Check if point is inside vector - c++

I want to check if a given point with x and yvalue is inside a vector of points:
bool inside_vector(int x, int y, vector<Point2i>& points)
{
for(vector<Point2i>::const_iterator it = points.begin();
it != points.end();
++it)
{
if(it->x == y && it->y == y)
{
return true;
}
}
return false;
}
Are there other approaches without the for loop?

You can use std::find or std::find_if with suitable functor to avoid writing your own loop. But you don't gain in terms of complexity: it is still O(n). For example,
bool operator==(const Point2i& lhs, const Point2i& rhs)
{
return lhs.x == rhs.x && lhs.y == rhs.y;
}
Point2Di somePoint = Point2Di(x,y); // point to find
auto it = std::find(points.begin(), points.end(), somePoint);
or, without the equality operator,
auto it = std::find_if(points.begin(),
points.end(), [&somePoint](const Point2Di& p)
{return somePoint.x == p.x && somePoint.y == p.y;});

Assumming you have an operator== defined for your Point2i structure, you can use std::find(), like this:
std::vector<Point2i>::const_iterator findIt = std::find(
points.begin(),
points.end(),
Point2i(x, y));
I also assume you have a Point2i constructor that takes the two coordinates.

Related

std::unique not working as expected with struct

I'm currently working on a 2D game where the level is defined by edges:
struct Edge
{
vec2int start;
vec2int end;
}
The struct vec2int is a vector with x, y coordinates and has all needed operators (in this particular case operator==) overloaded.
Because of a data structure that stores the edges inside of a grid, there can be duplicate edges in different cells inside the grid. When combining them back into a single std::vector<Edge> I tried to get rid of them like this:
auto it = std::unique(
edges.begin(),
edges.end(),
[&](const Edge& e1, const Edge& e2)
{
return e1.start == e2.start && e1.end == e2.end;
});
edges.resize(std::distance(edges.begin(), it));
For whatever reason this deletes only a few (or none) of the duplicate edges. I have no idea why. Is there something I am missing about std::unique?
The code:
#include <algorithm>
#include <iostream>
#include <vector>
template<class T>
struct v2d_generic
{
T x = 0;
T y = 0;
bool operator==(const v2d_generic& rhs) const
{
return (this->x == rhs.x && this->y == rhs.y);
}
bool operator!=(const v2d_generic& rhs) const
{
return (this->x != rhs.x || this->y != rhs.y);
}
};
typedef v2d_generic<int> vec2i;
struct Edge
{
vec2i start;
vec2i end;
};
int main(void)
{
std::vector<Edge> edges;
edges.push_back(Edge{vec2i{1, 1}, vec2i{1, 1}});
edges.push_back(Edge{vec2i{1, 1}, vec2i{1, 2}});
edges.push_back(Edge{vec2i{1, 1}, vec2i{1, 1}});
edges.push_back(Edge{vec2i{1, 1}, vec2i{1, 2}});
std::cout << edges.size() << std::endl;
auto it = std::unique(
edges.begin(),
edges.end(),
[&](const Edge& e1, const Edge& e2)
{
return e1.start == e2.start && e1.end == e2.end;
});
edges.resize(std::distance(edges.begin(), it));
std::cout << edges.size() << std::endl;
}
This outputs 4 both times.
std::unique removes consecutive equivalent elements. In your example, you do not have consecutive equal elements, so it should not remove anything.
If you do not care about the order of the elements in your range, you can sort it before calling std::unique.

What is wrong with my Operator <?

