返回局部变量C ++的指针 [英] Returning a pointer of a local variable C++

问题描述

我需要创建一个返回指向int的函数的函数。

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

像这样：

int * count()
{
    int myInt = 5;

    int * const p = &myInt;

    return p;
}

由于指针只是一个地址，变量myInt在此之后被销毁函数被调用。我如何声明一个int在这个方法中，将保​​持在内存中的一个地方，以便我以后通过返回的指针访问它？我知道我可以在函数之外声明这个int，但是我要在函数中声明它。

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.

先感谢任何帮助！

推荐答案

使用新运算符

int * count()
{
    int myInt = 5;

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

    return p;
}

正如其他答案所指出的，这通常是一个坏主意。如果你必须这样做，那么也许你可以使用智能指针。查看这个问题如何做这个
什么是智能指针，我应该什么时候使用？

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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