Memory alignment Mode

Example 1:

<span style="font-size:18px;">#include <iostream>using namespace std;struct Node1{    bool m1;    int m2;bool m3;double m4;bool m5;};// struct Node2{//     char m1;//     char m2;// int m3;// };int main(){cout << sizeof(Node1) << endl;/*cout << sizeof(Node2) << endl;*/}</span>

Output result: 32

Cause: Double (8)> int (4)> bool (1)

So Int Is used before the appearance of double. The memory allocation of each member variable of the original struct is: 1, 4, 1, 8, 1

Therefore, 1 extension to 4 and 4 remain unchanged, and the second 1 must be extended to 8, so that 1 (4) + 4 (4) + 1 (8) = 16 can be divisible by 8, the last 1 is extended to 8.

Total memory 4 + 4 + 8 + 8 + 8 = 32.

Example 2:

<span style="font-size:18px;">#include <iostream>using namespace std;struct Node1{    bool m1;bool m3;    int m2;double m4;bool m5;};// struct Node2{//     char m1;//     char m2;// int m3;// };int main(){cout << sizeof(Node1) << endl;/*cout << sizeof(Node2) << endl;*/}</span>

Output result: 24

Cause: see article 1

Example 3: Array

<span style="font-size:18px;">#include <iostream>using namespace std;// // struct Node1{//     bool m1;// bool m3;//     int m2;// double m4;// bool m5;// };struct Node2{    char m1;    char m2[6];};int main(){/*cout << sizeof(Node1) << endl;*/cout << sizeof(Node2) << endl;}</span>

Result: 7

Cause: the char array cannot be regarded as an alignment, and node2 alignment is based on 1. Therefore, the result is 7;

Example 4: struct in struct

<span style="font-size:18px;">#include <iostream>using namespace std;struct Node1{    char m1;    double m2;char m3;};struct Node2{char m1;Node1 m2;};int main(){/*cout << sizeof(Node1) << endl;*/cout << sizeof(Node2) << endl;}</span>

Output result: 32

Cause:

Node1 is 24. See example 1;

However, the alignment base in node1 is 8, because node2 only has char M1, and the base number is 1 less than the base number of node1 8;

Therefore, the alignment base of node1 is 8, but node1 is a whole. The content of the filled part cannot be allocated to M1 in node2. Hence

M1 must be allocated 8 bytes to him, total number of bytes; 8 + 24 = 32

Memory alignment Mode