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假設(shè)我們有一個(gè)整型變量，其大小為 4 字節(jié)，還有另一個(gè)指針變量，其大小為 8 字節(jié)。那么下面的輸出會(huì)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;
}

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解決這個(gè)問題，我們可以遵循以下一些重要點(diǎn)：

整數(shù)大小為4字節(jié)（32位），指針大小為8字節(jié)。如果我們將1與指針相加，它將指向下一個(gè)立即類型。

&a1的類型是int **，&a2是int ***，&a3的類型是int ****。這里都指向指針。如果我們加1，我們將添加8字節(jié)。

a [0] [0] [0]是一個(gè)整數(shù)，&a [0] [0] [0]是int *，a [0] [0]是int *，&a [0] [0]的類型是int（*）[6]，依此類推。所以&a的類型是int（*）[4] [5] [6]。

輸出

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

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