相关疑难解决方法(0)

C++标准(第8.5节)说:

如果程序要求对const限定类型T的对象进行默认初始化,则T应为具有用户提供的默认构造函数的类类型.

为什么？在这种情况下,我无法想到为什么需要用户提供的构造函数.

struct B{
  B():x(42){}
  int doSomeStuff() const{return x;}
  int x;
};

struct A{
  A(){}//other than "because the standard says so", why is this line required?

  B b;//not required for this example, just to illustrate
      //how this situation isn't totally useless
};

int main(){
  const A a;
}

最近为什么const对象需要用户提供的默认构造函数？被标记为重复为什么C++需要用户提供的默认构造函数来默认构造一个const对象？.我正在使用coliru和rextexter来测试各种版本的gcc(g ++ - 4.7,g ++ - 4.8,g ++ - 4.9)和clang(3.4和3.5)以查看是否在较新版本的编译器中引入了这种行为.这里我们分别从两个问题中提取了两个测试用例:

class A {
public:
    void f() {}

};

int main()
{
    A a;       // OK
    const A b; // ERROR

    a.f();
    return 0;
}

和:

struct B{
  B():x(42){}
  int doSomeStuff() const{return x;}
  int x;
};

struct A{
  A(){}//other than "because the standard says so", why is this line required?

  B b;//not required for this example, just to illustrate
      //how this situation isn't …