在 C# 中将 double 转换为 int

Converting a double to an int in C#(在 C# 中将 double 转换为 int)

问题描述

在我们的代码中，我们需要将 double 转换为 int.

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;

谁能解释一下为什么i1 != i2?

我得到的结果是:i1 = 9 and i2 = 8.

The result that I get is that: i1 = 9 and i2 = 8.

推荐答案

因为 Convert.ToInt32 回合:

Because Convert.ToInt32 rounds:

返回值:四舍五入到最接近的 32 位有符号整数.如果值在两个整数之间，返回偶数；即4.5转4，5.5转6.

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.

...而演员截断:

当您从 double 或 float 值转换为整数类型时，值被截断.

When you convert from a double or float value to an integral type, the value is truncated.

更新:有关其他差异，请参阅下面 Jeppe Stig Nielsen 的评论(但是，如果 score 是实数，就像这里的情况一样，这些差异就不会起作用).

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).

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