CODEWITHSUNDEEP
c
pr1m3x
pr1m3x

Reputation: 2087

Strange behavior when printing float value

Тhese are my calculations

float c = 5.0 * (12.0 - 32.0) / 9.0;
printf("%f.2", c);

Result is -11.111111.2 but I expected to be -11.11. What is wrong ?

Upvotes: 1

Views: 106

Answers (3)

Seeder
Seeder

Reputation: 113

You need to change the Format specification of float; as it is incorrect Instead of using 

printf("%f.2", c);

try using this:

float c = 5.0 * (12.0 - 32.0) / 9.0;
printf("%1.2f", c);

this should work! :)

Upvotes: 0

Sangeeth Saravanaraj
Sangeeth Saravanaraj

Reputation: 16597

printf("%f.2", c);

should be changed to 

printf("%.2f", c);

The update code is as follows:

$ cat f.c 
#include <stdio.h>

int main()
{
    float c = 5.0 * (12.0 - 32.0) / 9.0;
    printf("%.2f \n", c);

    return 0;
}
$ gcc f.c 
$ ./a.out 
-11.11 
$

Upvotes: 2

NPE
NPE

Reputation: 500167

The format specifier should read "%.2f".

In what you have right now, the .2 is misplaced. This is why you get more digits than expected, and why the .2 appears verbatim at the end of the output.

Upvotes: 5

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