C ++ const map元素访问 [英] C++ const map element access

问题描述

我试图使用operator []访问const C ++映射中的元素，但此方法失败。我也试图使用at（）做同样的事情。这次工作。但是，我找不到任何参考使用at（）访问元素在const C ++映射。是at（）一个新添加的函数在C ++映射？在哪里可以找到更多的信息？非常感谢！

I tried to use the operator[] access the element in a const C++ map, but this method failed. I also tried to use "at()" to do the same thing. It worked this time. However, I could not find any reference about using "at()" to access element in a const C++ map. Is "at()" a newly added function in C++ map? Where can I find more info about this? Thank you very much!

例如：

#include <iostream>
#include <map>

using namespace std;

int main()
{
        map<int, char> A;
        A[1] = 'b';
        A[3] = 'c';

        const map<int, char> B = A;

        cout << B.at(3) << endl; // it works
        cout << B[3] << endl;  // it does not work

}

]，它在编译期间返回以下错误：

For using "B[3]", it returned the following errors during compiling:

t01.cpp：14：error：passing'const
std： ：map，
std :: allocator>>'as'this''参数'_Tp&
std :: map< _Key，_Tp，_Compare，
_Alloc> :: operator []（const _Key&）[with _Key = int，_Tp = char，_Compare = std :: less，_Alloc =
std :: allocator>]'丢弃限定符

t01.cpp:14: error: passing ‘const std::map, std::allocator > >’ as ‘this’ argument of ‘_Tp& std::map<_Key, _Tp, _Compare, _Alloc>::operator[](const _Key&) [with _Key = int, _Tp = char, _Compare = std::less, _Alloc = std::allocator >]’ discards qualifiers

使用的编译器是g ++ 4.2.1

The compiler used is g++ 4.2.1

推荐答案

at（） 是C ++ 11中 std :: map 的一种新方法。

而不是插入一个新的默认构造的元素 operator [] 如果一个元素与给定的键不存在，它抛出一个 std :: out_of_range 异常。 （这类似于 at（） for deque 和向量。）

Rather than insert a new default constructed element as operator[] does if an element with the given key does not exist, it throws a std::out_of_range exception. (This is similar to the behaviour of at() for deque and vector.)

由于这种行为，有一个 const code> $ <$ p $ c> $

Because of this behaviour it makes sense for there to be a const overload of at(), unlike operator[] which always has the potential to change the map.

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