I was working on some 'determine the output' questions in C. I came across this question which looked simple on the face of it, but after running the code left me puzzled.
The output I expected was "True". However upon running, it was "False". And when I checked the value of f using printf(), it showed 0.1. Can someone please explain why f being assigned 0.1 doesn't return true for the IF statement?
There was no explanation for the answer from where I picked the question up, and I wasn't able to find an answer myself.
#include <stdio.h>
int main()
{
float f = 0.1;
if (f == 0.1)
printf("True");
else
printf("False");
}
The problem is that variable f defined as having type float while float constant 0.1 has type double. Type double has more precision than type float. it has more binary digits to represent fraction. So in this statement
float f = 0.1;
there is a truncation.
To get the expected result you should write at least
if (f == 0.1f)
Also that to be sure that there is no rounding the code should look as
#include <stdio.h>
int main( void )
{
float f = 0.1f;
if ( f == 0.1f )
printf("True\n");
else
printf("False\n");
}
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