C++ lambda default arguments compiler misbehavior?

Question

Which of the following C++ lambdas/statements are supposed to work according to the latest C++ specification?

Context in case this is relevant: see here.

I tested the following code snippets with -std=c++17 on Fedora 33 with clang 11.0.0 and gcc 10.2.1.

Update: Replace __PRETTY_FUNCTION__ with __func__ for standard compliance. The same behavior can be observed.

Update2: Example using const char * s = __func__ as default argument to verify that it should be valid within a function scope (thanks to @BenVoigt).

---

1. LLVM __func__ within lambda default argument

void clang() {
  [](const char* c = __func__) {std::cout << c << std::endl;}();
}

Expected behavior (CLANG):

• Print out clang\n (void clang() for __PRETTY_FUNCTION__)

Observed behavior (CLANG):

• Compiler warning: warning: predefined identifier is only valid inside function [-Wpredefined-identifier-outside-function]
• Print out \n (top level() for __PRETTY_FUNCTION__)

---

2. GCC ignores statements

template <typename L>
constexpr std::string_view methodName(L l) { return l(); }
#define __METHOD_NAME__ (\
  __func__, /* needed for pointer to work */ \
  methodName([](const char* c = __func__) {return std::string_view(c);}) \
)
void gcc1() {
  std::cout << [](const char* c = __func__) { return c; }() << std::endl;  // GCC: This statement doesn't do anything
  std::cout << [](const char* c = __func__) { return c; }("gcc") << std::endl;
  std::cout << __METHOD_NAME__ << std::endl;  // GCC: This statement somehow conflicts with the statements above
}
void gcc2() {
  std::cout << __METHOD_NAME__ << std::endl;  // GCC: This statement itself works
}

Expected output (GCC):

gcc1
gcc
gcc1
gcc2

Observed output (GCC):

gcc
gcc2

---

3. GCC Compile error

void gcc3() {
  std::string_view s = [](const char* c = __func__) { return std::string_view(c); }();
  std::cout << s << std::endl;
}

Expected behavior (GCC): Compiles without problems.

Observed behavior (GCC): error: internal compiler error: in finish_expr_stmt

Answer 1

[class.local] The local class is in the scope of the enclosing scope, and has the same access to names outside the function as does the enclosing function. [Note: A declaration in a local class cannot odr-use (6.2) a local entity from an enclosing scope. — end note]

A lambda is a local class, and as such it cannot use variables from the enclosing scope (e.g. __func__) other than in its capture clause.