argptr );

    return minData;
}

Ensure that all of your arguments in the argument list are floating point values of type double:

int main()
{
    double a, b, c;
    printf("Introduceti numerele: ");
    scanf_s("%lf%lf%lf", &a, &b, &c);

    double minVal = min( 10, 1.34, 4.34, 7.0, 5.23, 6.23, 2.0, 8.232, a, b, c)
                                       // ^^               ^^     
    printf("\nMinimul este %lf", minVal);
    return 0;
}

edited Mar 6, 2016 at 20:25

Jonathan Leffler

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answered Mar 6, 2016 at 15:32

Rabbid76

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1 Comment

Jonathan Leffler Over a year ago

You could improve the function declaration by insisting on one double argument: double min(int numargs, double v1, ...). This prevents you from calling it with 0 actual floating point arguments (but doesn't prevent you calling it with the number of arguments as 0 or even negative — but C is not a language for idiots to use). It would mean you only use va_arg() in the body of the loop.

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