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Unexpected Output when adding two float numbers

Tags:

c++

c

I wrote the following C++ code:

float a, b;
int c;

a = 8.6;
b = 1.4;
c = a + b;

printf("%d\n", c);

The output is 10.

But when I run the following code:

float a, b;
int c;

a = 8.7;
b = 1.3;
c = a + b;

printf("%d\n", c);

The output is 9.

What is the difference between the two, as they are giving different outputs?

like image 669
chinmayaposwalia Avatar asked Sep 10 '11 05:09

chinmayaposwalia


1 Answers

There is no such number as 8.7 or 1.3 in floating point. There is a number 10, and a number -6.5, and a number 0.96044921875... but no 8.7 or 1.3.

At best, your computer can round 8.7 to the nearest floating point number, and round 1.3 to the nearest floating point number as well. The computer adds these rounded numbers to each other, and then rounds the result.

Do not use floating point numbers for money.

#include <stdio.h>
int main(int argc, char *argv[])
{
    float a = 8.7, b = 1.3;
    printf("Looks like: %.1f + %.1f = %.1f\n", a, b, a+b);
    printf("The truth: %.20f + %.20f = %.20f\n", a, b, a+b);
    return 0;
}

On an x86 GCC/Linux computer, I get the result:

Looks like: 8.7 + 1.3 = 10.0
The truth: 8.69999980926513671875 + 1.29999995231628417969 = 9.99999976158142089844

On a PPC GCC/OS X computer, I get the result:

Looks like: 8.7 + 1.3 = 10.0
The truth: 8.69999980926513671875 + 1.29999995231628417969 = 10.00000000000000000000

Notice how 8.7 and 1.3 are both rounded down in this particular case. If you chose numbers that get rounded up, you might see a number larger than 10 on the right hand side.

See What Every Computer Scientist Should Know About Floating-Point Arithmetic, by David Goldberg (link).

like image 187
Dietrich Epp Avatar answered Oct 30 '22 01:10

Dietrich Epp