带指针的数组长度 [英] Array length with pointers

问题描述

在C ++中如何仅使用指针获取数组长度?我知道选项卡名称是指向第一个元素的指针，但是下一个是什么?

How in C++ get array length with pointers only ? I know that tab name is pointer to first element, but what next ?

推荐答案

您不能.指针只是一个内存位置，不包含任何可以确定大小的特殊内容.

You cannot. A pointer is just a memory location, and contains nothing special that could determine the size.

由于这是C ++，所以您可以做的是像这样通过引用传递数组:

Since this is C++, what you can do is pass the array by reference like so:

template <typename T, size_t N>
void handle_array(T (&pX)[N])
{
    // the size is N

    pX[0] = /* blah */;
    // ...
    pX[N - 1] = /* blah */;
}

// for a specific type:
template <size_t N>
void handle_array(int (const &pX)[N]) // const this time, for fun
{
    // the size is N

    int i = pX[0]; // etc
}

但是，否则您需要通过启动&结束并进行减法运算(如Alok建议的那样)，即开始与大小，如您建议的那样，或者抛弃静态数组并使用矢量，如Tyler建议的那样.

But otherwise you need to pass start & end and do a subtraction, like Alok suggests, a start & size, like you suggest, or ditch a static array and use a vector, like Tyler suggests.

如果知道要使用的数组的大小，则可以创建typedef:

If you know the size of the array you'll be working with, you can make a typedef:

typedef int int_array[10];

void handle_ten_ints(int_array& pX)
{
    // size must be 10
}

---

只是大小:

---

And just for the size:

template <typename T, size_t N>
size_t countof(T (&pX)[N])
{
    return N;
}

template <typename T, size_t N>
T* endof(T (&pX)[N])
{
    return &pX[0] + N;
}

// use
int someArray[] = {1, 2, 6, 2, 8, 1, 3, 3, 7};

size_t count = countof(someArray); // 9
std::for_each(someArray, endof(someArray), /* ... */);

我不时使用这些实用程序功能.

I use these utility functions from time to time.

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