C&C++:指向数组的指针和指向数组的地址有什么区别?

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C++11 代码:

int a[3]; auto b = a; // b is of type int* auto c = &a; // c is of type int(*)[1]

C 代码:

int a[3]; int *b = a; int (*c)[3] = &a;

b 和 c 的值是一样的.

The values of b and c are the same.

b 和 c 有什么区别?为什么它们不是同一类型?

What is the difference between b and c? Why are they not the same type?

更新:我将数组大小从 1 更改为 3.

推荐答案

sizeof 运算符的行为应该有所不同，其中之一，尤其是当您将 a 的声明更改为不同数量的整数，例如 int a[7]:

The sizeof operator should behave differently, for one, especially if you change the declaration of a to a different number of integers, such as int a[7]:

int main() { int a[7]; auto b = a; auto c = &a; std::cout << sizeof(*b) << std::endl; // outputs sizeof(int) std::cout << sizeof(*c) << std::endl; // outputs sizeof(int[7]) return 0; }

对我来说，这会打印:

4 28

那是因为两个指针是非常不同的类型.一个是整数指针，另一个是7个整数数组的指针.

That's because the two pointers are very different types. One is a pointer to integer, and the other is a pointer to an array of 7 integers.

第二个确实有指向数组的指针类型.如果你取消引用它，当然，在大多数情况下它会衰减到一个指针，但它是实际上不是指向 int 指针的指针.第一个是指向 int 的指针，因为衰减发生在赋值时.

The second one really does have pointer-to-array type. If you dereference it, sure, it'll decay to a pointer in most cases, but it's not actually a pointer to pointer to int. The first one is pointer-to-int because the decay happened at the assignment.

它会出现的其他地方是，如果您确实确实有两个指向数组类型的变量，并尝试将一个分配给另一个:

Other places it would show up is if you really did have two variables of pointer-to-array type, and tried to assign one to the other:

int main() { int a[7]; int b[9]; auto aa = &a; auto bb = &b; aa = bb; return 0; }

这让我收到错误消息:

xx.cpp: In function ‘int main()’: xx.cpp:14:8: error: cannot convert ‘int (*)[9]’ to ‘int (*)[7]’ in assignment aa = bb;

然而，这个例子是有效的，因为取消引用 bb 允许它衰减到指向 int 的指针:

This example, however, works, because dereferencing bb allows it to decay to pointer-to-int:

int main() { int a; int b[9]; auto aa = &a; auto bb = &b; aa = *bb; return 0; }

请注意，衰减不会发生在作业的左侧.这不起作用:

Note that the decay doesn't happen on the left side of an assignment. This doesn't work:

int main() { int a[7]; int b[9]; auto aa = &a; auto bb = &b; *aa = *bb; return 0; }

它为您赢得了这个:

xx2.cpp: In function ‘int main()’: xx2.cpp:14:9: error: incompatible types in assignment of ‘int [9]’ to ‘int [7]’ *aa = *bb;