Identical float values comparing as inequal [duplicate]

Answer 1

That is because of the precision of float value.

Check this:- What Every Computer Scientist Should Know About Floating-Point Arithmetic

You can better use double instead of float.

Also to note that(Correctly pointed by WhozCraig) that 0.1 is a double not a float. Change the float to double and you will get the expected output.

Answer 2

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");
}

Answer 3

Well, that's a problem float type has. Check that if you change float with double type it prints true instead of false. Also try this output:

#include <stdio.h>

int main()
{
    float f;
    for(f=.0;f<10.;f+=0.1)
        printf("%f \n",f);    
}

Answer 4

0.1 cannot be represented exactly in binary . 0.1 in binary would be 0.0r0011r which is recurring so internally it is rounded off which is not exact value. That is why it returns false. But if you try with number like 0.5 which is 0.1 in binary or 0.250 or 0.125 whichever can be perfectly represented it would return true. Also read http://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html . Better to use double for more accurate results.