int s = (.7 * 3) + (.3 * 3) is 2 ?
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I can't figure this out. I think it is from roundoff error, but I don't know how to correct it.
squicklid: I conqur.
By default I believe float to int conversion rounds down.
Therefore, you probably want "round to nearest" instead. Just add 0.5 to the end of the equasion, and you'll be set.
-Mike
By default I believe float to int conversion rounds down.
Therefore, you probably want "round to nearest" instead. Just add 0.5 to the end of the equasion, and you'll be set.
-Mike
Quote:Original post by MichaelT
.7 * 3 = 2.1
.3 * 3 = 0.9
(int)2.1 = 2;
(int)0.9 = 0;
2 + 0 = 2;
WRONG order of operations for his statement, sorry.
(.7 * 3) gets treated as a float or double (froget which by default).
Next, (.3 * 3) is evaluated (also float or double).
Then, addition is preformed. If (.3 * 3) had been an int, it would be implicitly cast to float. (float * int = float, int * float = int).
What's happening is:
.7 * 3 = ~2.1
.3 * 3 = ~0.9
~2.1 + ~0.9 = 2.99999
int(2.999999) = 2
Quote:
.7 * 3 = ~2.1
.3 * 3 = ~0.9
~2.1 + ~0.9 = 2.99999
int(2.999999) = 2
.7*3 equals exactly 2.1
and .3 *3 = exactly 0.9
so 2.1+0.9=3 so that's cant be the case
Quote:Original post by keen
.7*3 equals exactly 2.1
and .3 *3 = exactly 0.9
Nothing is 'exactly' in floating point math. If you want to write 0.9 exacly in FP mathy you need 'exactly' infinite digits. [grin]
int main(){ int a,b; float c; a=(.7 * 3) + (.3 * 3); b=((float)(.7 * 3)) + ((float)(.3 * 3)); c=(.7 * 3) + (.3 * 3); printf("%i,%i,%f\n",a,b,c);}output: 2,3,3.000000
proves my point
Quote:Original post by _DarkWIng_Quote:Original post by keen
.7*3 equals exactly 2.1
and .3 *3 = exactly 0.9
Nothing is 'exactly' in floating point math. If you want to write 0.9 exacly in FP mathy you need 'exactly' infinite digits. [grin]
But those numbers are doubles, not floats. I agree with MichaelT on this one.
/MindWipe
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