Welcome to ShenZhenJia Knowledge Sharing Community for programmer and developer-Open, Learning and Share
menu search
person
Welcome To Ask or Share your Answers For Others

Categories

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!

Question&Answers:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
thumb_up_alt 0 like thumb_down_alt 0 dislike
954 views
Welcome To Ask or Share your Answers For Others

1 Answer

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?


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
thumb_up_alt 0 like thumb_down_alt 0 dislike
Welcome to ShenZhenJia Knowledge Sharing Community for programmer and developer-Open, Learning and Share
...