Value category of const int variable captured by lambda

Question

I have been attempting to understand when and when not a lambda with a capture-default odr-uses a variable with automatic storage duration defined in its surrounding scope (prompted by this answer). While exploring around this I came across a little curiosity. GCC and Clang appear to disagree about the value category of the id-expression n in the following code:

template <typename T> void assert_is_lvalue(const T&) {} template <typename T> void assert_is_lvalue(const T&&) = delete; int main() { const int n = 0; [=] { assert_is_lvalue(n); }; }

Clang compiles the code successfully, while GCC does not (error: use of deleted function). Which one is correct? Or is this something that is unspecified or implementation-defined?

Binding a reference to an object is supposed to odr-use it, and this is confirmed by removing the lambda's capture-default and observing that both compilers then complain that n can not be implicitly captured without a capture-default.

Marking the lambda as mutable makes no appreciable difference to the compilers' output.

GCC does not complain about n not being captured if we remove the capture-default. Seems to just be an obvious bug. — Columbo
– Columbo, Commented Apr 29, 2017 at 20:26
@Columbo Well spotted, I missed that. I don't think it's a bug, though. I think gcc is treating n as a constant expression, which it may be allowed to do. — Oktalist
– Oktalist, Commented Apr 29, 2017 at 21:58
Yes, GCC is known to incorrectly do constant folding in lambda ultra-early. — T.C.
– T.C., Commented Apr 29, 2017 at 22:41
it looks like a premature optimization made by the compiler trying to make n constexpr, then rvalue gets deduced and fails in your example. Definitelly a GCC bug — Ivan Sanz Carasa
– Ivan Sanz Carasa, Commented Feb 8, 2018 at 13:08

Ali · Accepted Answer · 2023-05-19 23:56:39Z

It turns out gcc behavior has changed from gcc-7.5 upward! I used the following code to see how n is captured within lambda and which template is matched.

#include <iostream> #include <string> #include <typeinfo> #include <type_traits> #include <memory> #include <string> #include <cstdlib> template <class T> constexpr std::string_view type_name() { using namespace std; #ifdef __clang__ string_view p = __PRETTY_FUNCTION__; return string_view(p.data() + 34, p.size() - 34 - 1); #elif defined(__GNUC__) string_view p = __PRETTY_FUNCTION__; #if __cplusplus < 201402 return string_view(p.data() + 36, p.size() - 36 - 1); #else return string_view(p.data() + 49, p.find(';', 49) - 49); #endif #elif defined(_MSC_VER) string_view p = __FUNCSIG__; return string_view(p.data() + 84, p.size() - 84 - 7); #endif } template <typename T> void assert_is_lvalue(const T& param) { std::cout << " T is " << type_name<T>() << " param is " << type_name<decltype(param)>() << '\n'; } //template <typename T> void assert_is_lvalue(const T&&) = delete; template <typename T> void assert_is_lvalue(const T&& param) { std::cout << " T is " << type_name<T>() << " param is " << type_name<decltype(param)>() << '\n'; } int main() { const int n = 0; [=] { std::cout << " n is " << type_name<decltype(n)>() << '\n'; assert_is_lvalue(n); }(); return 0; }

and here are the results:

gcc-7.5

n is const int T is int param is const int&&

gcc-8.1

n is const int T is int param is const int&

you can play with the code here.

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