The analyzer has detected the '__func__' identifier in the body of the overloaded '()' operator.
Consider the following example:
class C { void operator()(void) { std::cout << __func__ << std::endl; } }; void foo() { C c; c(); }
This code will output the string 'operator()'. This behavior may seem reasonable in code like this, so let's take a look at a less trivial example:
void foo() { auto lambda = [] () { return __func__; }; std::cout << lambda() << std::endl; }
It is important to remember that '__func__' is not a typical variable, so the following versions will not work as intended and the program will be still outputting the string 'operator()':
void fooRef() { auto lambda = [&] () { return __func__; }; std::cout << lambda() << std::endl; } void fooCopy() { auto lambda = [=] () { return __func__; }; std::cout << lambda() << std::endl; }
In the case of lambdas, this can be fixed by passing '__func__' explicitly using a capture list:
void foo() { auto lambda = [func = __func__] () { return func; }; std::cout << lambda() << std::endl; }
To get full-fledged output of the function name even inside the overloaded 'operator()' or lambdas, you can use the platform/compiler-specific macros. The MSVC compiler provides three such macros:
Clang and GCC provide the following macros: