C++ 'new' keyword and memory management not necessarily with 'new'

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Hi,

Just a quick memory management question, let's say I have the following:

const std::vector<int> intArray = {1, 2, 3};
delete intArray;

This should allocate memory for the array then free it up again.

How does this work in this situation?

doSomething({1, 2, 3});

void doSomething(const std::vector<int> arr)
{
        // whatever
}

Here, there's no opportunity to delete the original array. Which of the following should I do?

1. Nothing, that memory will be managed already.

2.

const std::vector<int> arr = {1, 2, 3};
doSomething(arr);
delete arr;

void doSomething(const std::vector<int> arr)
{
        // whatever
}

3.

const std::vector<int>* arr = {1, 2, 3};
doSomething(arr);

void doSomething(const std::vector<int>& arr)
{
        delete arr;
}

I'm not sure if the syntax for 1 and 2 are correct either, please notify me of any issues you see there.

What I have tried:

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Posted 23-May-18 4:50am

User 13204940

Updated 23-May-18 5:09am

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Leo Chapiro · Answer 1 · 2018-05-23T05:03:00

Solution 1

Have you tried to compile this code:

const std::vector<int> intArray = {1, 2, 3};
delete intArray;

I think no, because you should get an error on "delete intArray" like "Expression must be a pointer".

You can only delete the objects created on the heap with "new", not such stack objects like intArray: take a look at operator delete.

Posted 23-May-18 5:03am

Leo Chapiro

Comments

[no name] 23-May-18 11:06am

Ah ok, this is where things differ to other languages I'm used to, something like

Object obj;

in another language would create the object but not construct it, and you'd have to have

obj = new Object();

afterwards. From what I understand, simply something like

Object obj;

in C++ does all of that already.

So doing that creates the variable on the stack, whereas

Object obj = new Object();

would create the variable on the heap and would therefore require

delete obj;

When using 'new' with no arguments, do you have to have the () after the object type or can you leave it as without, like this?

Object obj = new Object;

Rick York 23-May-18 12:41pm

One thing to keep in mind is with C++, the operator new creates an object and returns a pointer to it. Your statement should be:

Object * pobj = new Object;

The parenthesis are needed only if you are passing parameters to the object's constructor like:

Object * pobj = new Object( param1, param2 );

Jochen Arndt · Answer 2 · 2018-05-23T05:11:00

If you declare an object on the stack as in your case, it is destroyed automatically when going out of scope.

As already noted at your other question, you should pass it by reference to avoid creating a copy.

So it should be:

C++

{
    const std::vector<int> arr = {1, 2, 3};
    doSomething(arr);

    // arr goes out of scope here at the end of the block
    // The desctructor is called and any memory is freed
}

void doSomething(const std::vector<int>& arr)
{
    // do something
}

If you pass the array by value:

C++

// A temporary copy of the passed array is created and assigned to arr
void doSomething(const std::vector<int> arr)
{
    // do something

    // The destructor of arr is called here when returning from the function
    // The original array passed from the caller still exists
}