問題描述
我嘗試使用 operator[] 訪問 const map 中的元素，但此方法失敗.我也嘗試使用 at() 來做同樣的事情.這次成功了.但是，我找不到任何關(guān)于使用 at() 訪問 const map 中的元素的參考.at()是map新增的函數(shù)嗎?我在哪里可以找到有關(guān)此的更多信息?非常感謝！
I tried to use the operator[] access the element in a const 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 map. Is at() a newly added function in map? Where can I find more info about this? Thank you very much!
示例如下:
#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
}
使用B[3]"，編譯時返回如下錯誤:
For using "B[3]", it returned the following errors during compiling:
t01.cpp:14: 錯誤:傳遞‘conststd::map
t01.cpp:14: error: passing ‘const std::map<int, char, std::less, std::allocator<std::pair<const int, char> > >’ 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<std::pair<const int, char> >]’ discards qualifiers
使用的編譯器是g++ 4.2.1
The compiler used is g++ 4.2.1
推薦答案
at() 是 C++11 中 std::map 的新方法.
如果具有給定鍵的元素不存在，則不會像 operator[] 那樣插入新的默認(rèn)構(gòu)造元素，而是拋出 std::out_of_range 異常.(這類似于 at() 對于 deque 和 vector 的行為.)
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.)
由于這種行為，at() 的 const 重載是有意義的，不像 operator[] 總是有改變地圖的潛力.
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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