Function pointer to variadic lambda

Question

Why does the compiler deny conversion of variadic lamda to function pointer? The conversion works perfectly fine without the variadic args.

    auto lambda = [] (void *, const char *) {};
    auto variadicLambda = [] (void *, const char *, auto ...) {};

    auto ptrLambda = +[] (void *, const char *) {};
    auto ptrVariadicLambda = +[] (void *, const char *, auto ...) {};

Tried with gcc 12.2 and clang 15.0.0.

<source>:7:30: error: invalid argument type '(lambda at <source>:7:31)' to unary expression
    auto ptrVariadicLambda = +[] (void *, const char *, auto ...) {};
                             ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

https://godbolt.org/z/G6cnK6oxP

Using auto creates a generic lambda, which is a templated functor and templates are not objects. — Richard Critten
– Richard Critten, Commented Nov 23, 2022 at 14:45
because (void *, const char *, int) and (void *, const char *, float) are different function and cannot squeeze into same pointer. (and what's the pointer type is another problem) — apple apple
– apple apple, Commented Nov 23, 2022 at 15:03

Richard · Accepted Answer · 2022-11-23 14:48:08Z

2

Why does the compiler deny conversion of variadic lamda to function pointer?

Because there is no overloaded operator+ that takes a generic lambda as its argument. This is because the use of auto in the parameter of the lambda makes it a generic lambda but since there is no operator+ for a generic lambda we get the mentioned error. This is exactly what the error says:

no match for 'operator+' (operand type is '<lambda(void*, const char*, auto:17 ...)>')
   12 | auto ptrVariadicLambda = +[] (void *, const char *, auto ...) {};

answered Nov 23, 2022 at 14:48

Richard

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