可以使用条件运算符在两个 class 成员函数调用之间切换

Question

考虑一下：

int func1( int i );
int func2( int i );

条件运算符可以这样使用：

int res = (cond)?func1(4):func2(4);

或者，如果两者可以使用相同的参数：

int res = ((cond)?func1:func2)(4);

现在，a class:

的成员函数呢？

class T
{
public:
    T( int i ) : i(i) {}

    int memfunc1() { return 1*i; }
    int memfunc2() { return 2*i; }

private:
    int i;
};

我试过了，但没用：

T t(4);
int res2 = t.((cond)?memfunc1:memfunc2)();

...也尝试了其他语法 ((t.*((cond)?&(T::memfunc1):&(T::memfunc2)))()) 但没有成功...

这可行吗，那么好的语法是什么？一行代码答案更可取（使用临时自动变量存储指向函数的指针太容易了......;-)

Answer 1

§ 5.3.1 [expr.unary.op]/p4:

A pointer to member is only formed when an explicit & is used and its operand is a qualified-id not enclosed in parentheses. [ Note: that is, the expression &(qualified-id), where the qualified-id is enclosed in parentheses, does not form an expression of type “pointer to member.” Neither does qualified-id, because there is no implicit conversion from a qualified-id for a non-static member function to the type “pointer to member function” as there is from an lvalue of function type to the type “pointer to function” (4.3). Nor is &unqualified-id a pointer to member, even within the scope of the unqualified-id’s class. — end note ]

如果仍然没有帮助，您可以在下面找到正确的语法：

(t.*(cond ? &T::memfunc1 : &T::memfunc2))()

可以使用条件运算符在两个 class 成员函数调用之间切换

Can conditional operator be used to toggle between two class member function calls

c++11

conditional

member-function-pointers