Question

Possible Duplicate:
strange output in comparison of float with float literal

When I am trying to compare 2 same float values it doesn’t print “equal values” in the following code :

void main()
{
    float a = 0.7;
    clrscr();
    if (a < 0.7)
        printf("value :  %f",a);
    else if (a == 0.7)
        printf("equal values");
    else
        printf("hello");
    getch();
}

Thanks in advance.

Answer 1

While many people will tell you to always compare floating point numbers with an epsilon (and it’s usually a good idea, though it should be a percentage of the values being compared rather than a fixed value), that’s not actually necessary here since you’re using constants.

Your specific problem here is that:

float a = 0.7;

uses the double constant 0.7 to create a single precision number (losing some precision) while:

if (a == 0.7)

will compare two double precision numbers (a is promoted first).

The precision that was lost when turning the double 0.7 into the float a is not regained when promoting a back to a double.

If you change all those 0.7 values to 0.7f (to force float rather than double), or if you just make a a double, it will work fine – I rarely use float nowadays unless I have a massive array of them and need to save space.

You can see this in action with:

#include <stdio.h>
int main (void){
    float f = 0.7;    // double converted to float
    double d1 = 0.7;  // double kept as double
    double d2 = f;    // float converted back to double

    printf ("double:            %.30f\n", d1);
    printf ("double from float: %.30f\n", d2);

    return 0;
}

which will output something like (slightly modified to show difference):

double:            0.6999999|99999999955591079014994
double from float: 0.6999999|88079071044921875000000
                            \_ different beyond here.