为什么返回类型上的类型限定符没有意义？ [英] Why is a type qualifier on a return type meaningless?

问题描述

说我有这个例子：

char const * const
foo( ){
   /* which is initialized to const char * const */
   return str;
}

什么是正确的方法来避免编译器警告type qualifier返回类型是无意义的？

What is the right way to do it to avoid the compiler warning "type qualifier on return type is meaningless"?

推荐答案

但非类类型的值不可修改（继承自C），因此标准表示非类类型的值不会是const限定的（最右边的const被忽略，即使你指定的），因为const会有点冗余。一个不写它 - 例如：

The way you wrote it, it was saying "the returned pointer value is const". But non-class type rvalues are not modifiable (inherited from C), and thus the Standard says non-class type rvalues are never const-qualified (right-most const was ignored even tho specified by you) since the const would be kinda redundant. One doesn't write it - example:

  int f();
  int main() { f() = 0; } // error anyway!

  // const redundant. returned expression still has type "int", even though the 
  // function-type of g remains "int const()" (potential confusion!)
  int const g(); 

请注意，对于g类型，const ，但对于从类型 int const 生成的右值表达式，const将被忽略。所以下面是一个错误：

Notice that for the type of "g", the const is significant, but for rvalue expressions generated from type int const the const is ignored. So the following is an error:

  int const f();
  int f() { } // different return type but same parameters

我知道你可以观察到const，除了获得类型的g本身（并传递& f 到一个模板，并推断其类型，为例）。最后注意char const和const char表示相同的类型。我建议你用一个概念来解决，并在整个代码中使用。

There is no way known to me you could observe the "const" other than getting at the type of "g" itself (and passing &f to a template and deduce its type, for example). Finally notice that "char const" and "const char" signify the same type. I recommend you to settle with one notion and using that throughout the code.

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