template and std::forward in c++

Question

The first lines of code will return 112 but why? 1 is not supposed to be of type int&&? And why it is working in the second example of code i.e. returning 312?

void fun(int&) { cout << 1; }
void fun(int const&) { cout << 2; }
void fun(int&&) { cout << 3; }

template
<typename T>
void fun2(T&& t) { fun(t); }

int main() {
  int x{};
  int const y{};
  fun2(1);
  fun2(x);
  fun2(y);
}

void fun(int&) { cout << 1; }
void fun(int const&) { cout << 2; }
void fun(int&&) { cout << 3; }

template
<typename T>
void fun2(T&& t) { fun(std::forward<T>(t)); }

int main() {
  int x{};
  int const y{};
  fun2(1);
  fun2(x);
  fun2(y);
}

Cause it's no longer an rvalue and you don't pass it as one to `fun()` in `fun2()` in the first snippet. — Hatted Rooster, May 27 '19 at 09:06

score 0 · Answer 1 · answered May 27 '19 at 09:07

0

std::forward basically means that you want to preserve the lvalueness or rvalueness of the calling argument when passing/forwarding it to another function.

So in the second example you keep the rvalueness, while in the first example you pass t as a lvalue reference.

You might want to read a bit more on universal/forwarding reference for example here: https://isocpp.org/blog/2012/11/universal-references-in-c11-scott-meyers

answered May 27 '19 at 09:07

maxbachmann

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how 1 can be a lvalue then ? – Gauvain May 27 '19 at 09:16
A lvalue is a value that has an identifiable location in memory which you can get the address of. So t is in fact a lvalue – maxbachmann May 27 '19 at 09:24
you are saying that 1 has an identifiable location memory? – Gauvain May 27 '19 at 10:43
no 1 is a rvalue, but you could get the address of t – maxbachmann May 27 '19 at 10:43
ok but when you call foo2(1), why it is foo2(int&) that get called? – Gauvain May 27 '19 at 10:45
because you do not call foo2(1), but you call foo2(t) with t being a lvalue. – maxbachmann May 27 '19 at 10:47

template and std::forward in c++

1 Answers1