为什么 C++ 映射类型参数在使用 [] 时需要一个空的构造函数?

2021-12-10 00:00:00 dictionary c++

另见C++ 标准列表和默认构造类型

不是大问题，只是烦人，因为我不希望我的类在没有特定参数的情况下被实例化.

Not a major issue, just annoying as I don't want my class to ever be instantiated without the particular arguments.

#include <map>

struct MyClass
{
    MyClass(int t);
};

int main() {
    std::map<int, MyClass> myMap;
    myMap[14] = MyClass(42);
}

这给了我以下 g++ 错误:

This gives me the following g++ error:

/usr/include/c++/4.3/bits/stl_map.h:419: 错误:没有匹配的函数调用‘MyClass()’

/usr/include/c++/4.3/bits/stl_map.h:419: error: no matching function for call to ‘MyClass()’

如果我添加一个默认构造函数，这编译得很好；我确定这不是由错误的语法引起的.

This compiles fine if I add a default constructor; I am certain it's not caused by incorrect syntax.

推荐答案

此问题来自 operator[].引用自 SGI 文档:

This issue comes with operator[]. Quote from SGI documentation:

data_type&operator[](const key_type& k) - 返回对象的引用与特定的钥匙.如果地图还没有包含这样一个对象，operator[]插入默认对象data_type().

data_type& operator[](const key_type& k) - Returns a reference to the object that is associated with a particular key. If the map does not already contain such an object, operator[] inserts the default object data_type().

如果您没有默认构造函数，您可以使用插入/查找函数.以下示例工作正常:

If you don't have default constructor you can use insert/find functions. Following example works fine:

myMap.insert( std::map< int, MyClass >::value_type ( 1, MyClass(1) ) );
myMap.find( 1 )->second;

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