Evaluating mathematical operations

Question

I'm running this snippet of code:

int x, y; float val1, val2, val3; x = 20; y = 25; val1 = x - 10; val2 = y/val1; val3 = float(val2); val4 = float(y/(x-10));

The above outputs:

val1 = 10
val2 = 2.5
val3 = 2.5
val4 = 2

Can anyone explain to me why val3 and val4 evaluate differently? ~~, when in fact they're both doing the same calculations?~~

Possible duplicate of [Why can't I return a double from two ints being divided](http://stackoverflow.com/questions/7571326/why-cant-i-return-a-double-from-two-ints-being-divided) — , Oct 02 '16 at 17:55
"when in fact they're both doing the same calculations" - Wrong prerequisite. — too honest for this site, Oct 02 '16 at 18:01

score 0 · Answer 1 · answered Oct 02 '16 at 17:54

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(y/x-10) is dealing with integers. The float isn't applied until after this calculation, which would produce 2.

y and x are both integers. So, 20 (value of x) - 10 = 10. Then 25 / 10 (still dealing with integers - no floats yet) = 2.

answered Oct 02 '16 at 17:54

Gavin

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score 0 · Accepted Answer · answered Oct 02 '16 at 18:00

0

They are evaluated differently.

val3 = float(y / float(x - 10))
val4 = float(y / (x - 10)

For val3, the types are float(int / float) and val4 is float(int / int) which will cast the result of y / (x - 10) to int first and then cast it to float.

answered Oct 02 '16 at 18:00

Pan Long

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Evaluating mathematical operations

2 Answers2