CODEWITHSUNDEEP
coutput
yupe
yupe

Reputation: 127

Why can't i get the proper output by this code?

I'm working on a problem that troubling me for a while. Still can't understand. It makes sense to change the type of m, but I still can't understand what m went through in this process. How does its value change? Below is the code.

I've tried it on code blocks.

#include<stdio.h>
int main()
{
    int m=5;
    m = (float)m / (2.0);//There, shouldn't m be 2.500000?
    printf("%f\n", m);
    printf("%0.2f\n",(float)m/2.0);
    getchar();
    return 0;
}

Expected result:

2.500000
1.25

Actual result:

nan
1.00

Upvotes: 1

Views: 90

Answers (2)

Blaze
Blaze

Reputation: 16876

m is an int. So if you do m = (float)m / (2.0);, then even if (float)m / (2.0) is 2.5, it will be truncated to be stored in m as 2. So when you do this

printf("%f\n", m);

You have undefined behavior because it's an int and you say it's a float.

Later you do this

printf("%0.2f\n",(float)m/2.0);

and you get 1.0 because m was 2 and you divide it by 2.0. You can get the expected output by changing the type of m to float instead.

Upvotes: 5

pmg
pmg

Reputation: 108986

printf("%f\n", m); // m is int

Nope. "%f" does not go with int.

Either use "%d" or convert the value

printf("%d\n", m);
printf("%f\n", (double)m);

Note that converting to float forces a second automatic conversion to double

printf("%f\n", (float)m); // convert twice

Upvotes: 1

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