Arrays and Structures

一.数组

提出问题：

• Why should the first element of the array index from0开始？
• Multiple sequence of storage for variables local to is自下而上,还是自下而上分配的呢？
• Why the compiler to allocate12个4Bytes of memory？

猜测1: 用Against the conventionalMethods for计算速度
猜测2:The same way with arguments stack,First statement in the above

第i个元素的地址 = 首地址a + (i-1)×元素大小
第i个元素的地址 = 首地址a + i×元素大小

实证1：
C++代码

int main()
{
    
    int arr[10]{
    0};
    for (int i = 0; i < 10; i++) {
    
        arr[i] = i;
    }
}

汇编代码

003C1048 8B 55 D0             mov         edx,dword ptr [ebp-30h]  
003C104B 8B 45 D0             mov         eax,dword ptr [ebp-30h]  
003C104E 89 44 95 D4          mov         dword ptr [ebp+edx*4-2Ch],eax

实证2：First statement, first pressure stack,猜测2是错的

发现规律

• CArray of language in the global area,On the stack分配连续的内存空间;而且索引从零开始;
• The local variable memory layout isFirst statement, first pressure stack,The need in the future to further modify the validation;

一.结构体

提出问题：

• 编译器是把Person p当作一个整体还是单独的变量To allocate memory？答：整体
• 为什么结构体Person的内存大小占四个字节？What's the use of redundant one byte？
• Person p{}加了**{}为什么会影响局部变量的内存地址**呢

C++代码

struct Person {
    
    char c;
    short a;
};
int main()
{
    
    Person p;
    p.c = 0xffff;
    p.a = 0xffff;
}

00891004 C6 45 FC FF          mov         byte ptr [ebp-4],0FFh  
00891008 83 C8 FF             or          eax,0FFFFFFFFh  
0089100B 66 89 45 FE          mov         word ptr [ebp-2],ax

设计实验：From the table we can find the following pattern

• 当对齐字节Than the largest footprint bytes of the structure that a成员变量大时,aThe offset will not change again
• When the byte alignment for1时,The size of the structure will not change,没有字节填充

实验二：No matter how much byte alignment is,All the phenomenon of byte filling won't occur

struct Person {
    
    char a;
    char b;
    short c;
};

实验三：

struct Person {
    
    char a;
    char b;
    int c;
};

Speculation and infer(Based on the experimental phenomena)：Structure of the data storage to follow the following process

> First select a box,Then in order to field put into box,When the box can not let go,Again with another box to store,Until all fields are stored to complete.
> 盒子长度= min{
     max{
     sizeof(成员变量) }, 对齐长度 }
> Variable in the rules of the box：The head offset bytes from the box= n×sizeof(成员变量) （n=0,1,2,…）

实证：

struct Person {
    
    char a;
    short a2;
    char b;
    int c;
};

实际应用

待填充