如何通过范围解析操作员在使用指针到成员时获取类成员的地址？ [英] How Does Getting the Address of a Class Member Through a Scope Resolution Operator Work When Using a Pointer-to-Member?

问题描述

使用指向成员的指针（AKA点星形或箭头星形）访问类成员时，可以使用以下语法：

When using a pointer-to-member (AKA dot-star or arrow-star) to access a class member, we can use the following syntax:

A * pa;
int A::*ptm2 = &A::n;
std::cout << "pa->*ptm: " << pa->*ptm << '\n';

我的问题是& A :: n 语句工作？

在上面的例子中， n 是一个变量。如果不是成员变量， n 是一个函数（我们定义了一个指向成员函数而不是指向成员的指针）我可能会认为，由于类的函数可以有效地静态（参见Nemo的注释），我们可以通过& A :: some_function 找到一个类的函数地址。但是如何通过类范围解析来获得非静态类成员的地址？当我打印& A :: n 的值时，这更令人困惑，因为输出只是 1 。

In the above example n is a variable. If instead of a member variable, n was a function (and we defined a pointer-to-a-member-function instead of a pointer-to-a-member), I might reason that since a class's functions can be effectively static (see Nemo's comment), we could find a class's function's address via &A::some_function. But how can we get the address of a non-static class member through a class scope resolution? This is even more confusing when I print the value of &A::n because the output is merely 1.

推荐答案

当你声明一个指向成员数据的指针时，它不限于任何特定的实例。如果你想知道一个给定实例的数据成员的地址，你需要在执行。* 或 - >后输入结果的地址。 ; * 。例如：

When you declare a pointer to member data, it isn't bounded to any particular instance. If you want to know the address of a data member for a given instance you need to take the address of the result after doing .* or ->*. For instance:

#include <stdio.h>

struct A
{
  int n;
};

int main()
{
  A a  = {42};
  A aa = {55};
  int A::*ptm = &A::n;
  printf("a.*ptm: %p\n", (void *)&(a.*ptm));
  printf("aa.*ptm: %p\n", (void *)&(aa.*ptm));
}

打印为一个可能的输出：

prints as one possible output:

a.*ptm: 0xbfbe268c
aa.*ptm: 0xbfbe2688

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