I'm trying to create an std::set of GLM vectors (glm::vec3 specifically). Since C++ does not know how to perform < operation on the vectors, I must pass in a Compare funciton.
I can write my own by creating a structure like so:
struct compareVec
{
bool operator() (const glm::vec3& lhs, const glm::vec3& rhs) const
{
return lhs.x < rhs.x && lhs.y < rhs.y && lhs.z < rhs.z;
}
};
std::set< glm::vec3, compareVec > myset;
However, I'm sure that GLM includes their own vector comparison functions.
I found the following resource, but I'm unsure of how to use it:
https://glm.g-truc.net/0.9.4/api/a00137.html
How can I pass one of these comparison functions to my set?
Ok, almost there! The glm::lessThan returns a vector type, not a bool. This is why your comparator does not work. You could use glm::all on it to get a bool. From Documentation of glm::all
bool glm::all ( vecType< bool > const & v )
Returns true if all components of x are true.
Template Parameters
vecType Boolean vector types.
If this makes sense for you, you have to decide for yourself, even if I recommend against it as this, as far as I understand, will lead to the following issue:
Consider:
lhs = (1,2,3)
rhs = (0,1,4)
Than:
lhs < rhs ==> false, since lhs.x and lhs.y are larger than the corresponding components of rhs
rhs < lhs ==> false, since rhs.z component is larger than lhs.z
Since neither vector can be ordered to be less, this implies that they are equal. I doubt this is the behavior you have in mind (I was already warning you about this).
If you still decide to use it, here is a minimal working example tested on MSVC2010:
#include <set>
#include <glm/vec3.hpp>
#include <glm/detail/func_vector_relational.hpp>
struct compareVec
{
bool operator() (const glm::vec3& lhs, const glm::vec3& rhs) const
{
return glm::all(glm::lessThan(lhs, rhs));
}
};
int main()
{
std::set<glm::vec3, compareVec> myset;
return 0;
}
Maybe this helps.
By giving priority to early components of the vec, we can avoid (1,2,3) and (0,1,4) from both being less than each other. what we want is something that does the same thing as this:
lh.x != rh.x ?
lh.x < rh.x
: lh.y != rh.y ?
lh.y < rh.y
: lh.z < rh.z
I believe this should work as the compareVec:
struct compareVec {
bool operator() (const glm::vec3& lhs, const glm::vec3& rhs) const
{
glm::vec3 nequ = glm::notEqual(lhs, rhs);
return glm::lessThan(lhs, rhs)[nequ[0] ? 0 : nequ[1] ? : 1 : 2];
}
};
I couldn't find a glm function that returns the first index that is true, so I just used nequ[0] ? 0 : nequ[1]? : 1 : 2
Related
map<pair<int,int>,int>pairOfNumbers;
pairOfNumbers.insert(pair<pair<int,int>,int>({1,2},2));
this is working, but
map<pair<point,point>,int>PointsOnLine;
PointsOnLine.insert(pair<pair<point,point>,int>(make_pair(points[i],points[j]),count));
this doesn't.
point is just a structure of two ints x and y;
I keep getting the error 'Invalid Operands to binary expression(const point and const point' this is the structure of point.
struct point
{
int x;
int y;
public:
bool operator==(const point& p)
{
if(x==p.x && y==p.y)
return true;
else
return false;
}
bool operator!=(const point& p)
{
if(x==p.x &&y==p.y)
return false;
else
return true;
}
};
how do I insert two points and distance between them in the map?
in Xcode I get this error
Your point type does not support weak-ordering. It has no method of determining is-less-than. You may think you don't need that because your point is actually tucked into a std::pair<point,point> but you do.
std::pair<T1,T2> supports weak ordering only if T1 and T2 do. In your case, they're the same type, so for a std::pair<point,point> to be used as key in a std::map<std::pair<point,pint>,T>, point must support weak ordering.
To support weak ordering, you must either provide an operator< that compares two of your objects in question, or a comparator functor type that does the same thing. The easiest way for you to do this would be:
#include <tuple>
struct point
{
int x;
int y;
bool operator <(const point& p) const
{
return std::tie(x, y) < std::tie(p.x, p.y);
}
bool operator ==(const point& p) const
{
return !(*this < p || p < *this);
}
bool operator !=(const point& p) const
{
return *this < p || p < *this;
}
};
I took liberty to redefine operator == and operator != to utilize the weak order properties of the proper operator <. It wasn't necessary, but ultimately it's just easier if operators root to as basic code as possible. With the above change you should be able to use both point and std::pair<point,point> as key types in std::map and std::set. In truth, a strict weak ordering can define all of the basic comparators (<=, >, >=, !=, ==) as derivations from operator < in one form or another (or relative to something that does). I challenge you to consider them, try implementing them, and above all, writing some test harnesses that verify your implementation.
