Why Does Pushing Back Local Variable to Vector Works

The C++ vector stores pointers to the values it stores (i.e. vector of ints will store pointers to ints). In the following code, int i is a local variable in the for loop. Once the for loop is finished, the int i variable should be deleted from memory. Therefore, the vector pointers should be pointing to some garbage place in memory.

I plugged this code into XCode, yet it prints "30313233" – the ints that should have been erased from memory.

Why does it do this?

int main(int argc, const char * argv[]) {
std::vector<int> vec;
for(int i = 30; i < 34; i++)
{
    vec.push_back(i);
}
cout << vec[0];
cout << vec[1];
cout << vec[2];
cout << vec[3];

}

The C++ vector stores pointers to the values it stores

Nope, that's not true. Objects in C++ are truely objects, they aren't hidden references like in Java.1

int a = 1;
int b = a;
a = 3;
assert(b == 1); // b is a distinct object

In your example, you are pushing back i. What this means is that a copy of the object will get added to the vector, and not the variable itself.


1: Technically, it does store a pointer, but that pointer is to refer to the memory block where the array lies, where actual ints are stored. But that's an implementation detail that you shouldn't (at this point) be worried about.

Beginner questions - Marionette, Is there any way of hiding some of the variables? you can rename the node in the script which will not push that input out to the OIP of your Marionette object. I can't seem to find any kind of personal user library on my local hardrive. coming back to this thread cause I don't know where to go otherwise:. void push_back (const value_type& val); void push_back (value_type&& val); Add element at the end. Adds a new element at the end of the vector, after its current last element. The content of val is copied (or moved) to the new element. This effectively increases the container size by one, which causes an automatic reallocation of the allocated

The vector stores a pointer to the block of memory where the objects are stored, not the individual objects. When you insert into a vector, the object is copied into that block of memory.

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vector<int> stores values of type int. vector<int*> stores values of type int*, i.e., pointer to int.

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Comments
  • So it (1) makes a copy of the object and then (2) stores a pointer to that copy of the object in memory? (At this point) I would like to know the full info as I am applying to internships
  • @Y.Moondhra 1) yes, 2) no. The vector has a block of memory. When you use push_back, the vector creates a new object into that block of memory, which copies the object you passed into the new object being created. The only pointer it stores is to the start of the block of memory (like int a[] = {1, 2}; int *b = a;, b is a pointer to the block of memory of the array).
  • Now I completely understand it. Thank you.
  • gave u the upvote but it doesnt show because I don't have enough rep. thanks