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，该指针到底是什么?
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]，其中arr是指向第一个元素的指针，是否也一样?
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，那么指向数组的指针到底是什么?
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，无论指针指向什么，指针都是只是指针"，区别在于语义.与指向元素的指针相比，指向数组的指针(对于编译器)传达了不同的含义
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
在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.
即该代码将编译
int theArray[5];
int (*ptrToArray)[5];
ptrToArray = &theArray; // OK
但这会中断:
int anotherArray[10];
int (*ptrToArray)[5];
ptrToArray = &anotherArray; // ERROR!
在处理指向元素的指针时，您可以指向内存中具有匹配类型的任何对象. (它甚至不一定需要位于数组中；编译器不会做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
和..
int anotherArray[10];
int* ptrToElement = &anotherArray[0]; // Also OK!
总而言之，数据类型int*并不暗示对数组有任何了解，但是数据类型int (*)[5]隐含着一个数组，该数组必须恰好包含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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