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?
To capture the member variables inside lambda function, capture the “this” pointer by value i.e. std::for_each(vec. begin(), vec. end(), [this](int element){ //.... }
A lambda expression can't define any new scope as an anonymous inner class does, so we can't declare a local variable with the same which is already declared in the enclosing scope of a lambda expression. Inside lambda expression, we can't assign any value to some local variable declared outside the lambda expression.
Capturing Local Variables by value inside Lambda Function To capture the local variables by value, specify their name in capture list i.e. // Local Variables std::string msg = "Hello"; int counter = 10; // Defining Lambda function and // Capturing Local variables by Value auto func = [msg, counter] () { //... };
Lambda expression in java Using a local variable is as stated. However, when a lambda expression uses a local variable from its enclosing scope, a special situation is created that is referred to as a variable capture. In this case, a lambda expression may only use local variables that are effectively final.
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
}
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
});
#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
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