11

I need to create a function that returns a pointer to an int.

Like so:

int * count()
{
    int myInt = 5;

    int * const p = &myInt;

    return p;
}

Since a pointer is simply an address, and the variable myInt is destroyed after this function is called. How do I declare an int inside this method that will keep a place in the memory in order for me to access it later via the returned pointer? I know I could declare the int globally outside of the function, but I want to declare it inside the function.

Thanks in advance for any help!

user906357
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  • You cannot (as described in you question). Why not pass an integer pointer into the function (or better still use a reference)? – Ed Heal Sep 27 '13 at 04:12
  • @EdHeal he can use `new int(something)` but that's really dumb so ... – aaronman Sep 27 '13 at 04:13
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    To get a really meaningful answer, you're probably going to have to tell us more about how you intend to use the variable. For example, if I called the function twice in a row, should it return the address of the same variable both times, or a unique variable each time? – Jerry Coffin Sep 27 '13 at 04:21

5 Answers5

12

Use the new operator

int * count()
{
    int myInt = 5;

    int * p = new int;
    *p = myInt;

    return p;
}

As pointed out in other answers this is generally a bad idea. If you must do it this way then maybe you can use a smart pointer. See this question for how to do this What is a smart pointer and when should I use one?

Community
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Nathaniel Johnson
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    This will leak memory unless he remembers to free it. – Jonathan Potter Sep 27 '13 at 04:18
  • Yeah, I should really let someone who is a const pointer expert answer this. – Nathaniel Johnson Sep 27 '13 at 04:20
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    IMHO - The thing that creates memory should be responsible for freeing it. This answer breaks that pattern. – Ed Heal Sep 27 '13 at 05:08
  • How can I do this same thing with an array? i.e. myInt[10] – user906357 Sep 27 '13 at 17:41
  • You better ask a new question with sample code. An array is a pointer so it gets into a more complicated de-referencing issues. See http://stackoverflow.com/questions/3473438/c-return-array-in-a-function – Nathaniel Johnson Sep 27 '13 at 17:48
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    This doesn't answer the question, you're creating a new int on the heap every time this is called. The OP wants "an int inside this method that will keep a place in the memory in order for me to access it later via the returned pointer?". I.e. a _single_ int in memory. @JonathanPotter's answer should be the accepted answer. – Steve Folly Oct 27 '17 at 08:50
9

You could use smart pointers.

For example:

unique_ptr<int> count()
{
   unique_ptr<int> value(new int(5));
   return value;
}

Then you can access the integer as follows:

cout << "Value is " << *count() << endl;
Arnav Borborah
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Ed Heal
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8

You can do this by making the variable static:

int* count()
{
    static int myInt = 5;
    return &myInt;
}
Jonathan Potter
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  • Why would you give this advice – aaronman Sep 27 '13 at 04:16
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    It answers the question. Why would you pretend it's not possible when it is? – Jonathan Potter Sep 27 '13 at 04:16
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    just because it's true doesn't mean it's helpful, if he doesn't understand static vars this will confuse him for hours – aaronman Sep 27 '13 at 04:17
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    Read the question: "How do I declare an int inside this method that will keep a place in the memory in order for me to access it later via the returned pointer?" This is exactly what the `static` keyword does. – Jonathan Potter Sep 27 '13 at 04:17
  • I guess you might actually have a point though it's hard to know if your giving him good advice without knowing how he plans to use the var, BTW I did not downvote – aaronman Sep 27 '13 at 04:22
  • I think the people downvoting either do not understand what `static` means or do not know how to read and understand the question... – Jonathan Potter Sep 27 '13 at 04:23
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    It's technically true but dangerous to just say without any more context (like the risks when mixed with threads). – nobody Sep 27 '13 at 04:30
  • This seems like a good answer, as long as you remember that it will always return a pointer to the same variable – memo1288 Sep 27 '13 at 04:30
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    Given that the OP's other option was a global variable this is no more dangerous or worrisome. It gives him a global var with private scope, and I assume since that's what he asked for that's what he actually wants. I think people tend to over-analyse questions a bit on here at times. – Jonathan Potter Sep 27 '13 at 04:31
  • There is some truth to that. – Nathaniel Johnson Sep 27 '13 at 04:37
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    its not that we over analyze, but we can see the flaws better than we can see the answers – Ryhan Feb 18 '15 at 09:14
2

It is an error to return a pointer to a local variable. x points to a variable allocated on the heap:

link x = new node(a[m]);
Thus x isn't pointing to a local variable.

The reason that returning a pointer to a local variable is an error is that such a variable exists for only as long as the function is active (i.e. between it is entered and exited). Variables allocated on the heap (e.g. with the use of the new operator) exist until they are deallocated (e.g. with the delete operator).

Anne Staley
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2

If you want to return a pointer of a variable correctly you have to do something like.

int * myInt = new int(5);

This is not a local variable BTW, meaning it does not have automatic storage and you have to delete the memory yourself

However using pointers like this is generally unnecessary and unadvised. It's better to create an int outside the function and have the function take a reference.

void count(int & i)
{
    i = 5;
}

BTW I don't know how you are planning to use the variable but since you also suggested using a global variable you may want to use a static var which @JonathanPotter suggested first. In many ways a static variable is similar to a global variable (both have static storage durations)

aaronman
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