是否不可能将 STL 映射与结构一起用作键?
我有以下代码:
struct Node
{
int a;
int b;
};
Node node;
node.a = 2;
node.b = 3;
map<int, int> aa;
aa[1]=1; // OK.
map<Node, int> bb;
bb[node]=1; // Compile error.
当我尝试将结构 Node
的实例映射到 int
时,出现编译错误.为什么?
When I tried to map an instance of my struct Node
to an int
, I got a compile error. Why?
推荐答案
对于一个可以用作地图键的东西,你必须能够使用 operator<()
来比较它.您需要将这样的运算符添加到您的节点类中:
For a thing to be usable as a key in a map, you have to be able to compare it using operator<()
. You need to add such an operator to your node class:
struct Node
{
int a;
int b;
bool operator<( const Node & n ) const {
return this->a < n.a; // for example
}
};
当然,真正的操作符做什么取决于比较对你的结构的实际意义.
Of course, what the real operator does depends on what comparison actually means for your struct.
相关文章