C& C ++:指向数组的指针和指向数组的地址有什么区别? [英] C & C++: What is the difference between pointer-to and address-of array?

问题描述

C ++ 11代码:

C++11 code:

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的值相同.

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运算符的行为应有所不同，特别是如果将a的声明更改为不同数量的整数，例如:

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，它打印:

4
28

那是因为两个指针是非常不同的类型.一个是指向整数的指针，另一个是指向由7个整数组成的数组的指针.

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.

第二个确实具有指针数组类型.如果您取消引用它，可以肯定，它在大多数情况下会衰减到指针，但这是实际上不是指向int的指针.第一个是指向int的指针，因为衰减是在赋值时发生的.

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.

其他可能出现的地方是，如果您确实有两个指针数组类型的变量，并试图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;
}

这为我赢得了错误消息:

This earns me the error message:

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

但是，此示例有效，因为取消引用bb允许它衰减为指向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:

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

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