The given code snippet involves a `std::function` instance and checks...

The given code snippet involves a std::function instance and checks its validity before throwing an exception. Let's break it down:

Code Explanation:

1. Declare a std::function<void()> object:

   std::function<void()> checkIsUsableFunction;
   

   This creates a std::function object capable of storing any callable entity (e.g., a lambda, function pointer, etc.) that matches the signature void(). When declared like this, checkIsUsableFunction is uninitialized and defaults to an empty state (i.e., it does not point to any callable entity).

2. Call operator bool() to check its validity:

   if (checkIsUsableFunction.operator bool())
   

   • The std::function class provides an operator bool() to check if it holds a valid callable object.
   • If the std::function is empty (i.e., does not store any callable object), operator bool() will return false.
   • If the std::function stores a callable object, operator bool() will return true.

3. Throw an exception if the std::function is valid:

   throw "asdasdas";
   

   This means that if checkIsUsableFunction contains a callable object (i.e., its operator bool() evaluates to true), the program will throw an exception with the message "asdasdas".

What it does:

• Since checkIsUsableFunction is default-constructed and thus empty, the condition:

  if (checkIsUsableFunction.operator bool())
  

  evaluates to false.

• As a result, the code inside the if block (throw "asdasdas";) will not execute.

Summary:

The snippet declares an empty std::function<void()>, checks if it is valid (non-empty), and, if valid, throws an exception with the string "asdasdas". However, as written, the std::function is empty, so nothing happens when this code runs.