std::map::find()
I have a simple struct which i'm using as a key in a std::map
struct PpointKey{
unsigned int xp,yp; //pixel coordinates
unsigned int side;
PpointKey(unsigned xp,unsigned yp,unsigned side=5):xp(xp),yp(yp),side(side)
{}
bool operator==(const PpointKey& other) const{
const unsigned int x = other.xp;
const unsigned int y = other.yp;
return ((x>=xp && x<=xp+side) && (y>=yp && y<=yp+side));
}
bool operator<(const PpointKey& other) const{
const unsigned int x = other.xp;
const unsigned int y = other.yp;
const unsigned other_distance_2 = x*x + y*y;
const unsigned this_distance_2 = this->xp*this->xp + this->yp * this->yp;
return this_distance_2 < other_distance_2;
}开发者_JS百科
};
What I would like to achieve is to use the find() to access the map with a key that has its xp,yp attributes within a side
distance. In other words, if I have an (x,y) tuple, I would like to find inside the map the first PpointKey that fulfils the condition inside the operator== function
return ((x>=xp && x<=xp+side) && (y>=yp && y<=yp+side));
Is this possible using find? I'm getting map.end(), so I would like to check wheter the find () function uses the operator==. Maybe the search algorithm would be better?
Thanks in advance.
The find
function of map
does not use the operator==
.
However you can use std::find
, passing in the begin()
and end()
iterator of map
. It will simply iterate through the sequence one at a time and yield the first object that matches (complexity is linear).
The issue you encounter is due to the fact that you have abused operator overload. The problem here is that the common definition of operator==
is:
T operator==(T lhs, T rhs)
{
return !(lhs < rhs) && !(rhs < lhs);
}
And this is not the case with your definition, thus you cannot substitute one for the other.
It would be best if you used traditional functions with expressive names rather than operator overloading, it would be less misleading. Note that map
and std::find
allow you to pass suitable predicate objects, you don't need to overload the operators to use them.
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