C float decimal not being memorized

Question

I have a piece of code that looks like this:

float nb = 100 / 42;
printf("%.2f", nb);

which I expect to print out 2.38, but instead it prints out 2.00.

The 42 is just an example. In the original code it's a variable.

Here's a related question: http://stackoverflow.com/questions/3602827/what-is-the-behavior-of-integer-division-in-c — Mark Westling, Jul 25 '14 at 00:57
Memorized? Whatcha mean? And what "float decimal"? Both 100 and 42 are int literals. — Jim Balter, Jul 25 '14 at 01:28
@Engine - it's just those that lack the integrity to help others learn by explaining there down-votes. We could all benefit from understanding had the down-vote been explained. Unfortunately, the down-vote is all too frequently used by those who feel their important down-vote says it all... — David C. Rankin, Jul 25 '14 at 03:48

score 0 · Accepted Answer · answered Jul 25 '14 at 00:55

0

You need to specify the numbers as floating point themselves. Try this:

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char** argv)
{
    float nb = 100.0/42.0;
    printf("%.2f\n", nb);
    return 0;
}

answered Jul 25 '14 at 00:55

CDahn

Note that `100.0/42.0` is a `double`, which is then downcast to `float`, and then upcast back to `double` for `printf()`. So the conversion may be unnecessary. – Dietrich Epp Jul 25 '14 at 01:10

1 Answers1