问题
I am trying to get an explicit type of a range (I may want to store it as a field in a class in the future). However, for some reason, it evaluates to void
?
#include <iostream>
#include <set>
#include <range/v3/view/transform.hpp>
class Alpha {
public:
int x;
};
class Beta : public Alpha {
};
class Foo {
public:
std::set<Alpha*> s;
using RangeReturn = decltype(std::declval<std::set<Alpha*>>() | ranges::v3::view::transform(std::function<Beta*(Alpha*)>()));
RangeReturn r();
};
Foo::RangeReturn Foo::r() {
return s | ranges::v3::view::transform([](Alpha* a) { return static_cast<Beta*>(a); });
}
int main() {
}
When compiling with g++ -std=c++17, it gives
main.cpp:24:88: error: return-statement with a value, in function returning 'void' [-fpermissive]
(g++ version g++ (Ubuntu 7.3.0-27ubuntu1~18.04) 7.3.0)
I get an error of similar kind on Visual Studio 2017, v. 15.9
This question is a continuation of my other question: How to store a range as a field in a class? but is more specific and I believe it deserves to be separate.
回答1:
Your code doesn't work because:
range/v3 view disables view from rvalue, because that will cause dangling reference. Thus in your
declval()
, you should also use an lvalue:std::declval<std::set<Alpha*>&>() // ^ here should be lvalue
view transformation information is encoded inside the template parameter. So if you use
view::transform(std::function<Beta*(Alpha*)>())
to represent the type, your expression should have exactly the same type. A lambda is not ok.
A working version would be:
class Foo {
public:
std::set<Alpha*> s;
using RangeReturn = decltype(std::declval<std::set<Alpha*>&>() | ranges::v3::view::transform(std::function<Beta*(Alpha*)>()));
RangeReturn r();
};
Foo::RangeReturn Foo::r() {
return s | ranges::v3::view::transform(std::function<Beta*(Alpha*)>{
[](Alpha* a) { return static_cast<Beta*>(a); }
});
}
But actually, it's not a good idea to store a view this way.
来源:https://stackoverflow.com/questions/53664253/explicit-range-v3-decltype-evaluates-to-void