Is []<typename>(){} a valid lambda definition?

Question

I was experimenting with lambdas and compilers because of another question here on SO.
I've just realized (and it's perfectly normal indeed) that the following code is valid:

int main() {     auto l = [](){};     l.operator()(); } 

Actually the standard says that the closure type has a public inline function call operator and so on, thus it makes sense to be able to invoke it.

What I can't explain by looking at the standard (well, the working draft) is the fact that GCC (6.1) compiles the following snippet (clang 3.9 does not):

int main() {     auto l = []<typename>(){};     l.operator()<void>(); } 

No warnings, no errors. Is it valid code or should it be rejected by the compiler?

Answer 1

In N4140 5.1.2 [expr.prim.lambda], a Lambda expression is defined as

lambda-introducer lambda-declarator_opt compound-statement

where a "lambda-introducer" is the [], enclosing an optional "lambda-capture" and "lambda-declarator_opt" is the stuff starting with "( parameter-declaration-clause )".

[]<typename>(){} 

does not meet that requirement because there is something between the lambda introducer and the lambda declarator, so it is not a valid lambda expression.

Thus, your example code is not valid C++ and should be rejected by the compiler.

---

As this is also tagged gcc, I clicked through the list of GNU C++ extensions. I did not find any extension that would make the syntax in question legal in GNU C++.

However, according to Section 4 of this proposal (P0428R0), which proposes to add templated lambdas to C++, gcc got an experimental implementation of the aforementioned paper in 2009. This probably explains why gcc doesn't complain here.