为什么条件运算符总是返回在C＃中一个int？ [英] Why does the conditional operator always return an int in C#?

问题描述

可能重复：结果
的有条件的经营者不能隐式转换？

在一份书面陈述使用href=\"http://msdn.microsoft.com/en-us/library/ty67wk28.aspx\" rel=\"nofollow\">条件运算的 INT 类型。这就需要分配使用该运营商短变量投。

When writing a statement using the conditional operator, if the either of the expressions are numeric values they are always interpreted as int type. This makes a cast necessary to assign a short variable using this operator.

bool isTrue = true;
int intVal = isTrue ? 1 : 2;
short shortVal = isTrue ? 1 : 2;  // Compile error: Cannot implicitly convert type 'int' to 'short'. 

不应该编译器能够知道这两个值都是有效的短值，就像它在一个典型的赋值语句（短shortVal = 1; ）？

Shouldn't the compiler be able to know that both values are valid short values just as it would in a typical assignment statement(short shortVal = 1;)?

推荐答案

这是因为你的1和2是 INT 秒。 ：返回相同类型的第二和第三个操作数你的情况

This is because your 1 and 2 are ints. ?: returns the same type as 2nd and 3rd operands in your case.

编辑：在我的VS2008这个作品：

At my VS2008 this works:

short x = true ? 1 : 2;

难道我做错了什么？

Did I do something wrong?

编辑：的确，不同的是，真正是一个编译时间常数。对于非常量表达式，我得到了相同的错误消息。

Indeed, the difference was that true was a compile-time constant. For non-constant expressions I got the same error message.

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