A variable not detected as not used

Question

I am using g++ 4.3.0 to compile this example :

#include 

int main()
{
  std::vector< int > a;
  int b;
}

If I compile

Answer 1

a is not a built-in type. You are actually calling the constructor of std::vector<int> and assigning the result to a. The compiler sees this as usage because the constructor could have side effects.

Answer 2

In theory, the default constructor for std::vector<int> could have arbitrary side effects, so the compiler cannot figure out whether removing the definition of a would change the semantics of the program. You only get those warning for built-in types.

A better example is a lock:

{
    lock a;
    // ...
    // do critical stuff
    // a is never used here
    // ...
    // lock is automatically released by a's destructor (RAII)
}

Even though a is never used after its definition, removing the first line would be wrong.

Answer 3

a is actually used after it is declared as its destructor gets called at the end of its scope.