返回局部变量 C++ 的指针

2021-12-07 00:00:00 function pointers c++

我需要创建一个函数来返回一个指向 int 的指针.

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，调用这个函数后变量myInt就被销毁了.如何在此方法中声明一个 int 以在内存中保留一个位置，以便我稍后通过返回的指针访问它?我知道我可以在函数外部全局声明 int，但我想在函数内部声明它.

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.

在此先感谢您的帮助！

推荐答案

使用新运算符

int * count()
{
    int myInt = 5;

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

    return p;
}

正如其他答案中所指出的，这通常是一个坏主意.如果您必须这样做，那么也许您可以使用智能指针.请参阅此问题以了解如何执行此操作什么是智能指针，什么时候应该使用一个?

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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