Float addition promoted to double?

Posted by Andreas Brinck on Stack Overflow See other posts from Stack Overflow or by Andreas Brinck
Published on 2009-12-03T11:01:48Z Indexed on 2010/05/04 20:48 UTC
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I had a small WTF moment this morning. Ths WTF can be summarized with this:

float x = 0.2f;
float y = 0.1f;
float z = x + y;
assert(z == x + y); //This assert is triggered! (Atleast with visual studio 2008)

The reason seems to be that the expression x + y is promoted to double and compared with the truncated version in z. (If i change z to double the assert isn't triggered).

I can see that for precision reasons it would make sense to perform all floating point arithmetics in double precision before converting the result to single precision. I found the following paragraph in the standard (which I guess I sort of already knew, but not in this context):

4.6.1. "An rvalue of type float can be converted to an rvalue of type double. The value is unchanged"

My question is, is x + y guaranteed to be promoted to double or is at the compiler's discretion?

UPDATE: Since many people has claimed that one shouldn't use == for floating point, I just wanted to state that in the specific case I'm working with, an exact comparison is justified.

Floating point comparision is tricky, here's an interesting link on the subject which I think hasn't been mentioned.

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