I get an assertion failure about my Vec2's Operator <, I have no idea what is wrong though.
bool Vec2::operator<( const Vec2& v ) const
{
if(x < v.x)
return true;
else
return y < v.y;
}
Invalid Operator < for std set insert
template<class _Pr, class _Ty1, class _Ty2> inline
bool __CLRCALL_OR_CDECL _Debug_lt_pred(_Pr _Pred, const _Ty1& _Left, const _Ty2& _Right,
const wchar_t *_Where, unsigned int _Line)
{ // test if _Pred(_Left, _Right) and _Pred is strict weak ordering
if (!_Pred(_Left, _Right))
return (false);
else if (_Pred(_Right, _Left))
_DEBUG_ERROR2("invalid operator<", _Where, _Line);
return (true);
}
Thanks
The problem is that this operator does not satisfy the weak ordering. For example consider two points
( 2, 1 ) and ( 1, 2 )
( 2, 1 ) is less ( 1, 2 ) because the second value 1 is less than 2.
At the same time ( 1, 2 ) is also less than ( 2, 1 ) because the first value 1 is less than the first value 2.
Look how thsi operator is defined for standard class std::pair and use the same operator.
The corrected way to satisfy the ordering is :
bool Vec2::operator<( const Vec2& v ) const
{
if(x < v.x)
return true;
if(x > v.x)
return false;
else
return y < v.y;
}
Or (code golf mode) :
bool Vec2::operator<( const Vec2& v ) const
{
return (x != v.x)? (x < v.x)
: (y < v.y) ;
}
Generally for a comparison like this, you want to compare the first pair of items, then if and only if those are equal, compare the second pair (and so on).
if (x < v.x)
return true;
if (x > v.x)
return false;
return y < v.y;
operator < should satisfy the Strict weak ordering.
A short way to do the job:
bool Vec2::operator< (const Vec2& v) const
{
return std::tie(x, y) < std::tie(v.x, v.y);
}

How to find an element in a vector?

I have defined a structure Coord as
struct Coord {
int x;
int y;
int z;
};
Overloaded operator!= for Coord
bool Coord::operator !=(Coord& crd)const {
if(this->x != crd.x)
{
if(this->y != crd.y)
{
if(this->z != crd.z)
return true;
else
return false;
}
else
return false;
return true;
}
else
return false;
}
Then initialized a vector variable as
vector<Coord> vcoord;
Now I am using following code to get index of vector having a perticular Coord object
int Index::getVIndex(Coord crd) {
vector<Coord>::iterator it;
int indx;
it = vcoord.begin();
while(it != vcoord.end() && (*it) != crd)
++it;
indx = distance(vcoord.begin(),it);
cerr << (*it).x << " " << (*it).y << " " << indx << endl;
return indx;
}
But the value of indx is always 0. Kindly help to get correct result.
You need a not-equals operator for your Coord struct in order to be able to do this:
(*it) != crd
The logic of your not-equals operator is incorrect. The best and easiest option is to provide an equality comparison and use std::find:
struct Coord {
int x;
int y;
int z;
};
bool operator == (const Coord& lhs, const Coord& rhs)
{
return lhs.x==rhs.x && lhs.y==rhs.y && lhs.z==rhs.z;
}
You can then implement != in terms of ==, but you don't need it if you use std::find, which uses == by default:
vector<Coord>::iterator it = std::find(vcoord.begin(), vcoord.end(), crd);
Your != operator returns true only if all coordinates differ; it should return true if any differ. This means your function will return zero if any coordinate of the first element matches the function argument's.
Your version is a long-winded way of writing:
return x != crd.x && y != crd.y && z != crd.z;
when it should be:
return x != crd.x || y != crd.y || z != crd.z;
It may be easier to get the logic correct by implementing it in terms of ==:
bool operator==(Coord const & lhs, Coord const & rhs) {
return lhs.x == rhs.x && lhs.y == rhs.y && lhs.z == rhs.z;
}
bool operator!=(Coord const & lhs, Coord const & rhs) {
return !(lhs == rhs);
}
Also, given a definition of ==, you can use std::find rather than rolling your own loop:
auto found == std::find(vcoord.begin(), vcoord.end(), crd);
if (found == vcoord.end()) {
// Decide what to do if not found.
// Returning zero is a bad idea, since there's no way distinguish that
// from a successful outcome.
} else {
return std::distance(vcoord.begin(), found);
}
You incorrectly implemented the logic in the inequality operator.
It should be
bool Coord::operator !=(const Coord& crd)const {
return x != crd.x || y != crd.y || z != crz.z;
}
Your implementation is logically equivalent to
return x != crd.x && y != crd.y && z != crz.z;

