How to capture variable inside lambda - c++

Question

I have code from here:
std::sort(begin(v), end(v), [](auto const &t1, auto const &t2) {
return get<0>(t1) < get<0>(t2); // or use a custom compare function
});
I wanted to sort tuple multiple times so I wrote this code:
int k = 10;
while(k--){
std::sort(begin(v), end(v), [](auto const &t1, auto const &t2) {
return get<k>(t1) < get<k>(t2); // or use a custom compare function
});
}
but I get error error: ‘k’ is not captured. I tried to do it in this way:
int k = 10;
while(k--){
std::sort(begin(v), end(v), [&k](auto const &t1, auto const &t2) {
return get<k>(t1) < get<k>(t2); // or use a custom compare function
});
}
but it is not the proper way and error error: the value of ‘k’ is not usable in a constant expression occurs.
How to capture k variable?

Answer 1

As mch said in the comment, the problem is that k is not a compile time constant.
For a compile time constant, to iterate from N to 0, you may need template and recursion:
#include <algorithm>
#include <tuple>
#include <type_traits>
#include <vector>
using namespace std; // just for simplify, and not recommended in practice
template <size_t N, typename Iterator, enable_if_t<N == 0, int> = 0>
void foo(Iterator begin, Iterator end)
{
sort(begin, end,
[](const auto &t1, const auto &t2) {
return get<0>(t1) < get<0>(t2);
}
);
}
template <size_t N, typename Iterator, enable_if_t<N != 0, int> = 0>
void foo(Iterator begin, Iterator end)
{
sort(begin, end,
[](const auto &t1, const auto &t2) {
return get<N>(t1) < get<N>(t2);
}
);
foo<N - 1>(begin, end);
}
int main()
{
vector<tuple<int, int>> v{{0, 1}, {0, 0}, {1, 1}};
foo<1>(v.begin(), v.end());
// posible results:
// {0, 0}, {0, 1}, {1, 1}
// {0, 1}, {0, 0}, {1, 1} // impossible if use std::stable_sort instead
}

Answer 2

std::get only accepts a template argument which is an expression whose value that can be evaluated at compiling time.
You can't use k, because it is a variable that changes value.
std::sort(begin(v), end(v), [](auto const &t1, auto const &t2) {
const int k = 3;
return std::get<k>(t1) < std::get<k>(t2); // or use a custom compare function
});
As I wrote in the comments, I know const int = 3 will shadow the k value outside of the lambda expression, but this example shows that get will work when it it receives a compiling time constant value.
For example, if you try to set k = 5, for example where v only has 4 tuple parameters, the compiler will give an error because it knows that this is out of range.
The following code will give an error, but if k is set to 3, it will work
std::vector<std::tuple<int, int, int, int>> v;
std::sort(begin(v), end(v), [](auto const &t1, auto const &t2) {
const int k = 5;
return std::get<k>(t1) < std::get<k>(t2); // or use a custom compare function
});

Answer 3

#include <tuple>
#include <utility>
#include <cstddef>
#include <algorithm>
#include <cstddef>
#include <iterator>
template <std::size_t I, typename It, typename F>
void sort_tuple(It it, It end, F f)
{
std::stable_sort(it, end, [f](const auto& t1, const auto& t2)
{
return f(std::get<I>(t1), std::get<I>(t2));
});
}
template <typename It, typename F, std::size_t... Is>
void sort_tuple(It it, It end, F f, std::index_sequence<Is...>)
{
int dummy[] = { 0, (sort_tuple<sizeof...(Is) - Is - 1>(it, end, f), 0)... };
static_cast<void>(dummy);
}
template <typename It, typename F>
void sort_tuple(It it, It end, F f)
{
sort_tuple(it, end, f, std::make_index_sequence<
std::tuple_size<typename std::iterator_traits<It>::value_type>::value
>{});
}
Test:
std::vector<std::tuple<int, int, int>> v{ {2,1,2}, {2,2,2}, {3,2,1}
, {1,1,1}, {1,2,1}, {2,2,1} };
sort_tuple(begin(v), end(v), [](const auto& t1, const auto& t2)
{
return t1 < t2;
});
for (auto& t : v)
{
std::cout << std::get<0>(t) << " " << std::get<1>(t) << " " << std::get<2>(t) << std::endl;
}
DEMO

Answer 4

You are using k as a template argument so it must be available at compile time. One way to do this by making it a constexpr:
constexpr int k = 1;
int j = k;
while(j--){
std::sort(begin(v), end(v), [&k](auto const &t1, auto const &t2) {
return get<k>(t1) < get<k>(t2); // or use a custom compare function
})
}

Answer 5

Template non-type arguments must be compile time constants. int k=0; is not a compile time constant.
template<std::size_t...Is>
auto index_over(std::index_sequence<Is...> ={}){
return [](auto&&f)->decltype(auto){
return f( std::integal_constant<std::size_t,Is>{}... );
};
}
template<std::size_t N>
auto index_upto(std::integral_constant<N> ={}){
return index_over(std::make_index_sequence<N>{});
}
these are helpers to get efficient compile time values from 0 up to N-1.
auto foreacher=[](auto&&f){
return [f](auto&&...args){
using discard=int[];
(void)discard{0,(void(
f(decltype(args)(args))
),0)...};
};
};
the above is obscure c++14 to replace short c++17 code. foreacher(f) returns a function g. g(a,b,c) does f(a) then f(b) then f(c).
Now glue it together:
auto sort_v_by_k=[&](auto k){
std::sort(begin(v), end(v), [](auto const &t1, auto const &t2) {
return get<k>(t1) < get<k>(t2); // or use a custom compare function
});
};
index_upto<11>()( foreacher( [&](auto k){
sort_v_by_k( std::integral_constant<std::size_t, 11-k >{} );
}));
and ignoring typos, done.