c++ - 在构造函数参数列表中构造 boost::function 对象时出现错误 C2228

Question

以下代码无法在 Visual C++ 2005 中编译。

class SomeClass {
public: boost::function<void()> func;
        SomeClass(boost::function<void()> &func): func(func) { }
};

void someFunc() {
    std::cout << "someFunc" << std::endl;
}

int main() {
    SomeClass sc(boost::function<void()>(&someFunc));
    sc.func(); // error C2228: left of '.func' must have class/struct/union
    return 0;
}

如果我在 SomeClass 构造函数的参数周围加上括号，或者在参数列表之外构造 boost::function 对象，它可以正常编译。

    SomeClass sc((boost::function<void()>(&someFunc)));
    // or
    boost::function<void()> f(&someFunc);
    SomeClass sc(f);

之前的代码有什么问题？

Answer 1

它是一个函数的函数声明，它引用了一个 boost:function <void()>。并返回 SomeClass .您可以记住以下规则，事实证明它适用于许多其他此类消歧案例。您可以在 8.2 部分找到这些案例的描述。 C++ 标准。

Any construct that could possibly be a declaration will be taken as a declaration

也就是说，下面会被当作参数声明，括号多余

boost::function<void()>(&someFunc)

如果你去掉括号，这将变得清晰

boost::function<void()> &someFunc

因此，整个声明将不再声明一个对象，而是一个函数

SomeClass sc(boost::function<void()> &someFunc);

要修复它，请使用转换符号

SomeClass sc((boost::function<void()>)&someFunc);

或者像您一样在整个表达式周围加上括号。

这是来自 8.2 的所有荣耀标准:

The ambiguity arising from the similarity between a function-style cast and a declaration mentioned in 6.8 can also occur in the context of a declaration. In that context, the choice is between a function declaration with a redundant set of parentheses around a parameter name and an object declaration with a function-style cast as the initializer. Just as for the ambiguities mentioned in 6.8, the resolution is to consider any construct that could possibly be a declaration a declaration. [Note: a declaration can be explicitly disambiguated by a nonfunction-style cast, by a = to indicate initialization or by removing the redundant parentheses around the parameter name. ]

请注意，为了控制优先级，您可以在任何地方引入括号，如下所示

int (((((((a))))))) = 3;
int (*(pa)) = &a;