add a non-comparable object to a set

i have a class named Point (in an external library, i can't modify the code) that is used to represent a point in a 3d space:
int x = 0; int y = 0; int z = 0;
Point my_point p(x,y,z);
It overloads the == and != operators but not < or > ones. I need to store them in an efficient way (no double element, no repetition). I thought my data structure is set, but if i try to use, i get this error:
error: no match for ‘operator<’ in ‘__x < __y’
some advice?
Write a comparison operator, and instantiate the set with that:
struct ComparePoints
{
bool operator()( Point const& lhs, Point const& rhs ) const
{
if ( lhs.x != rhs.x ) {
return lhs.x < rhs.x;
} else if ( lhs.y != rhs.y ) {
return lhs.y < rhs.y;
} else {
return lhs.z < rhs.z;
}
}
};
std::set <Point, ComparePoints> mySet;
You can define a comparison functor and pass it as second template argument to std::set. See here, look at Compare. You can also define bool operator<(const Point& lhs, const Point& rhs), but if you cannot touch the class, this requires that the comparison can be implemented via the public interface of Point.
You can define operator < yourself. It doesn't have to be inside the Point class if x, y and z are available from Point's public interface.
bool operator<(const Point& lhs, const Point& rhs)
{
if( lhs.x != rhs.x ) return lhs.x < rhs.x;
if( lhs.y != rhs.y ) return lhs.y < rhs.y;
return lhs.z < rhs.z;
}

