C++“错误：将‘const std::map<int, std::basic_string<char> >’作为……的‘this’参数”
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C++ "error: passing 'const std::map<int, std::basic_string<char> >' as 'this' argument of ..."
提问by WXB13
With the following code (excerpted for brevity):
使用以下代码（为简洁起见摘录）：
color.h:
颜色.h：
class color {
public:
color();
enum colorType {
black, blue, green, cyan, red,
magenta, brown, lightgray, nocolor
};
colorType getColorType();
void setColorType(colorType cColortype);
string getColorText() const;
private:
colorType cColortype = nocolor;
map<int, string> colors = {
{black, "black"},
{blue, "blue"},
{green, "green"},
{cyan, "cyan"},
{red, "red"},
{magenta, "magenta"},
{brown, "brown"},
{lightgray, "lightgray"},
{nocolor, "nocolor"}};
};
color.cpp:
颜色.cpp：
color::color() {
}
color::colorType color::getColorType() {
return cColortype;
}
void color::setColorType(colorType cColortype) {
this->cColortype = cColortype;
}
string color::getColorText() const {
return colors[cColortype];
}
I get the following error:
我收到以下错误：
color.cpp:16:29: error: passing 'const std::map >' as 'this' argument of 'std::map<_Key, _Tp, _Compare, _Alloc>::mapped_type& std::map<_Key, _Tp, _Compare, _Alloc>::operator[](std::map<_Key, _Tp, _Compare, _Alloc>::key_type&&) [with _Key = int; _Tp = std::basic_string; _Compare = std::less; _Alloc = std::allocator > >; std::map<_Key, _Tp, _Compare, _Alloc>::mapped_type = std::basic_string; std::map<_Key, _Tp, _Compare, _Alloc>::key_type = int]' discards qualifiers [-fpermissive]
color.cpp:16:29: 错误：将 'const std::map >' 作为 'std::map<_Key, _Tp, _Compare, _Alloc>::mapped_type& std::map<_Key, _Tp 的 'this' 参数传递, _Compare, _Alloc>::operator[](std::map<_Key, _Tp, _Compare, _Alloc>::key_type&&) [with _Key = int; _Tp = std::basic_string; _Compare = std::less; _Alloc = std::allocator > >; std::map<_Key, _Tp, _Compare, _Alloc>::mapped_type = std::basic_string; std::map<_Key, _Tp, _Compare, _Alloc>::key_type = int]' 丢弃限定符 [-fpermissive]
The error refers to "return colors[cColortype];" in getColorText.
错误是指“返回颜色[cColortype];” 在 getColorText 中。
I'm writing this for a class project and I can get it to work for the sake of the assignment by removing the const declaration in the getColorText signature but I'm trying to learn/adopt good practices and adhere to the recommendation to use const for member functions that don't modify data so I want to know how to deal with this going forward.
我正在为一个班级项目编写这个，我可以通过删除 getColorText 签名中的 const 声明来使其工作以便分配，但我正在尝试学习/采用良好做法并遵守使用 const 的建议对于不修改数据的成员函数，所以我想知道如何处理这个问题。
I'm usually really good at debugging/troubleshooting but the error message is so convoluted that it's not much help.
我通常非常擅长调试/故障排除，但错误消息非常复杂，没有太大帮助。
Any help is appreciated.
任何帮助表示赞赏。
回答by Dave S
string color::getColorText() const {
return colors[cColortype];
}
The issue is that you've marked the function as const. The operator[]on std::mapis marked as non-const, and cannot be used in a const function like this. You need to manually use std::map::find(or other mechanism) to search for the input type and handle the case where it's not found.
问题是您已将该函数标记为const. 的operator[]上std::map被标记为非const，并且不能在这样一个const函数一起使用。您需要手动使用std::map::find（或其他机制）来搜索输入类型并处理找不到的情况。
If you're using C++11, you can instead use std::map::at, which IS allowed to be used on a constant map, and throws an exception if the requested element is not present.
如果您使用的是 C++11，则可以改用std::map::at，它允许在常量映射上使用，如果请求的元素不存在，则抛出异常。
回答by Pete Becker
The key is near the end: "discards qualifiers". getColorTextis a constmember function, so colorsis const. But map::operator[]()is not const.
关键是接近尾声：“丢弃限定符”。getColorText是const成员函数，所以colors也是const。但map::operator[]()不是const。
回答by Brainless
First : the map map<int, string> colorsmust be a map from cColorTypeto string instead of int :
首先：映射map<int, string> colors必须是从cColorType到 string 而不是 int的映射：
map<cColorType, string> colors
Second : As some people already answered : map::operator[]()is not const. The reason for that is that this operator returns a reference, which allows you to modify its value.
第二：正如一些人已经回答：map::operator[]()不是const。原因是这个操作符返回一个引用，它允许你修改它的值。
Let me suggest the following solution : You can create a second private attribute : the color in string format. Therefore you'll have 2 Get functions (one for each type of color), and one Set function (which will modify the 2 color attributes).
让我建议以下解决方案：您可以创建第二个私有属性：字符串格式的颜色。因此，您将有 2 个 Get 函数（每种颜色一个）和一个 Set 函数（将修改 2 个颜色属性）。
