它有没有需要声明一个类常量实例的所有属性常量？ [英] It there a need to declare const instance of a class with all attributes const?

问题描述

这是一个后续到<一个href=\"http://stackoverflow.com/questions/23449597/does-a-class-with-all-attributes-const-need-to-have-member-function-declared-con/23449626?noredirect=1#23449626\">Does所有的一类属性常量需要有成员函数声明为const呢？。

所以我一类 PermutationGroup ，其所有属性都是常量。编译器仍然让const和非const实例之间的区别：

So I've a class PermutationGroup whose all attribute are const. The compiler still make the distinction between const and non-const instance:

struct Foo {
  const int bar;
  void meth();
};
int main() {
   Foo foo {2};
   foo.meth();          // correct

   const Foo cfoo {1};
   cfoo.meth();         // wrong
};

正如提到的问题注意到了@nosid一个不能调用非常量成员函数一个const实例：

As noticed by @nosid in the referred question One cannot call a non const member function a const instance:

bla.cpp: In function ‘int main()’:
bla.cpp:10:14: error: passing ‘const Foo’ as ‘this’ argument of ‘void Foo::meth()’ discards qualifiers [-fpermissive]
    cfoo.meth();

所以，问题是：为什么它可以声明一个类，其属性都是常量的非const的实例。有没有合理的利用呢？

So the question is: why is it possible to declare non const instance of a class whose attribute are all const. Is there any reasonable use of that ?

推荐答案

好了，的可能的，为什么它应该被允许申报一个非推理 - 常量例如在一个类，其成员都是常量很简单，就是你不能写以下code的事实：

Well, a possible reasoning on why it should be allowed to declare a non-const instance in a class whose members are all const is simply the fact that you can't write the following code:

class Foo { Foo(void) const; };

这引发了：

error: constructors may not be cv-qualified

这意味着，至少有一个成员 - 构造函数，当然析构函数 - 始终是非 - 常量

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