Math.Round为什么不返回int? [英] Why doesn't Math.Round return an int?

问题描述

在C#中，为什么舍入数学函数Floor，Ceiling和Round不返回int?考虑到函数的结果将始终是整数，为什么它返回float，double或decimal?

In C#, why don't the rounding math functions Floor, Ceiling and Round return an int? Considering the result of the function will always be an integer, why does it return a float, double or decimal?

推荐答案

double的范围为±5.0×10 ⁻³²⁴ 至±1.7×10 ³⁰⁸ 和

double has the range of ±5.0 × 10⁻³²⁴ to ±1.7 × 10³⁰⁸ and long has the range of –9,223,372,036,854,775,808 to 9,223,372,036,854,775,807. Unfortunately not all integral floating point values can be represented by an integer.

例如，1e19已经超出了64位带符号整数的范围.

For example, 1e19 has already exceeded the range of a 64-bit signed integer.

(long)Math.Round(1e19) == -9223372036854775808L // WTF?

虽然单参数重载Math.Round(double)和Math.Round(decimal)始终会返回整数值，但这些重载仍然无法返回整数类型.

While it is true that the single-argument overloads Math.Round(double) and Math.Round(decimal) will always return an integral value, these overloads still cannot return an integer value type.

如果您知道，传递给函数的值将返回一个可以用整数类型表示的值，则可以自己进行强制转换.该库不会这样做，因为它需要考虑一般情况.

If you know that the value passed to the function will return a value representable by an integer value type, you can cast it yourself. The library won't do that because it needs to consider the general case.

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