C ++对象泄漏

Question

I am currently learning C++, and I'm experiencing with local scope of objects.

If I understand correctly, with this code :

void stepOne() {
TestClass t1;
t1.thisIsAInt = 2;}

Once we exit the scope of the stepOne method, the t1 object should be cleared out of memory.

Now, I wanted to test this, so with a debugger I got the address of the object, and once I got back into main , I made a pointer on the address that I got from the debugger.

I was surprised to find that the pointer was still pointing to the same object I created earlier.

Normally, it is my understanding that the object would have been cleared of memory since we left the scope.

Now,

• Do I not understand it properly?
• Did getting the address of it made it not leave the scope?
• Is it free memory for the OS, and just not cleared yet? (Filled with zeroes)
• If I passed the pointer to a function from stepOne, would it still have been cleared once I left the scope? ( I understand I could have used a smart pointer, to make sure it only survives the local scope)

Answer 1

Once an object has gone out of scope and is destructed (add a breakpoint in the destructor to make sure it is) you should not reference the memory that was owned by that object.

Just because the object is destructed doesn't mean the memory it occupied will just "disappear". The memory still exists, and until it is reused by another object will not change its contents. Clearing memory takes (a little) time, doing it perhaps thousands of times each second will suddenly become very expensive.