I'm using a c++ STL set and I want to know if it's present in the set an equivalent instance. To retrive the instance I'm using the find set method. The problem is that it doesn't work. I think the problem is in my comparator object:
bool SetComparator::operator ()( const Point* i1, const Point* i2 ) const {
if ( *i1 == *i2 )
return false;
return true;
}
The operator == is redefined for the class Point in a simple way:
bool Point::operator ==( const Point& p ) const {
if (x == p.x && y == p.y)
return true;
return false;
}
After a debugging I can see that the find method calls operator() but it doesn't find the same instance so the find returns end() but I know that there is an equal object. I think the problem is related to the set internal order. How can I do?
std::set uses partial ordering (i.e. the operator<), so when you pass in an operator that can only decide equality, you break the assumption of the implementation of std::set. Your SetComparator has to behave similar to std::less.
For example std::pair (utility) implements relational operators for two items, e.g. for operator<:
template <class T1, class T2>
bool operator< (const std::pair<T1,T2>& lhs, const std::pair<T1,T2>& rhs) {
return lhs.first<rhs.first || (!(rhs.first<lhs.first) && lhs.second<rhs.second);
}
note that (!(rhs.first<lhs.first) && lhs.second<rhs.second) is a workaround for (rhs.first == lhs.first && lhs.second < rhs.second) using only operator<
If you only want to check for equality maybe using std::set is the wrong decision. If you can hash your objects, you could use a std::unordered_set (C++11 and later).
I have a vector of structs. I need to check if the struct is or is not in the vector. The entire struct, not any specific member. It throws me this error upon compile time:
binary '==' : no operator found which takes a left-hand operand of type 'NavigationNode'
(or there is no acceptable conversion)
My struct:
struct NavigationNode{
int x, y; //their x and y position on the grid
float f, g, h;
int parentGCost;
int value;
};
NavigationNode currentNode;
The vector
vector<NavigationNode> openList;
My find:
if (find(closedList.begin(), closedList.end(), currentNode) == closedList.end() )
{
}
You need to overload operator==.
As global function:
bool operator==( const NavigationNode& lhs, const NavigationNode& rhs )
{
// compare lhs and rhs
}
Or as member function:
bool operator==( const NavigationNode& other ) const
{
// compare this & other
}
You will have to write an equality operator for your custom type. Assuming all variables have to be the same for two NavigationNode objects to be the same, it should look something like this:
bool floatEqual(float a, float b)
{
// adapt this comparison to fit your use-case - see the notes below
static const int EPSILON = 0.00001; // arbitrarily chosen - needs to be adapted to the occuring values!
return std::abs(a – b) <= EPSILON;
}
bool operator==(NavigationNode const & a, NavigationNode const & b)
{
return a.x == b.x &&
a.y == b.y &&
floatEqual(a.f, b.f) &&
floatEqual(a.g, b.g) &&
floatEqual(a.h, b.h) &&
a.parentGCost == b.parentGCost &&
a.value == b.value;
}
Even if you could also do it as a member function of NavigationNode, the recommended way is to implement the operator== as a free function (that way, both parameters can take advantage of any possible implicit conversions).
Note on float comparison: Due to how floating point numbers are represented, it is not a trivial task to compare them. Just checking for equality might not give the desired results. See e.g. this question for details:
What is the most effective way for float and double comparison?
You need to overload the comparison operator.
If your intention of "==" is "are each of the values contained in my struct equal to the corresponding members in this other struct" then you can write that.
bool operator==(const NavigationNode& lhs, const NavigationNode& rhs)
{
return /* compare each member in here */
}
I have a struct with members x,y,z and w. How do I sort efficiently
first by x, then by y, by z and finally by w in C++?
If you want to implement a lexicographical ordering, then the simplest way is to use std::tie to implement a less-than or greater-than comparison operator or functor, and then use std::sort on a collection of your structs.
struct Foo
{
T x, y, z, w;
};
....
