向量的const对象给出编译错误

问题描述

我已在我的代码中声明了以下内容

I have declared the following in my code

vector <const A> mylist;

我得到以下编译错误 -

I get the following compile error -

new_allocator.h:75: error: `const _Tp* __gnu_cxx::new_allocator<_Tp>::address(const _Tp&) const \[with _Tp = const A]' and `_Tp* __gnu_cxx::new_allocator<_Tp>::address(_Tp&) const [with _Tp = const A]' cannot be overloaded

但如果声明 -

vector <A> mylist;

我的代码编译。

不允许在此上下文中？

我复制我的代码在这里供大家参考 -

I am copying my code here for everyone'e reference -

#include <iostream>
#include <vector>

using namespace std;
class A
{
public:
    A () {cout << "default constructor\n";}
    A (int i): m(i) {cout << "non-default constructor\n";}

private:
    int m;
};

int main (void)
{
    vector<const A> mylist;

    mylist.push_back(1);

    return 0;
}

推荐答案

可分配。 const 对象不可分配，因此尝试将它们存储在向量中将失败（或至少可失败 - 代码无效，但是编译器可以自由地接受它，如果它这样选择，虽然大多数程序员通常喜欢无效的代码被拒绝）。

Items in a vector must be assignable. const objects aren't assignable, so attempting to store them in a vector will fail (or at least can fail -- the code is invalid, but a compiler is free to accept it anyway, if it so chooses, though most programmers would generally prefer that invalid code be rejected).

编辑：我想真正拙劣，我应该添加一个纯理论的可能性。如果你想要足够糟糕，你可以定义一个可分配类型，尽管 const ，类似这样：

I suppose for the truly pedantic, I should probably add a purely theoretical possibility. If you wanted to badly enough, you could define a type that was assignable despite being const, something like this:

class ugly {
    mutable int x;
public:
    ugly const &operator=(ugly const &u) const {
        x = u.x;
        return *this;
    }
};

至少在理论上，我相信你可能即使它们 const ，这种类型的项目也会在向量中。用VC ++创建这些成功向量的快速测试，但失败与gcc（g ++ 4.8.1 / MinGW）。从技术上来说，我认为这可能是g ++（或其库）中的一个错误，但我不能说它是一个我认为是主要问题。恰恰相反，我不知道（有一点）它是否可能不是实际的故意，试图保存程序员做一些可怕和愚蠢的事情。

At least in theory, I believe you probably should be able to store items of this type in a vector even though they're const. A quick test of creating a vector of these succeeds with VC++, but fails with gcc (g++ 4.8.1/MinGW). Technically, I think this probably qualifies as a bug in g++ (or its library), but I can't say it's one I see as as major problem. Rather the contrary, I wonder (a little) whether it may not actually be intentional, in an attempt at saving the programmer from doing something horrible and stupid.

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