如何从'char const *'中删除const [英] How to remove the const from 'char const*'

问题描述

看起来std :: remove_const不能删除 const char * 的常数。考虑下面的代码：

It appears that std::remove_const isn't able to remove the const-ness of const char*. Consider the following code:

#include <iostream>
#include <type_traits>
#include <typeinfo>

template< typename T >
struct S
{
    static void foo( ) {
        std::cout << typeid(T).name() << std::endl;
        std::cout << typeid( std::remove_const<T>::type ).name() << std::endl;
    }
};


int main( )
{
    S<char const*>::foo();
}

此程序的输出（在Visual Studio 2010上）：

Output of this program (on Visual Studio 2010):

char const *
char const *

在gcc中，我们有可读输出（代码这里）：

And in gcc we have the readable output (code here):

PKc
PKc

我希望在微软编译器的第二行获得 char * ，并且在gcc上获得任何（但不同于第一行）。我做错了什么？如何将 char const * 更改为 char * ？

I would hope to get char * on the second line of Microsoft compiler, and whatever (but different than 1st line) on gcc. What am I doing wrong? How do I turn char const* to char*?

推荐答案

如果你想删除所有的const限定符，你需要一个解决方案递归地从所有级别中删除const：

If you want to remove all const qualifiers you need a solution that recursively removes const from all levels:

template<typename T> struct remove_all_const : std::remove_const<T> {};

template<typename T> struct remove_all_const<T*> {
    typedef typename remove_all_const<T>::type *type;
};

template<typename T> struct remove_all_const<T * const> {
    typedef typename remove_all_const<T>::type *type;
};

int main() {
    std::cout << typeid(remove_all_const<int const * * const>::type).name() << '\n';
}

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