#include <tuple> // for std::tie
bool operator<(const Foo& lhs, const Foo& rhs)
{
// assumes there is a bool operator< for T
return std::tie(lhs.x, lhs.y, lhs.z, lhs.w) < std::tie(rhs.x, rhs.y, rhs.z, rhs.w);
}
....
#include <algorithm> // for std::sort
std::vector<Foo> v = ....;
std::sort(v.begin(), v.end());
If there is not a natural ordering for Foo, it might be better to define comparison functors instead of implementing comparison operators. You can then pass these to sort:
bool cmp_1(const Foo& lhs, const Foo& rhs)
{
return std::tie(lhs.x, lhs.y, lhs.z, lhs.w) < std::tie(rhs.x, rhs.y, rhs.z, rhs.w);
}
std::sort(v.begin(), v.end(), cmp_1);
If you do not have C++11 tuple support, you can implement this using std::tr1::tie (use header <tr1/tuple>) or using boost::tie from the boost.tuple library.
You can turn a struct into a std::tuple using std::tie, and use the lexicographic comparison std::tuple::operator<. Here's an example using a lambda to std::sort
#include <algorithm>
#include <tuple>
#include <vector>
struct S
{
// x, y, z, w can be 4 different types!
int x, y, z, w;
};
std::vector<S> v;
std::sort(std::begin(v), std::end(v), [](S const& L, S const& R) {
return std::tie(L.x, L.y, L.z, L.w) < std::tie(R.x, R.y, R.z, R.w);
});
This example supplies std:sort with a comparison operator on-the-fly. If you always want to use lexicographic comparison, you could write a non-member bool operator<(S const&, S const&) that would automatically be selected by std::sort, or by ordered associative containers like std::set and std::map.
Regarding efficiency, from an online reference:
All comparison operators are short-circuited; they do not access tuple
elements beyond what is necessary to determine the result of the
comparison.
If you have a C++11 environment, prefer std::tie over hand-written solutions given here. They are more error-prone and less readable.
This solution has at most 4 comparisons per element-compare and does not require construction of other objects:
// using function as comp
std::sort (quadrupleVec.begin(), quadrupleVec.end(), [](const Quadruple& a, const Quadruple& b)
{
if (a.x != b.x)
return a.x < b.x;
if (a.y != b.y)
return a.y < b.y;
if (a.z != b.z)
return a.z < b.z;
return a.w < b.w;
});
If you roll your own comparison operator, then you can freely throw objects into std::maps or invoke std::sort. This implementation's designed to be simple so you can easily verify and modify it if needed. By only using operator< to compare x, y, z and w, it minimises the number of operators you may need to implement if those variables are not already comparible (e.g. if they're your own structs rather than ints, double, std::strings etc.).
bool operator<(const Q& lhs, const Q& rhs)
{
if (lhs.x < rhs.x) return true;
if (rhs.x < lhs.x) return false;
if (lhs.y < rhs.y) return true;
if (rhs.y < lhs.y) return false;
if (lhs.z < rhs.z) return true;
if (rhs.z < lhs.z) return false;
if (lhs.w < rhs.w) return true;
return false;
}
Sometimes types will define a comparison function that returns -1, 0 or 1 to indicate less-than, equal or greater-than, both as a way to support the implementation of <, <=, ==, !=, >= and > and also because sometimes doing a < then a != or > would repeat a lot of work (consider comparing long textual strings where only the last character differs). If x, y, z and w happen to be of such types and have a higher-performance compare function, you can possibly improve your overall performance with:
bool operator<(const Q& lhs, const Q& rhs)
{
int c;
return (c = lhs.x.compare(rhs.x) ? c :
(c = lhs.y.compare(rhs.y) ? c :
(c = lhs.z.compare(rhs.z) ? c :
lhs.w < rhs.w;
}
I have a STL container full of billions of the following objects
pair<SomeClass*, SomeClass*>
I need some function of the following form
/*returns items sorted biggest first */
bool sortPredicate (pair<SomeClass*, SomeClass*>two, pair<SomeClass*, SomeClass*> one)
{
return ???;
}
Is there some trick I can use to very quickly compare pairs of pointers?
Edit 1: A clarification
In the end I just want to sort the list of pointer-pairs such that all of the duplicates are next to each other. Assume that there is no clear method in SomeClass that can be used for this purpose---I only have pointer pairs, and I want to find all identical pairs (in parallel). I thought a sort would do the trick, but if you can think of a better parallel method, let me know.
