Why 0.1 + 0.2 == 0.3 in D?

Question

assert(0.1 + 0.2 != 0.3); // shall be true 

is my favorite check that a language uses native floating point arithmetic.

C++

#include <cstdio>  int main() {    printf("%d\n", (0.1 + 0.2 != 0.3));    return 0; } 

Output:

1 

http://ideone.com/ErBMd

Python

print(0.1 + 0.2 != 0.3) 

Output:

True 

http://ideone.com/TuKsd

Other examples

• Java: http://ideone.com/EPO6X
• C#: http://ideone.com/s14tV

Why is this not true for D? As understand D uses native floating point numbers. Is this a bug? Do they use some specific number representation? Something else? Pretty confusing.

D

import std.stdio;  void main() {    writeln(0.1 + 0.2 != 0.3); } 

Output:

false 

http://ideone.com/mX6zF

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UPDATE

Thanks to LukeH. This is an effect of Floating Point Constant Folding described there.

Code:

import std.stdio;  void main() {    writeln(0.1 + 0.2 != 0.3); // constant folding is done in real precision     auto a = 0.1;    auto b = 0.2;    writeln(a + b != 0.3);     // standard calculation in double precision } 

Output:

false true 

http://ideone.com/z6ZLk

Answer 1

_{(Flynn's answer is the correct answer. This one addresses the problem more generally.)}

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You seem to be assuming, OP, that the floating-point inaccuracy in your code is deterministic and predictably wrong (in a way, your approach is the polar opposite of that of people who don't understand floating point yet).

Although (as Ben points out) floating-point inaccuracy is deterministic, from the point of view of your code, if you are not being very deliberate about what's happening to your values at every step, this will not be the case. Any number of factors could lead to 0.1 + 0.2 == 0.3 succeeding, compile-time optimisation being one, tweaked values for those literals being another.

Rely here neither on success nor on failure; do not rely on floating-point equality either way.

Answer 2

It's probably being optimized to (0.3 != 0.3). Which is obviously false. Check optimization settings, make sure they're switched off, and try again.