C++ STL Range Container

I'm looking for a container that maps from a double to object pointers. However, each key is simply a range of doubles that would correspond to that object.
For example, there could be a key/value pair that's <(0.0 3.0), ptr>, or <(3.5 10.0), ptr2>
container[1.0] should return ptr, container[3.0] should also return ptr, and container[-1.0] should be undefined.
Is there any object with similar behaviour by default or will I have to implement it myself?
Edit
Here's the actual code that I've written, it might be easier to debug/offer advice on it.
// Behavior: A range is defined mathematically as (min, max]
class dblRange
{
public:
double min;
double max;
dblRange(double min, double max)
{
this->min = min;
this->max = max;
};
dblRange(double val)
{
this->min = val;
this->max = val;
};
int compare(const dblRange rhs)
{
// 1 if this > rhs
// 0 if this == rhs
//-1 if this < rhs
if (rhs.min == rhs.max && min == max)
{
/*if (min > rhs.min)
return 1;
else if (min == rhs.min)
return 0;
else
return -1;*/
throw "You should not be comparing values like this. :(\n";
}
else if (rhs.max == rhs.min)
{
if (min > rhs.min)
return 1;
else if (min <= rhs.min && max > rhs.min)
return 0;
else // (max <= rhs.min)
return -1;
}
else if (min == max)
{
if (min >= rhs.max)
return 1;
else if (min < rhs.max && min >= rhs.min)
return 0;
else // if (min < rhs.min
return -1;
}
// Check if the two ranges are equal:
if (rhs.min == min && rhs.max == max)
{
return 0;
}
else if (rhs.min < min && rhs.max <= min)
{
// This is what happens if rhs is fully lower than this one.
return 1;
}
else if (rhs.min > min && rhs.min >= max)
{
return -1;
}
else
{
// This means there's an undefined case. Ranges are overlapping,
// so comparisons don't work quite nicely.
throw "Ranges are overlapping weirdly. :(\n";
}
};
int compare(const dblRange rhs) const
{
// 1 if this > rhs
// 0 if this == rhs
//-1 if this < rhs
if (rhs.min == rhs.max && min == max)
{
/*if (min > rhs.min)
return 1;
else if (min == rhs.min)
return 0;
else
return -1;*/
throw "You should not be comparing values like this. :(\n";
}
else if (rhs.max == rhs.min)
{
if (min > rhs.min)
return 1;
else if (min <= rhs.min && max > rhs.min)
return 0;
else // (max <= rhs.min)
return -1;
}
else if (min == max)
{
if (min >= rhs.max)
return 1;
else if (min < rhs.max && min >= rhs.min)
return 0;
else // if (min < rhs.min
return -1;
}
// Check if the two ranges are equal:
if (rhs.min == min && rhs.max == max)
{
return 0;
}
else if (rhs.min < min && rhs.max <= min)
{
// This is what happens if rhs is fully lower than this one.
return 1;
}
else if (rhs.min > min && rhs.min >= max)
{
return -1;
}
else
{
// This means there's an undefined case. Ranges are overlapping,
// so comparisons don't work quite nicely.
throw "Ranges are overlapping weirdly. :(\n";
}
};
bool operator== (const dblRange rhs ) {return (*this).compare(rhs)==0;};
bool operator== (const dblRange rhs ) const {return (*this).compare(rhs)==0;};
bool operator!= (const dblRange rhs ) {return (*this).compare(rhs)!=0;};
bool operator!= (const dblRange rhs ) const {return (*this).compare(rhs)!=0;};
bool operator< (const dblRange rhs ) {return (*this).compare(rhs)<0;};
bool operator< (const dblRange rhs ) const {return (*this).compare(rhs)<0;};
bool operator> (const dblRange rhs ) {return (*this).compare(rhs)>0;};
bool operator> (const dblRange rhs ) const {return (*this).compare(rhs)>0;};
bool operator<= (const dblRange rhs ) {return (*this).compare(rhs)<=0;};
bool operator<= (const dblRange rhs ) const {return (*this).compare(rhs)<=0;};
bool operator>= (const dblRange rhs ) {return (*this).compare(rhs)>=0;};
bool operator>= (const dblRange rhs ) const {return (*this).compare(rhs)>=0;};
};
Right now I'm having trouble having the map accept a double as a key, even though the comparison operators are defined.
Here's some driving code that I'm using to test if it would work:
std::map<dblRange, int> map;
map[dblRange(0,1)] = 1;
map[dblRange(1,4)] = 2;
map[dblRange(4,5)] = 3;
map[3.0] = 4;
I mostly agree with Earwicker in that you can define a range. Now, I am in favor of implementing operators with the real meaning (do what basic types do: two ranges compare equal if both ranges ARE equal). Then you can use the third map parameter to pass it a comparison functor (or function) that solves your particular problem with this map.
// Generic range, can be parametrized for any type (double, float, int...)
template< typename T >
class range
{
public:
typedef T value_type;
range( T const & center ) : min_( center ), max_( center ) {}
range( T const & min, T const & max )
: min_( min ), max_( max ) {}
T min() const { return min_; }
T max() const { return max_; }
private:
T min_;
T max_;
};
// Detection of outside of range to the left (smaller values):
//
// a range lhs is left (smaller) of another range if both lhs.min() and lhs.max()
// are smaller than rhs.min().
template <typename T>
struct left_of_range : public std::binary_function< range<T>, range<T>, bool >
{
bool operator()( range<T> const & lhs, range<T> const & rhs ) const
{
return lhs.min() < rhs.min()
&& lhs.max() <= rhs.min();
}
};
int main()
{
typedef std::map< range<double>, std::string, left_of_range<double> > map_type;
map_type integer; // integer part of a decimal number:
integer[ range<double>( 0.0, 1.0 ) ] = "zero";
integer[ range<double>( 1.0, 2.0 ) ] = "one";
integer[ range<double>( 2.0, 3.0 ) ] = "two";
// ...
std::cout << integer[ range<double>( 0.5 ) ] << std::endl; // zero
std::cout << integer[ range<double>( 1.0 ) ] << std::endl; // one
std::cout << integer[ 1.5 ] << std::endl; // one, again, implicit conversion kicks in
}