Edit 2: A clarification
Fixed my code (the arguments to the sort predicate were wrong--they should be pairs).
It is a quirk of C++ that arbitrary pointers of the same type are not (necessarily) comparable with <, but are comparable with std::less.
Unfortunately, the operator< for std::pair is defined in terms of operator< on the components, not std::less.
So, assuming that you want two pairs to fall in the same sort position if and only if they point to the same two objects, you need:
// "less than"
template<typename T>
bool lt(const T &lhs, const T &rhs) {
return std::less<T>()(lhs, rhs);
}
typedef std::pair<SomeClass*, SomeClass*> mypair;
bool sortPredicate(const mypair &lhs, const mypair &rhs) {
return lt(lhs.first, rhs.first)
|| (!lt(rhs.first, lhs.first) && lt(lhs.second, rhs.second));
}
On pretty much any system you can name, this should compile to the same code as return lhs < rhs;, but that is not formally correct. If the referands of the pointers are all subobjects of the same object (for instance if you have a huge array and all the pairs point to elements of that one array), then operator< is OK for the pointers and hence OK for std::pair<pointer,pointer>.
If you want to pairs to fall in the same sort position if and only if the objects they point to sort the same, then you'd add the extra dereference:
bool sortPredicate(const mypair &lhs, const mypair &rhs) {
return lt(*lhs.first, *rhs.first)
|| (!lt(*rhs.first, *lhs.first) && lt(*lhs.second, *rhs.second));
}
and perhaps you'd also add checks for null pointers, if those are permitted. Of course if you know that SomeClass really is a class type, not a pointer type, then you don't need to use std::less in the version above, just define operator< for SomeClass and:
inline bool lessptr(const SomeClass *lhs, const SomeClass *rhs) {
if (lhs == 0) return rhs != 0;
if (rhs == 0) return false;
return *lhs < *rhs;
}
bool sortPredicate(const mypair &lhs, const mypair &rhs) {
return lessptr(lhs.first, rhs.first)
|| (!lessptr(rhs.first, lhs.first) && lessptr(lhs.second, rhs.second));
}
You may or may not be able to optimise that a bit, since there are some repeated null checks performed in both the first and second calls to lessptr. If you care that much, see what the compiler does with it.
Assuming your class has comparison operators:
bool sortPredicate (SomeClass *two, SomeClass *one)
{
return *two > *one;
}
If you just want to compare the pointer addresses, use std::greater<T>:
sort(container.begin(), container.end(), std::greater<SomeClass *>());
EDIT: OK, I really have no idea what you are trying to do now, with your most recent edit. Why not just use the default sort, if all you want to do is find duplicates?
If I understand correctly Your predicate should have the following signature
bool sortPredicate(pair<SomeClass*, SomeClass*>& lhs, pair<SomeClass*, SomeClass*>& rhs);
I know nothing about Your class and if there is any natural order for it, so it's hard to guess how You want to sort it. In The comment You write that the biggest items should be first. I assume there is < operator for the class. How about this?
bool sortPredicate(pair<SomeClass*, SomeClass*>& lhs, pair<SomeClass*, SomeClass*>& rhs)
{
if(!(*(lhs.first) < *(rhs.first) || *(rhs.first) < *(lhs.first))) // If there is == operator use it.
{
return *(rhs.second) < *(lhs.second);
}
else
{
return *(rhs.first) < *(lhs.first);
}
}
EDIT: Ok thx for clarifying. How about this?
bool sortPredicate(pair<SomeClass*, SomeClass*>& lhs, pair<SomeClass*, SomeClass*>& rhs)
{
if(lhs.first == rhs.first)
{
return rhs.second < lhs.second;
}
else
{
return rhs.first < lhs.first;
}
}
You should define an operator<on your pair class. I assume that your pair holds item1 and item2. So:
template <class T>
class pair{
private:
T item1;
T item2
public:
// [...] other stuff goes here
// here the comparing
bool operator<(pair p){
return (item1 < p.item1 || (item1 == p.item1 && item2 < p.item2));
}
};
This solution assumes that the items have defined the < and the == operators.
I suppose I didn't meet what you were exactly looking for, but I recommend to overload the <, >, and == operators in your pair class.