Converting a double to an int in C#
c#, double, int
Solution
Because `Convert.ToInt32` rounds:
Return Value: rounded to the nearest 32-bit signed integer. If value is halfway between two whole numbers, the even number is returned; that is, 4.5 is converted to 4, and 5.5 is converted to 6.
...while the cast truncates:
When you convert from a double or float value to an integral type, the value is truncated.
Update: See Jeppe Stig Nielsen's comment below for additional differences (which however do not come into play if `score` is a real number as is the case here).
Problem
In our code we have a double that we need to convert to an int. ``` double score = 8.6; int i1 = Convert.ToInt32(score); int i2 = (int)score; ``` Can anyone explain me why `i1 != i2`? The result that I get is that: `i1 = 9` and `i2 = 8`.