16

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!

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3
  • You cannot (as described in you question). Why not pass an integer pointer into the function (or better still use a reference)? Commented Sep 27, 2013 at 4:12
  • @EdHeal he can use new int(something) but that's really dumb so ... Commented Sep 27, 2013 at 4:13
  • 2
    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? Commented Sep 27, 2013 at 4:21

5 Answers 5

Reset to default
14

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;
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Comments

11

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?

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6 Comments

This will leak memory unless he remembers to free it.
Yeah, I should really let someone who is a const pointer expert answer this.
IMHO - The thing that creates memory should be responsible for freeing it. This answer breaks that pattern.
How can I do this same thing with an array? i.e. myInt[10]
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.
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10

You can do this by making the variable static:

int* count()
{
    static int myInt = 5;
    return &myInt;
}
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11 Comments

It answers the question. Why would you pretend it's not possible when it is?
just because it's true doesn't mean it's helpful, if he doesn't understand static vars this will confuse him for hours
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.
It's technically true but dangerous to just say without any more context (like the risks when mixed with threads).
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.
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4

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).

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Comments

4

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)

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1 Comment

@JonathanPotter my bad typo

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