I’ve used std::tie to unbind the underlying data from a std::tuple in the past:
auto t = std::make_tuple( 1, 42, 100 );
int a, b, c;
std::tie( a, b, c ) = t;
But I hadn’t seen this handy trick for using std::tie to implement operator<
#include <tuple>
struct S
{
S(int a_, int b_, int c_) : a(a_), b(b_), c(c_)
{}
int a;
int b;
int c;
};
bool operator<( const S& lhs, const S& rhs )
{
return std::tie( lhs.a, lhs.b, lhs.c )
< std::tie( rhs.a, rhs.b, rhs.c );
}
int main()
{
S s1( 1, 2, 3), s2 ( 1, 2, 4 ), s3 ( 0, 2, 3);
std::cout << "s1 should be less than s2: " << (s1 < s2) << "\n"
<< "s3 should be less than s1: " << (s3 < s1) << "\n";
return 0;
}
Notice that std::tie does not copy the variables (as per the earlier example, it takes them by reference). You can also mimic the use of use of underscore in F#
let f i = (i, i*2, i*4)
let a, _, c = f 5
using std::ignore
auto f = []( int i){ return std::tuple(i, i*2, i*4); }
int a, b;
std::tie( a, std::ignore, b ) = f( 5 );
musingstudio 