Problem regarding type in function with variable parameters in C

Question

I’m having trouble solving a problem in a function that receives variable parameters. I have a function that sums a variable amount of numbers, its first parameter is the amount of numbers to be summed, and the others are the numbers themselves, it works when working with double numbers, but not when using numbers of type int, compiling this code the output will be 60.00, as expected:

#include <stdio.h>
#include <stdarg.h>

double somaVariaveis(int qtd, ...)
{
    va_list args;
    int i;
    double soma = 0;

    va_start(args, qtd);

    for(i = 0; i < qtd; i++)
        soma += va_arg(args, double);

    va_end(args);

    return soma;
}

int main()
{
    printf("soma de parametros variaveis: %.2lf", somaVariaveis(3, 10.0, 20.0, 30.0));

    return 0;
}

Now, in case I try to print the sum of integers:

printf("soma de parametros variaveis: %.2lf", somaVariaveis(3, 10, 20, 30));

The result will be 0.00;

Where did I go wrong? Shouldn’t I have cast when I pass variables of type int and specify that I want double variables? I noticed some difficulty when working with variable parameters in relation to the type, in another example, if I try to add parameters of type float:

#include <stdio.h>
#include <stdarg.h>

float somaVariaveis(int qtd, ...)
{
    va_list args;
    int i;
    float soma = 0;

    va_start(args, qtd);

    for(i = 0; i < qtd; i++)
        soma += va_arg(args, float);

    va_end(args);

    return soma;
}

int main()
{
    printf("soma de parametros variaveis: %.2f", somaVariaveis(3, 10.0f, 20.0f, 30.0f));
    return 0;
}

I get the following Warning, and I have a running problem.:

Warning: 'float' is promoted to 'double' when passed through '...'

note: (so you should pass 'double' not 'float' to 'va_arg')

note: if this code is reached, the program will abort

Answer 1

Face with explicit function cast.

#include <stdio.h>
#include <stdarg.h>

double somaVariaveis(int qtd, ...)
{
    va_list args;
    int i;
    double soma = 0;

    va_start(args, qtd);

    for(i = 0; i < qtd; i++)
        soma += va_arg(args, double);

    va_end(args);

    return soma;
}

int main()
{
    printf("soma de parametros variaveis: %.2lf", somaVariaveis(3, (double) 10, (double) 20, (double) 30));

    return 0;
}

Answer 2

You have to cast the int for double.

With this prototype double somaVariaveis(int qtd, ...) this function accepts parameters of any type in place of the 3 dots. In the general case there is no way for the compiler to know how the parameters will be interpreted by the function called.

As for the second program, the message is very explanatory: when you pass parameters float for a function, the compiler "promotes" the float for double when the function is called. However, its function considers parameters float, and as the sizes of float and double are different, your program will probably abort as the execution stack (stack) will get corrupted.

The two cases are similar.

This should also happen if your function with va_args deal with char, and you pass char in the call, because in function call (if I am not mistaken) parameters char are promoted to int.

Answer 3

If working only with numerical values, casting works well with array.

#include <stdio.h>

double soma(int qtd, double s[]){
    double ret=0;
    for(int i=0;i<qtd;i++){
        ret+=s[i];
    }
    return ret;
}

int main()
{   
    int arg1=8;
    float arg2=4.3,arg3=5;
    double arg4=2.2,arg5=4;
    double n[100]={arg1,arg2,arg3,arg4,arg5,1,16.4};
    printf("\nsoma de parametros variaveis: %.2lf\n\n", soma(7, n));
    n[7]=200.0;
    printf("\nsoma de parametros variaveis: %.2lf\n\n", soma(8, n));
    return 0;
}