How does the C++ runtime system know when objects go out of scope

Question

I was wondering how the C++ runtime system detects when an object goes out of scope so that it calls the destructor accordingly to free up the occupied memory.

Thanks.

Answer 1

The runtime doesn't - the compiler keeps tabs on scope and generates the code to call the destructor. If you make a simple test application and look at the generated disassembly you'll see explicit destructor calls.

Disassembly snippet from MSVC:

int main() {
    std::string s1;
...
00971416  call        dword ptr [__imp_std::basic_string<char,std::char_traits<char>,std::allocator<char> >::basic_string<char,std::char_traits<char>,std::allocator<char> > (979290h)] 
...
    {
        std::string s2;
00971440  call        dword ptr [__imp_std::basic_string<char,std::char_traits<char>,std::allocator<char> >::basic_string<char,std::char_traits<char>,std::allocator<char> > (979290h)] 
...    
    }
00971452  call        dword ptr [__imp_std::basic_string<char,std::char_traits<char>,std::allocator<char> >::~basic_string<char,std::char_traits<char>,std::allocator<char> > (979294h)] 
...
}
0097146B  call        dword ptr [__imp_std::basic_string<char,std::char_traits<char>,std::allocator<char> >::~basic_string<char,std::char_traits<char>,std::allocator<char> > (979294h)] 

Answer 2

This is known statically at compile time

{
  string s; /* ctor called here */
} /* dtor called here */

Sometimes, it's more difficult

{
  again:
  {
    string s; /* ctor called here */
    goto again; /* now dtor of s is called here */
    string q; /* ctor not called. not reached. */
  } /* dtor of s and q would be called here. but not reached */
}

Answer 3

It has nothing to do with runtime. The compiler keeps track of scope of every lexical variable, and adds destructor calls.