{ int a[]={1,2,3,4,5,6}; intp; p=a; printf("%d\n",p); prin

来源：学生作业帮编辑：拍题作业网作业帮分类：综合作业时间：2024/04/27 22:29:32

{ int a[]={1,2,3,4,5,6}; int*p; p=a; printf("%d\n",*p); printf("%d\n",*(++p)); printf("%d\n",*++p);
{ int a[]={1,2,3,4,5,6};
int*p;
p=a;
printf("%d\n",*p);
printf("%d\n",*(++p));
printf("%d\n",*++p);
printf("%d\n",*(p--));
P+=3;
printf("%d%d",*p,*(a+3));
}
求程序输出结果（详细解及思路）

首先你得明白操作符的优先级；
*的优先级小于++；
*P=1;
*(++)p=2;
这时P=a+1;
*++p=3;
*(p--)=3;//此处--为后减,先执行完*P,再P－－；
经过 P+=3后,P＝a＋4；
＊P＝5；
＊（a＋3）＝4；
是对的吗

{ int a[]={1,2,3,4,5,6}; int*p; p=a; printf("%d\n",*p); prin Int a=1; Int *p; p=&a; printf(“%d\n”,*p); 和Int a=1; Int*p; * int a[5]={2,4,6,8,10},*P,* *k; p=a; k=&p; printf("%d",*(p++) main() { int a【】={1,2,3,4,5,6,7,8,9,0},*p; p=a; printf("%d\n main() {int a[5]={2,4,6,8,10},*p,**k; p=a; k=&p; printf("%d" int a[]={1,3,5,7}; int*p[3]=(a+2,a+1,a); int **q=p; printf(" main() { int a=2;a%=4-1; printf("＼n%d,",a); a+=a*=a-=a*=3; p int a[3][3]={{1,2,3},{4,5,6},{7,8,9}}; int **p; p=(int**)a; int a[3][8],*p; printf("\n a+0=%-10p",a+0); int d=1; fun(int p) {int d=5; d+=p++; printf("%d",d); } main C语言 int a;int b; }d[3] ={{1,4},{2,5},{6,7}}; printf("%d\n",d # include void fun (int p) { int d=2; p=d++; printf("%d",p);