C++ wrong result of mathematical expression

Question

I have to calculate some simple mathematical expression, but when I do it in one row, the result always will be zero. But the correct result is obviously not zero. And its interesting but when I separate the parts of expression, I get the correct answer. Later I will divide with this result, so it should not be 0.

The expression is like this:

(X-X1)/(X2-X1)

In this case the delta: 0

double delta = (x - x1) / (x2 - x1);

But this way the delta will be correct:

double top = x - x1;
double bottom = x2 - x1;
double delta = top/bottom;

Do you have any idea, how could this happen?

I take it x, x1 and x2 are integer values, and `x2-x1` is greater than `x-x1`? — Mats Petersson, Mar 21 '13 at 14:47
Possible duplicate of [I didn't bother learning C before hacking on the Linux kernel](http://google.com/search?q=integer+division+c) — , Mar 21 '13 at 14:49
Yes, they are unsigned long int and yes, x2-x1 is the greater. — Steve M. Bay, Mar 21 '13 at 14:49

score 11 · Accepted Answer · edited Mar 21 '13 at 15:22

11

This is a typical case of "integer division used to make a float value". If x, x1 and x2 are all integers, then the compiler should, according to the rules of C and C++ use integer calculations to calculate x-x1 and x2-x1, and then divide the differences of those as integer division. If x2-x1 is greater than x-x1, then the result is zero [assuming positive values like 7/14. -14 / -7 is obviously 2].

The easy fix is to convert the type of one expression to a floating point type:

double delta = static_cast<double>(x - x1) / (x2 - x1);

edited Mar 21 '13 at 15:22

Bill

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answered Mar 21 '13 at 14:50

Mats Petersson

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And by "integer division" it is meant that the answer is truncated, that is, the whole part of the answer is kept and the remainder is discarded. So `1/2` gives 0 with a remainder of 1 which is thrown out, giving `1 / 2 == 0`. – Reinstate Monica -- notmaynard Mar 21 '13 at 14:53
Thanks a lot, I understood the problem. (I did not know how to describe my problem well, thats why I asked the question.) – Steve M. Bay Mar 21 '13 at 14:56

score 2 · Answer 2 · answered Mar 21 '13 at 14:49

2

In the first case, integer / integer, result is still integer, then convert integer to double.

In the second case, double / double, then you get correct result.

answered Mar 21 '13 at 14:49

TieDad

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score 1 · Answer 3 · answered Mar 21 '13 at 14:50

In the first case, you are performing int / int which will result in a int value. So, if your denominator is larger than the numerator, the result will be truncated to 0.

In the second case, you are evaluating a double/double expression which will evaluate to a double value, hence preserving the fractional and integer parts.

score 0 · Answer 4 · answered Mar 21 '13 at 15:52

0

You have to cast the integer division to double. type casting

answered Mar 21 '13 at 15:52

andre_lamothe

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C++ wrong result of mathematical expression

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