std::pair of references - c++

Question

Is it valid to have a std::pair of references ? In particular, are there issues with the assignment operator ? According to this link, there seems to be no special treatment with operator=, so default assignement operator will not be able to be generated.
I'd like to have a pair<T&, U&> and be able to assign to it another pair (of values or references) and have the pointed-to objects modified.

Answer 1

In C++11 you can use std::pair<std::reference_wrapper<T>, std::reference_wrapper<U>> and the objects of that type will behave exactly as you want.

Answer 2

No, you cannot do this reliably in C++03, because the constructor of pair takes references to T, and creating a reference to a reference is not legal in C++03.
Notice that I said "reliably". Some common compilers still in use (for GCC, I tested GCC4.1, #Charles reported GCC4.4.4) do not allow forming a reference to a reference, but more recently do allow it as they implement reference collapsing (T& is T if T is a reference type). If your code uses such things, you cannot rely on it to work on other compilers until you try it and see.
It sounds like you want to use boost::tuple<>
int a, b;
// on the fly
boost::tie(a, b) = std::make_pair(1, 2);
// as variable
boost::tuple<int&, int&> t = boost::tie(a, b);
t.get<0>() = 1;
t.get<1>() = 2;

Answer 3

I think it would be legal to have a std::pair housing references. std::map uses std::pair with a const type, after all, which can't be assigned to either.
I'd like to have a pair<T&, U&> and be able to assign to it another pair
Assignment won't work, since you cannot reset references. You can, however, copy-initialize such objects.

Answer 4

You are right. You can create a pair of references, but you can't use operator = anymore.

Answer 5

I was thinking along the same lines as you, I think. I wrote the following class to scratch this particular itch:
template <class T1, class T2> struct refpair{
T1& first;
T2& second;
refpair(T1& x, T2& y) : first(x), second(y) {}
template <class U, class V>
refpair<T1,T2>& operator=(const std::pair<U,V> &p){
first=p.first;
second=p.second;
return *this;
}
};
It allows you to do horrible things like:
int main (){
int k,v;
refpair<int,int> p(k,v);
std::map<int,int>m;
m[20]=100;
m[40]=1000;
m[60]=3;
BOOST_FOREACH(p,m){
std::cout << "k, v = " << k << ", " << v << std::endl;
}
return 0;
}
(remember the relevant includes).
The nastiness is of course that the references to k and v that I am assigning to are hidden inside p.
It almost becomes pretty again if you do something like this:
template <class T1,class T2>
refpair<T1,T2> make_refpair (T1& x, T2& y){
return ( refpair<T1,T2>(x,y) );
}
Which allows you to loop like this:
BOOST_FOREACH(make_refpair(k,v),m){
std::cout << "k, v = " << k << ", " << v << std::endl;
}
(All comments are welcome as I am in no way a c++ expert.)

Answer 6

I don't know what is "wrong" with std::pair in C++03 but if I reimplement it naively, I don't have any problem with it, (using the same compiler gcc and clang).
double a = 1.;
double b = 2.;
my::pair<double, double> p1(5., 6.);
my::pair<double&, double&> p2(a, b);
p2 = p1; // a == 5.
So a workaround could be to (1) reimplement pair (in a different namespace), or (2) specialize for std::pair<T&, T&>, or (3) simply use C++11 (where std::pair for refs works out of the box)
(1) Here it is the naive implementation
namespace my{
template<class T1, class T2>
struct pair{
typedef T1 first_type;
typedef T2 second_type;
T1 first;
T2 second;
pair(T1 const& t1, T2 const& t2) : first(t1), second(t2){}
template<class U1, class U2> pair(pair<U1, U2> const& p) : first(p.first), second(p.second){}
template<class U1, class U2>
pair& operator=(const pair<U1, U2>& p){
first = p.first;
second = p.second;
return *this;
}
};
template<class T1, class T2>
pair<T1, T2> make_pair(T1 t1, T2 t2){
return pair<T1, T2>(t1, t2);
}
}
(2) And here it is an specialization of std::pair (some people may complain that I am messing around overloading/specializing with the std namespace, but I think it is ok if it is to extend the capabilities of the class)
namespace std{
template<class T1, class T2>
struct pair<T1&, T2&>{
typedef T1& first_type; /// #c first_type is the first bound type
typedef T2& second_type; /// #c second_type is the second bound type
first_type first;
second_type second;
pair(T1& t1, T2& t2) : first(t1), second(t2){}
template<class U1, class U2> pair(pair<U1, U2> const& p) : first(p.first), second(p.second){}
template<class U1, class U2>
pair& operator=(const pair<U1, U2>& p){
first = p.first;
second = p.second;
return *this;
}
};
}
Maybe I am missing something obvious, I can edit the answer if some obvious flaws, are pointed.

Answer 7

Post c++14, you can do:
int a, b;
auto const p(std::make_pair(std::ref(a), std::ref(b)));
Using std::cref() is also possible.

Answer 8

I ended up solving a similar problem by just building a really simple structure. I didn't even worry about the assignment operator since the default one should work fine.
template<class U, class V>
struct pair
{
pair(U & first, V & second): first(first), second(second) {}
U & first;
V & second;
}