c++ - Local variable scope question

Question

Why is the following code prints "xxY"? Shouldn't local variables live in the scope of whole function? Can I use such behavior or this will be changed in future C++ standard?

I thought that according to C++ Standard 3.3.2 "A name declared in a block is local to that block. Its potential scope begins at its point of declaration and ends at the end of its declarative region."

#include <iostream>
using namespace std;

class MyClass
{
public:
  MyClass( int ) { cout << "x" << endl; };
  ~MyClass() { cout << "x" << endl; };
};

int main(int argc,char* argv[])
{
  MyClass  (12345);
// changing it to the following will change the behavior
//MyClass m(12345);
  cout << "Y" << endl;

  return 0;
}

---

Based on the responses I can assume that MyClass(12345); is the expression (and scope). That is make sense. So I expect that the following code will print "xYx" always:

MyClass (12345), cout << "Y" << endl;

And it is allowed to make such replacement:

// this much strings with explicit scope
{
  boost::scoped_lock lock(my_mutex);
  int x = some_func(); // should be protected in multi-threaded program
} 
// mutex released here

//    

// I can replace with the following one string:
int x = boost::scoped_lock (my_mutex), some_func(); // still multi-thread safe
// mutex released here

Answer 1

The object created in your

MyClass(12345);

is a temporary object which is only alive in that expression;

MyClass m(12345);

is an object which is alive for the entire block.