C++ 类构造函数限定为 __attribute__((pure)) 或 __attribute__((const))

Question

如果 C++ 类构造函数只能通过其参数访问数据，那么它们是否可以并且应该声明 __attribute__((pure))？在什么情况下它们应该被限定为__attribute__((const))？

Answer 1

当您将构造函数限定为 pure 或 const 时，GCC 会发出警告。这是因为构造函数不返回任何内容(返回 void)，并且在此类上拥有 pure 或 const 属性没有多大意义。功能。

查看 godbolt 演示 here .

<source>:3:30: warning: 'pure' attribute on function returning 'void' [-Wattributes]
     A()  __attribute__((pure));

                              ^
<source>:8:31: warning: 'const' attribute on function returning 'void' [-Wattributes]
     B()  __attribute__((const));                               ^

来自海湾合作委员会documentation :

const
...
Because a const function cannot have any side effects it does not make sense for such a function to return void. Declaring such a function is diagnosed.

pure
...
Because a pure function cannot have any side effects it does not make sense for such a function to return void. Declaring such a function is diagnosed.