指针与二维数组

xt11_1
/*p149 5.18*/
#include <stdio.h>
main()
{
int a[][3]={1,2,3,4,5,6,7,8,9,10,11,12},*p,(*q)[3];
p=*a; q=a;
printf("%d,%d\n",*(p+2),*(*(q+2)+0));
}

xt11_2
#include <stdio.h>
main()
{
int a[][3]={1,2,3,4,5,6,7,8,9,10,11,12},(*p)[3];
p=a;
printf("%d,%d\n",*(*(p+2)),*(p[1]+1));
}

xt11_3
#include <stdio.h>
main()
{
char *ps[]={"abc","ABC"};
printf("%s,%s\n",ps[0],ps[1]);
printf("%2c%2c\n",*(*(ps+0)+1),*(*(ps+1)+2));
}

xt11_4

#include <stdio.h>
main()
{
char a[][7]={"for","switch","do"},*str[]={"for","switch","do"};
int i;
printf("%d,%d\n",sizeof(a),sizeof(str));
for(i=0;i<3;i++) printf("%s%c",a[i],i<3?' ':'\n');
}

xt11_5

#include <stdio.h>
main()
{
char *a[]={"for","switch","if","while"};
char **p;
printf("\n");
for(p=a;p<a+4;p++)
printf("%s%c",*p,p<a+4?' ':'\n');
}

xt11_6

#include <stdio.h>
struct stu
{
int x,*y;
}*p,a[4];
main()
{
int dt[]={1,2,3,4},i;
a[0].x=5;
a[0].y=&dt[0];
a[1].x=6;
a[1].y=&dt[1];
a[2].x=7;
a[2].y=&dt[2];
a[3].x=8;
a[3].y=&dt[3];
p=a;
printf("%d,",p->x); //a[0].x
printf("%d,",++p->x);//a[0].x=a[0].x+1
printf("%d\n",++(*(p->y)));//p->y与&dt[0]等价
for(i=0;i<4;i++) printf("%2d",dt[i]);
printf("\n");
for(i=0;i<4;i++)printf("%2d",a[i].x);
}

xt11_7

#include <stdio.h>
main()
{
union val
{
char cval;
short ival;
}x;
char *p;
p=&x.cval+1;
scanf("%c,%c",&x.cval,p);
printf("%d,%d,%x\n",x.cval,*p,x.ival);
}

xt11_8

#include <stdio.h>
main()
{
int *p,**q,i=10;
p=&i;
q=&p;
printf("%d,%d,%d\n",i,*p,**q);
*p*=2;
i/=10;
printf("%d,%d,%d\n",i,*p,**q);
}

xt11_9

/*带参的main函数*/
#include <stdio.h>
main(int argc,char *argv[])
{
int i;
for(i=1;i<argc;i++)
printf("%s%c",argv[i],(i<argc-1)?' ':'\n');
}
/*步骤：1.编译、运行;
2.将debug文件夹中的xx.exe文件移入D区;
3.点击开始|运行->打开框中输入cmd|确定;
4.进入DOS,输入xx I am student,即可得到程序结果.
解释其结果*/