Pour obtenir une clé
d'essai remplissez le formulaire ci-dessous
Demandez des tariffs
Nouvelle licence
Renouvellement de licence
--Sélectionnez la devise--
USD
EUR
RUB
* En cliquant sur ce bouton, vous acceptez notre politique de confidentialité

Free PVS-Studio license for Microsoft MVP specialists
To get the licence for your open-source project, please fill out this form
** En cliquant sur ce bouton, vous acceptez notre politique de confidentialité.

I am interested to try it on the platforms:
** En cliquant sur ce bouton, vous acceptez notre politique de confidentialité.

Message submitted.

Your message has been sent. We will email you at


If you haven't received our response, please do the following:
check your Spam/Junk folder and click the "Not Spam" button for our message.
This way, you won't miss messages from our team in the future.

>
>
>
V1056. The predefined identifier '__fun…
Analyzer diagnostics
General Analysis (C++)
General Analysis (C#)
General Analysis (Java)
Diagnosis of micro-optimizations (C++)
Diagnosis of 64-bit errors (Viva64, C++)
Customer specific requests (C++)
MISRA errors
AUTOSAR errors
OWASP errors (C#)
Problems related to code analyzer
Additional information
Contents

V1056. The predefined identifier '__func__' always contains the string 'operator()' inside function body of the overloaded 'operator()'.

16 Avr 2020

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:

  • '__FUNCTION__' – outputs the function name including its namespace. For example, this is what we will get for a lambda inside the main function: 'main::<lambda_....>::operator ()';
  • '__FUNCSIG__' – outputs the full function signature. Similarly, it can be helpful when combined with a lambda: 'auto __cdecl main::<lambda_....>::operator ()(void) const';
  • '__FUNCDNAME__' – outputs the decorated name of the function. This information is quite specific, so it cannot fully replace '__func__'.

Clang and GCC provide the following macros:

  • '__FUNCTION__' – outputs the same name that the standard '__func__' does;
  • '__PRETTY_FUNCTION__' – outputs the full function signature. For example, you will get the following output for a lambda: 'auto main()::(anonymous class)::operator()() const'.
Unicorn with delicious cookie
Nous utilisons des cookies pour améliorer votre expérience de navigation. En savoir plus
Accepter