why in o/p the values of i is different

Question

Why i am getting the different values of i.

#include <stdio.h>
void pri(int,int);
int main()
{
    float a=3.14;
    int i=99;
    pri(a,i);
    getch();
}
void pri(int a,int i)
{
    printf("a=%f i=%d\n",a,i);
    printf("i=%d a=%f\n",i,a);
}

Please tell us a bit more, I can't understand your question. — joar, Sep 27 '12 at 10:30

score 3 · Answer 1 · edited May 23 '17 at 11:48

3

You declared a as an int, but you are using %f so it should be declared as a float:

void pri(float a, int i)
{
    printf("a=%f i=%d\n", a, i);
    printf("i=%d a=%f\n", i, a);
}

If you have an incorrect type you get undefined behaviour. The specification for printf reads (7.19.6.1 paragraph 9):

If a conversion speciﬁcation is invalid, the behavior is undeﬁned. If any argument is not the correct type for the corresponding conversion speciﬁcation, the behavior is undeﬁned.

Emphasis mine.

Source

edited May 23 '17 at 11:48

Community

1
1

answered Sep 27 '12 at 08:07

Mark Byers

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ya i know that's the correction for dis program.....but i am not getting why both consecutive printf statements giving different values of i. – Tushar Gaurav Sep 27 '12 at 08:23
@TusharGaurav: Because as you can see from my edit, it's specified as **undefined behaviour**. Anything can happen when you invoke UB, including incorrect results being displayed. – Mark Byers Sep 27 '12 at 08:37

score 0 · Accepted Answer · answered Sep 27 '12 at 08:35

to explain your comment's question:

printf("a=%f i=%d\n", a, i);

will evaluate to something like:

printf("a=");
printf(/*next arguments up to size of float*/ a, i);
printf(" i=");
printf(/*...decimal*/ /*whatever is on the stack, because a and i are already consumed*/);

you can rescue your function with: 'printf("a=%f i=%d\n", (float)a, i);'

why in o/p the values of i is different

2 Answers2