Why does a lambda convert to a bool whose value is true?
#include <iostream>
void IsTrue(const bool value) {
if (value) {
std::cout << "value is True!\n";
}
}
int main()
{
IsTrue([]() { ; /* some lambda */ });
return 0;
}
Output:
value is True!
Why does the lambda evaluate to true
on GCC & Clang? MSVC cannot build this (cannot convert lambda to bool).
Is it a compiler bug? Or which paragraph of the standard allows this?
The C++14 standard (§5.1.2) says:
The closure type for a non-generic lambda-expression with no lambda-capture has a public non-virtual non-explicit const conversion function to pointer to function with C++ language linkage (7.5) having the same parameter and return types as the closure type’s function call operator. The value returned by this conversion function shall be the address of a function that, when invoked, has the same effect as invoking the closure type’s function call operator.
Since a function pointer is implicitly convertible to bool
, you get the result you have shown. This is perfectly legal.
MSVC doesn't compile this because this conversion operator is overloaded with different calling conventions (__stdcall
, __cdecl
, etc).
When compiling for x64
all those calling conventions are not used, so there's just one conversion operator and it compiles fine.