数组？指针？
听说c++打算废弃指针了，谁让指针这么难呢！
我的环境：
>uname -a CYGWIN_NT-10.0-WOW DESKTOP-499IG24 2.10.0(0.325/5/3) 2018-02-02 15:21 i686 Cygwin
可见为32bit内核，也就是指针所占空间的sizeof为4.
那么这段程序：
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
/*main函数*/
int main(int argc, char**argv)
{
char s1[] = "abcdefg";
char s2[] = {'a','b','c','d','e','f','g'};
char s3[] = {"abcdefg"};
char *s4 = "abcdefg";
//char *s5 = {'a','b','c','d','e','f','g'};//error
char *s6 = {"abcdefg"};
printf("s:%s, length:%2d, sizeof:%2d\n",s1,strlen(s1),sizeof(s1));
printf("s:%s, length:%2d, sizeof:%2d\n",s2,strlen(s2),sizeof(s2));
printf("s:%s, length:%2d, sizeof:%2d\n",s3,strlen(s3),sizeof(s3));
printf("s:%s, length:%2d, sizeof:%2d\n",s4,strlen(s4),sizeof(s4));
//printf("s:%s, length:%2d, sizeof:%2d\n",s5,strlen(s5),sizeof(s5));//error
printf("s:%s, length:%2d, sizeof:%2d\n",s6,strlen(s6),sizeof(s6));
s1[0] = 'z';
s2[0] = 'z';
s3[0] = 'z';
//s4[0] = 'z';//error
//s5[0] = 'z';//error
//s6[0] = 'z';//error
return 1;
}
的输出为：
>gcc string.c >a.exe s:abcdefg, length: 7, sizeof: 8 s:abcdefgabcdefg, length:14, sizeof: 7 s:abcdefg, length: 7, sizeof: 8 s:abcdefg, length: 7, sizeof: 4 s:abcdefg, length: 7, sizeof: 4
那么这几个的sizeof为什么不一样呢？
32bit系统内核的指针大小为4，后面的两个4不难解释，上面的两个8可以解释为自动补齐（类似于struct，struct是按照最大字节数补齐），而大小为7的则是对应的数组。
sizeof(结构体)
一下均采用如下程序打印：
printf("%d+%d+%d+%d = %d\n",
sizeof(int),sizeof(char),sizeof(float),sizeof(double),sizeof(S));
需要搞明白的是struct的自动对齐机制：
例1：
struct{
int a;
char ch;
float f;
double d;
}S;
输出结果为：
>gcc string.c >a.exe 4+1+4+8 = 24
例2：
struct{
int a;
char ch;
float f;
}S;
结果：
>gcc string.c >a.exe 4+1+4+8 = 12
例3：
struct{
float f;
double d;
}S;
结果：
>gcc string.c >a.exe 4+1+4+8 = 16
例4：
struct{
int a;
double d;
}S;
结果：
>gcc string.c >a.exe 4+1+4+8 = 16
