为什么我不能这么做(错误C2275:“A”:非法使用此类型作为表达式)

Why I can't do this? compiler says

为什么我不能这么做?编译器说

warning C4832: token '.' is illegal after UDT 'A : see declaration of 'A' error C2275: 'A' : illegal use of this type as an expression : see declaration of 'A'

警告C4832:令牌”。在UDT 'A:看到'A'错误C2275: 'A':非法使用这种类型作为表达式:参见'A'声明

class A{
public:
    std::string getString(){return "olalal";}
};

class B{
public:
    std::string getStringFromA() {return A.getString();}
};


int main()
{
    B b;
    cout << b.getStringFromA();
    return 0;
}

3 个解决方案

#1

You can only call non-static member functions on an object.

您只能在对象上调用非静态成员函数。

E.g.

如。

std::string B::getStringFromA()
{
    A object_of_type_A;
    return object_of_type_A.getString();
}

Alternatively, you can create a unnamed temporary A and call getString() on that, E.g.

或者，您可以创建一个未命名的临时a，并在其上调用getString()。

class B {
public:
    std::string getStringFromA() { return A().getString(); }
};

If getString() were declared static in A, you could call the function without an object, but the syntax uses ::, not ..

如果getString()在A中被声明为静态的，那么您可以在没有对象的情况下调用该函数，但是语法使用的是::not ..

E.g.

如。

std::string B::getStringFromA()
{
    return A::getString();
}

With a function name like getStringFromA, though, I would expect to pass a parameter of the particular A that I wanted to get the string from, perhaps passing the A by reference.

不过，对于一个函数名，比如getStringFromA，我希望通过一个特定a的参数来获得字符串，或者通过引用传递a。

class B {
public:
    std::string getStringFromA(A& a) { return a.getString(); }
};

If getString in A doesn't actually modify the A object it would be better to declare the function const so that it can be called on const instances of A. In this case, a const reference could be used.

如果A中的getString实际上并没有修改一个对象，那么最好声明函数const，这样就可以调用A的const实例。在这种情况下，可以使用const引用。

class A {
public:
    std::string getString() const { return "olalal"; }
};

class B {
public:
    std::string getStringFromA(const A& a) { return a.getString(); }
};

#2

getString() is not a static member function; you must call it on a specific instance of A, e.g. A a; a.getString();.

getString()不是静态成员函数;你必须在a的特定实例上调用它，例如a;a.getString();。

#3

Since A::getString() isn't a static method you need to call it on an instance of A.

因为::getString()不是一个静态方法，您需要在一个实例上调用它。