我想用距离变量对“mystruct”进行排序,这样做的最快方法是什么?
struct MyStruct {
int scale;
bool pass;
float distance;
};
vector<MyStruct> mystruct;
...
sort (mystruct.begin(), mystruct.begin() + mystruct.size());
//this doesn't work since is trying to sort by "MyStruct" and not by a number
如果我有
vector<float> myfloat;
...
sort (myfloat.begin(), myfloat.begin() + myfloat.size());
然后将完美地运作。
答案 0 :(得分:6)
您需要为您的结构编写自己的operator<
。
应该是
bool operator<( const MyStruct& s1, const MyStruct& s2 )
{
// compare them somehow and return true, if s1 is less than s2
// for your case, as far as I understand, you could write
// return ( s1.distance < s2.distance );
}
另一个选择是编写一个功能对象,但这里没有必要,编写operator<
更容易(初学者)
答案 1 :(得分:5)
您需要为sort函数提供一个仿函数,或者为less运算符提供:
struct MyStruct_Compare {
bool operator()(const MyStruct& a, const MyStruct& b) {
return a.distance < b.distance;
}
}
std::sort(mystruct.begin(), mystruct.end(), MyStruct_Compare());
OR:
bool operator<(const MyStruct& a, const MyStruct& b) {
return a.distance < b.distance;
}
std::sort(mystruct.begin(), mystruct.end());