指向数组的指针和指向数组第一个元素的指针之间的区别 [英] difference between pointer to an array and pointer to the first element of an array

问题描述

int (*arr)[5] 表示 arr 是一个指向 5 个整数的数组的指针.现在这个指针到底是什么?

int (*arr)[5] means arr is a pointer-to-an-array of 5 integers. Now what exactly is this pointer?

如果我声明 int arr[5] 是否相同，其中 arr 是指向第一个元素的指针?

Is it the same if I declare int arr[5] where arr is the pointer to the first element?

两个示例中的 arr 是否相同?如果不是，那么究竟什么是指向数组的指针?

Is arr from both example are same? If not, then what exactly is a pointer-to-an-array?

推荐答案

在运行时，指针只是一个指针"，不管它指向什么，区别是语义上的；与指向元素的指针相比，指向数组的指针传达了不同的含义(对编译器而言)

At runtime, a pointer is a "just a pointer" regardless of what it points to, the difference is a semantic one; pointer-to-array conveys a different meaning (to the compiler) compared with pointer-to-element

处理指向数组的指针时，您指向的是指定大小的数组 - 编译器将确保您只能指向该大小的数组.

When dealing with a pointer-to-array, you are pointing to an array of a specified size - and the compiler will ensure that you can only point-to an array of that size.

即这段代码会编译

int theArray[5];
int (*ptrToArray)[5];
ptrToArray = &theArray;    // OK

但这会中断:

int anotherArray[10];
int (*ptrToArray)[5];
ptrToArray = &anotherArray;    // ERROR!

处理元素指针时，您可以指向内存中具有匹​​配类型的任何对象.(它甚至不一定需要在数组中；编译器不会做出任何假设或以任何方式限制您)

When dealing with a pointer-to-element, you may point to any object in memory with a matching type. (It doesn't necessarily even need to be in an array; the compiler will not make any assumptions or restrict you in any way)

即

int theArray[5];
int* ptrToElement = &theArray[0];  // OK - Pointer-to element 0

还有……

int anotherArray[10];
int* ptrToElement = &anotherArray[0];   // Also OK!

总而言之，数据类型int*并不意味着任何数组知识，但是数据类型int(*)[5]意味着一个数组，它必须正好包含 5 个元素.

In summary, the data type int* does not imply any knowledge of an array, however the data type int (*)[5] implies an array, which must contain exactly 5 elements.

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