c++ - sizeof运算符为VOID运算符返回1个字节

Question

我在看C++中的sizeof运算符。我遇到了意外的语义错误。我读过无效数据类型没有内存大小。但是，在我的程序中，sizeof运算符对于无效数据类型返回1。这怎么可能？

#include<iostream>
int main()
{
     std::cout<<"Void: "<<sizeof(void)<<" bytes\n";
     return 0;
}

我正在使用CodeBlocks编写代码，并且我的操作系统是Windows 10 x64。

Answer 1

这是一个GCC extension:

In GNU C, addition and subtraction operations are supported on pointers to void and on pointers to functions. This is done by treating the size of a void or of a function as 1.

A consequence of this is that sizeof is also allowed on void and on function types, and returns 1.

The option -Wpointer-arith requests a warning if these extensions are used.