Convert.ToInt32 和 (int) 有什么区别? [英] What is the difference between Convert.ToInt32 and (int)?

问题描述

以下代码会引发类似的编译时错误

The following code throws an compile-time error like

无法将类型 'string' 转换为 'int'

Cannot convert type 'string' to 'int'

string name = Session["name1"].ToString();
int i = (int)name;

而下面的代码编译并执行成功:

whereas the code below compiles and executes successfully:

string name = Session["name1"].ToString();
int i = Convert.ToInt32(name);

我想知道:

1. 为什么第一个代码会产生编译时错误?

1. Why does the the first code generate a compile-time error?

这两个代码片段有什么区别?

What's the difference between the 2 code snippets?

推荐答案

(int)foo 只是对 Int32 (int 在 C# 中) 类型.这是内置在 CLR 中的，并且要求 foo 是数字变量(例如 float、long 等).从这个意义上说，它是非常类似于 C 中的演员表.

(int)foo is simply a cast to the Int32 (int in C#) type. This is built into the CLR and requires that foo be a numeric variable (e.g. float, long, etc.) In this sense, it is very similar to a cast in C.

Convert.ToInt32 被设计成一个通用的转换函数.它比铸造更重要.也就是说，它可以从 any 原始类型转换为 int(最值得注意的是，解析 string).您可以在 在 MSDN 上查看此方法的完整重载列表.

Convert.ToInt32 is designed to be a general conversion function. It does a good deal more than casting; namely, it can convert from any primitive type to a int (most notably, parsing a string). You can see the full list of overloads for this method here on MSDN.

正如 Stefan Steiger 提到 在评论中:

另外，请注意，在数字级别上，(int) foo 会截断 foo (ifoo = Math.Floor(foo))，而 Convert.ToInt32(foo) 使用 半到偶数舍入(将 x.5 舍入到最接近的 EVEN 整数，意思是 ifoo = Math.Round(foo)).因此，结果不仅在实现方面，而且在数值上不相同.

Also, note that on a numerical level, (int) foo truncates foo (ifoo = Math.Floor(foo)), while Convert.ToInt32(foo) uses half to even rounding (rounds x.5 to the nearest EVEN integer, meaning ifoo = Math.Round(foo)). The result is thus not just implementation-wise, but also numerically not the same.

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