I am using algorithms from std::ranges
( max
and max_element
) with a projection. Is it possible for the result to also be the projected value? Currently I have to call the projection function again on the returned value.
Example : Here I want the size of the longest string, but the algorithms return only the string or an iterator to it.
int main()
{
const std::vector<std::string> vec = {
"foo",
"hello",
"this is a long string",
"bar"
};
//r1 is a string. r2 is an iterator
const auto r1 = std::ranges::max(vec, {}, &std::string::size);
const auto r2 = std::ranges::max_element(vec, {}, &std::string::size);
//I have to call size() again
std::cout << r1 << '\n' << *r2 << '\n';
std::cout << r1.size() << '\n' << r2->size() << std::endl;
}
You're using an algorithm ( max
/ max_element
) on the original range, which can't do anything but give you an element/iterator into the range.
If you want just the projected values, do the projection (via a views::transform
) to get the lengths first, and then find the maximum of that
auto const lens = std::views::transform(vec, &std::string::size);
const auto r1 = std::ranges::max(lens);
const auto r2 = std::ranges::max_element(lens);
std::cout << r1 << '\n' << *r2 << '\n'; // prints 21 21
Here's a demo .
As mentioned in this answer , taking the address of std::string::size
is not permitted, so you should use a lambda instead. In general though, taking a projection based on a member function works just fine, so long as it's not a std function.
Here I want the size of the longest string, but the algorithms return only the string or an iterator to it.
In fact, according to [namespace.std#6] :
Let
F
denote a standard library function ([global.functions]), a standard library static member function, or an instantiation of a standard library function template. UnlessF
is designated an addressable function, the behavior of a C++ program is unspecified (possibly ill-formed) if it explicitly or implicitly attempts to form a pointer toF
.
This is unspecified behavior since you are forbidden to extract the address of the string::size
.
You can use views::transform
to convert the original range to a range whose elements are the size value and then take the maximum value.
auto size = std::ranges::max(
vec | std::views::transform([](auto& s) { return s.size(); }));
std::cout << size << '\n';
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