One float variable shows expected result but other doesn't

Question

float a = 0.5;
float b = 0.7;
printf("%f",a);
printf("\n");
printf("%f",b);
if(a == .5)
  printf("\nOK\n");
else
   printf("NOTOK\n");

if(b == .7)
  printf("OK");
else
  printf("NOTOK");

It seems that output should be OK for both a and b but the output is OK for a but NOTOK for b. Why?

You've declared `b` as `float`, which is smaller than the default floating point type (`double`), so you're comparing `(float) 0.7` to `(double) 0.7)`. Since `0.7` cannot be exactly represented, they are not the same. Try replacing all your `float` variables with `double`. — Tom Karzes, Oct 18 '17 at 03:54
https://stackoverflow.com/questions/588004/is-floating-point-math-broken — Retired Ninja, Oct 18 '17 at 03:55
Also https://stackoverflow.com/questions/1839422/strange-output-in-comparison-of-float-with-float-literal — n. m. could be an AI, Oct 18 '17 at 04:15
Another option [Make C floating point literals float (rather than double)](https://stackoverflow.com/questions/32266864/make-c-floating-point-literals-float-rather-than-double) — David C. Rankin, Oct 18 '17 at 04:15

score 1 · Answer 1 · edited Oct 18 '17 at 04:13

1

For float numbers you need to represent it as example: 0.5f .Below example is tested and working fine.

float a = 0.5;
float b = 0.7;
printf("%f",a);
printf("\n");
printf("%f",b);

if(a == 0.5f && b == 0.7f)
{
  printf("\nOK\n");
}
else
{
   printf("NOTOK\n");
}

edited Oct 18 '17 at 04:13

David C. Rankin

81,885
6
58
85

answered Oct 18 '17 at 04:07

User

31
5

score 0 · Answer 2 · answered Oct 18 '17 at 05:37

0

.5 and .7 are constants and they are translated to double. Add "f" symbol at the end to tell compiler that .5f and .7f are floats.

if(a == .5f)
  printf("\nOK\n");
else
   printf("NOTOK\n");

if(b == .7f)
  printf("OK");
else
  printf("NOTOK");

answered Oct 18 '17 at 05:37

Gor Asatryan

904
7
24

One float variable shows expected result but other doesn't

2 Answers2