一个C/C++指针谜题？

假设我们有一个整型变量，其大小为 4 字节，还有另一个指针变量，其大小为 8 字节。那么下面的输出会是什么？

示例

#include<iostream>
using namespace std;
main() {
   int a[4][5][6];
   int x = 0;
   int* a1 = &x;
   int** a2 = &a1;
   int*** a3 = &a2;
   cout << sizeof(a) << " " << sizeof(a1) << " " << sizeof(a2) << " " << sizeof(a3) << endl;
   cout << (char*)(&a1 + 1) - (char*)&a1 << " ";
   cout << (char*)(&a2 + 1) - (char*)&a2 << " ";
   cout << (char*)(&a3 + 1) - (char*)&a3 << " ";
   cout << (char*)(&a + 1) - (char*)&a << endl;
   cout << (char*)(a1 + 1) - (char*)a1 << " ";
   cout << (char*)(a2 + 1) - (char*)a2 << " ";
   cout << (char*)(a3 + 1) - (char*)a3 << " ";
   cout << (char*)(a + 1) - (char*)a << endl;
   cout << (char*)(&a[0][0][0] + 1) - (char*)&a[0][0][0] << " ";
   cout << (char*)(&a[0][0] + 1) - (char*)&a[0][0] << " ";
   cout << (char*)(&a[0] + 1) - (char*)&a[0] << " ";
   cout << (char*)(&a + 1) - (char*)&a << endl;
   cout << (a[0][0][0] + 1) - a[0][0][0] << " ";
   cout << (char*)(a[0][0] + 1) - (char*)a[0][0] << " ";
   cout << (char*)(a[0] + 1) - (char*)a[0] << " ";
   cout << (char*)(a + 1) - (char*)a;
}

解决这个问题，我们可以遵循以下一些重要点：

• 整数大小为4字节（32位），指针大小为8字节。如果我们将1与指针相加，它将指向下一个立即类型。

• &a1的类型是int **，&a2是int ***，&a3的类型是int ****。这里都指向指针。如果我们加1，我们将添加8字节。

• a [0] [0] [0]是一个整数，&a [0] [0] [0]是int *，a [0] [0]是int *，&a [0] [0]的类型是int（*）[6]，依此类推。所以&a的类型是int（*）[4] [5] [6]。

输出

480 8 8 8
8 8 8 480
4 8 8 120
4 24 120 480
1 4 24 120

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