指针const vs常用指针（用于函数） [英] pointer to const vs usual pointer (for functions)

问题描述

函数的const和常用指针之间有什么区别吗？当适合使用const限定词用于独立函数？

Is there any difference between pointer to const and usual pointer for functions? When it is suitable to use const qualifier for stand alone functions?

我写了简短的例子来说明我的问题：

I wrote short sample to illustrate my question:

#include <iostream>
using namespace std;

int sum( int x, int y ) { return x + y; }
typedef int sum_func( int, int );

int main()
{
    const sum_func* sum_func_cptr = &sum; // const function
    sum_func* sum_func_ptr = &sum;        // non-const function ?

    // What is the difference between sum_func_cptr and sum_func_ptr

    int x = sum_func_cptr( 2, 2 );
    cout << x << endl;

    int y = sum_func_ptr( 2, 2 );
    cout << y << endl;

    sum_func_cptr = 0;
    sum_func_ptr = 0;

    return 0;
}

g ++不提供警告。这是我问的原因。

g++ gives no warnings. That's why I ask.

推荐答案

您的代码格式不正确 。您可以不构建一个const（或volatile）限定的函数类型。

Your code is ill-formed with regard to C++03. You can not ever construct a const (or volatile) qualified function type. Whenever you do, your program becomes ill-formed.

此规则已更改 C ++ 1x，以使编译器忽略 const / volatile 。 C ++编译器通常已经实现了这个规则，即使在C ++ 03模式。因此，以下两个将定义相同的函数两次，并导致编译错误。

This rule has been changed for C++1x, to make the compiler ignore the const / volatile. C++ compilers will usually already implement this rule even in C++03 mode. Thus, the following two will define the same function twice, and results in a compilation error.

typedef void Ft();


void f(Ft const*) { }
void f(Ft *) { } // another definition!

这是我的索赔证明。 C ++ 03， 8.3.5 / 1

Here is the proof of my claim. C++03, 8.3.5/1

A cv-qualifier- seq只能是非静态成员函数的函数类型的一部分，指向成员的指针的函数类型，或者函数typedef声明的顶层函数类型。 cv-qualifier-seq在函数声明符中的作用与在函数类型之上添加cv-qualification不同，即它不创建cv限定的函数类型。事实上，如果在确定类型的任何时候形成cv限定的函数类型，则程序是不成形的。

A cv-qualifier-seq shall only be part of the function type for a nonstatic member function, the function type to which a pointer to member refers, or the top-level function type of a function typedef declaration. The effect of a cv-qualifier-seq in a function declarator is not the same as adding cv-qualification on top of the function type, i.e., it does not create a cv-qualified function type. In fact, if at any time in the determination of a type a cv-qualified function type is formed, the program is ill-formed.

这里是C ++ 1x的文本， 8.3.5 / 7 n2914：

Here is that text for C++1x, 8.3.5/7 n2914:

cv-quali fi er-seq只能是非静态成员函数的函数类型的一部分，函数类型指向成员的指针，或者函数typedef声明的顶层函数类型。函数声明符中的cv-qualiter-seq的效果不同于在函数类型之上添加cv-qualial。在后一种情况下，忽略cv-qualifier。

A cv-qualifier-seq shall only be part of the function type for a non-static member function, the function type to which a pointer to member refers, or the top-level function type of a function typedef declaration. The effect of a cv-qualifier-seq in a function declarator is not the same as adding cv-qualification on top of the function type. In the latter case, the cv-qualifiers are ignored.

上面说的是下面的内容是有效的，并为一个可以声明一个const成员函数的函数创建函数类型。

The above says that the below is valid, though, and creates the function type for a function that can declare a const member function.

typedef void Ft() const;
struct X { Ft cMemFn; };
void X::cMemFn() const { }

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