You must be careful with equality and comparisons among double values. Different ways of getting to the same value (in the real world) can yield slightly different floating point results.
Create a class DoubleRange to store the double range, and implement the comparison operators on it. That way, std::map will do the rest for you, with the DoubleRange class as the key.
It is better to use Interval tree data structure. Boost has an implementation in Interval Container Library
One approach would be to calculate the "break points" before hand:
typedef vector< tuple<double, double, foo*> > collisionlist_t;
const collisionlist_t vec;
vec.push_back(make_tuple(0.0, 3.0, ptr));
vec.push_back(make_tuple(3.5, 10.0, ptr2));
// sort
std::map<double, foo*> range_lower_bounds;
for(collisionlist_t::const_iterator curr(vec.begin()), end(vec.end()); curr!=end; ++curr)
{
/* if ranges are potentially overlapping, put some code here to handle it */
range_lower_bounds[curr->get<0>()] = curr->get<2>();
range_lower_bounds[curr->get<1>()] = NULL;
}
double x = // ...
std::map<double, foo*>::const_iterator citer = range_lower_bounds.lower_bound(x);
Another suggestion: Use a mathematical transform to map the index from REAL to INT which can be directly compared.
If these ranges are multiple and dense there's also a structure known as an "interval tree" which may help.
Are the intervals open or closed or half open?
I will assumed closed. Note that the intervals cannot overlap by the definition of a map. You will also need splitting rules for when one inserts an over lapping interval. the rules need to decide where the split takes place and must take into account floating point epsilon.
this implementation uses map::lower_bound and does NOT use a class as the domain of the map
map::lower_bound returns an iterator to the first element in a map with a key value that is equal to or greater than that of a specified key. (ie the least key greater than or equal to K. An unfortunate choice of STL method names as it is the least upper bound of K.)
template <class codomain>
class RangeMap : private std::map<double,std::pair<double,codomain>{
public:
typedef double domain;
typedef std::map<double,std::pair<double,codomain>:: super;
typename super::value_type value_type;
protected:
static domain& lower(const value_type& v){
return v.first;
}
static domain& upper(const value_type& v){
return v.second.first;
}
static codomain& v(const value_type& v){
return v.second.second;
}
public:
static const domain& lower(const value_type& v){
return v.first;
}
static const domain& upper(const value_type& v){
return v.second.first;
}
static const codomain& v(const value_type& v){
return v.second.second;
}
static bool is_point(const value_type& vf) {
return lower(v) == upper(v);
}
static bool is_in(const domain& d,const value_type& vf) {
return (lower(v) <= d) && (d <= upper(v));
}
const_iterator greatest_lower_bound(const domain& d)const {
const_iterator j = super::lower_bound(d);
if(j!=end() && j->first==d) return j;//d is the lh side of the closed interval
//remember j->first >= d because it was lower but its the first
if(j==begin()) return end();//d < all intervals
--j; //back up
return j;
}
const_iterator find(domain& d) {
const_iterator j = greatest_lower_bound(d);
if (is_in(j,d)) return j;
return end();
}
iterator greatest_lower_bound(const domain& d) {
iterator j = super::lower_bound(d);
if(j!=end() && j->first==d) return j;//d is the lh side of the closed interval
//remember j->first >= d because it was lower but its the first
if(j==begin()) return end();//d < all intervals
--j; //back up
return j;
}
const_iterator find(domain& d) const{
iterator j = greatest_lower_bound(d);
if (is_in(j,d)) return j;
return end();
} //so much for find(d)
iterator find(domain& d){
iterator j = greatest_lower_bound(d);
if (is_in(j,d)) return j;
return end();
} //so much for find(d)
struct overlap: public std::exception{
};
bool erase(const double lhep,const double rhep);
//you have a lot of work regarding splitting intervals erasing when overlapped
//but that can all be done with erase, and insert below.
//erase may need to split too
std::pair<iterator,bool>
split_and_or_erase_intervals(const double lhep,
const double rhep,
const codomain& cd);
//the insert method - note the addition of the overwrtite
std::pair<iterator,bool>
insert(const double lhep,const double rhep,const codomain& cd,bool overwrite_ok){
if( find(lhep)!=end() || find(rhep)!=end() ) {
if(overwrite_ok){
return split_and_or_erase_intervals(const double lhep,
const double rhep,
const codomain& cd);
}
throw overlap();
}
return insert(value_type(lhep,pair<double,codomain>(rhep,cd)));
}
};
If your intervals must be non-overlapping, you must add some extra code to verify this property at insertion-time. Specifically, the property you wish to assert is that your new interval lies entirely within a range that was previously empty. An easy way to do this is to allow two types of ranges: "occupied" and "empty". You should begin by creating a single "empty" entry which covers the entire usable range. Insertion of a new "occupied" range requires:
(1) lookup some value within the new range.
(2) ensure that the returned range is empty, and wholly encompasses your new range. (This was the required assertion, above)
(3) modify the returned empty range so its end lies at the start of your new range.
(4) insert a new empty range that begins at the end of your new range, and ends at the old end of the returned range.
(5) insert your new range, confident that it is surrounded by empty-ranges.
(6) There may be additional corner-cases when inserting a new occupied range which has no empty space separating it from other occupied ranges.