typeof（）表达式中的副作用 [英] Side effects within a typeof() expression

问题描述

在GNUC C中，你可以使用 typeof（expression），并且在里面使用带有副作用的表达式是合法的。因此，例如，你可以有这个C代码：

  int x = 0; 
 typeof（x ++）y; 
  

在这种情况下，副作用被忽略，之后x仍然为零（这有意义的类型是一个只有编译时间的东西）。

然而， GCC文档说：

typeof的操作数的副作用为
，当且仅当它是可变修饰类型
的表达式或这种类型的名称。

这句话是什么意思？是否真的有可能编写带有副作用的 typeof ，并且副作用实际上是在运行时执行的？对我来说，这句话似乎表明了这一点。

解决方案

是的，在某些情况下， code> typeof 已评估。正如文件所述，它需要处于可变更改类型之内。这是一种依赖于某些运行时值的类型，例如 int [x] 。

所以下面的代码是合法的，声明y的类型为 int [1] ，x的值为1之后：

  int x = 0; 
 typeof（int [++ x]）y; 
  

当然，一般来说，对于可变修改类型同样如此，这只对局部变量声明合法功能）。

In GNUC C, you can use typeof(expression), and it is legal to use an expression with side effects inside. So for example you can have this C code:

int x = 0;
typeof(x++) y;

In this case, the side effect is ignored and x is still zero afterwards (this makes sense as types are a compile-time-only thing).

However, the GCC documentation says:

The operand of typeof is evaluated for its side effects if and only if it is an expression of variably modified type or the name of such a type.

What does this sentence mean? Is it really possible to write typeof with a side effect and have the side effect actually be executed at runtime? For me, this sentence seems to indicate this.

解决方案

Yes, it is possible in certain cases to have side-effects inside typeof evaluated. As the documentation says, it needs to be within a "variably modified type". This is a type which depends on some runtime value, such as int[x].

So the following code is legal, declares y to be of type int[1], and x has the value 1 afterwards:

int x = 0;
typeof(int[++x]) y;

Of course, equally to variably modified types in general, this is only legal for local variable declarations (inside function).

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