| Reporter |
Michael Heyman (Michael.Heyman)
|
|---|---|
| Created | Mar 28, 2012 12:26:30 AM |
| Updated | Mar 28, 2012 12:26:30 AM |
| Priority | Normal |
| Type | Feature |
| Fix versions | No Fix versions |
| State | Submitted |
| Assignee | Evgeny Pasynkov (pasynkov) |
| Subsystem | Code Analysis - Find Code Issues |
| Affected versions | No Affected versions |
| Fixed in build | No Fixed in build |
An error that I've seen multiple times regards using Convert.ToInt32() as if it were truncation - typically when rounding a floating point number to an integer by doing something like:
i = Convert.ToInt32(f + .5);
This, of course, is wrong because Convert.ToInt32() rounds to an even number so for 5.0 <= f < 5.5, the code sets i = 6. The coder obviously doesn't expect this.
The answer is to use (int)(f + .5) to rounding with the midpoint rounding up (what was probably expected). Better would be midpoint rounding away from zero with (int)Math.Round() (although a coworker said he had noticeably poorer performance with (int)Math.Round() then (int)(f + (f > 0. ? .5 : -.5)) in a tight loop).
i = Convert.ToInt32(f + .5);
This, of course, is wrong because Convert.ToInt32() rounds to an even number so for 5.0 <= f < 5.5, the code sets i = 6. The coder obviously doesn't expect this.
The answer is to use (int)(f + .5) to rounding with the midpoint rounding up (what was probably expected). Better would be midpoint rounding away from zero with (int)Math.Round() (although a coworker said he had noticeably poorer performance with (int)Math.Round() then (int)(f + (f > 0. ? .5 : -.5)) in a tight loop).