CODEWITHSUNDEEP
c
user4284784
user4284784

Reputation:

C function variable argument length

I want to pass variable argument to a function but don't want set the number in argument. so I check the end of argument with 0 as below:

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

void foo(char* str, ... )
{
    printf("str: %s\n",str);
    va_list arguments;                     
    va_start(arguments,str);
    while(1) {
        double data = va_arg(arguments,double); 
        if (data == 0) {
            break;
        }
        printf("data: %.2f\n",data);
    }
    va_end (arguments);
    return;
}

int main()
{
    foo("hello", 5.5, 4.5, 3.5);
}

But it seems the out put not correct:

str: hello
data: 5.50
data: 4.50
data: 3.50
data: 34498135662005122837381407988349324615424289861359674446209831441373336786008345008385190019649501215393603210517859097842586295826918562939434091794406737728862534747430060032.00

The last data output is random value, it should not there!

Upvotes: 0

Views: 82

Answers (2)

sth
sth

Reputation: 229593

When you call the function, you don't provide a 0.0 argument. The function tries to print everything up to a zero argument, but there is not such argument.

The function will go on to look at memory where it would expect a 5th, 6th, ... parameter until one such location randomly contains a zero or the program crashes.

Upvotes: 3

Weather Vane
Weather Vane

Reputation: 34585

You could solve it by adding a sentinel value to the values passed.

int main()
{
    foo("hello", 5.5, 4.5, 3.5, 0.0);
}

This had to be 0.0 in my test, not 0.

Upvotes: 4

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