int aa[3][3]={{2},{4},{6}};
main()
{int i,*p=&aa[0][0];
 for (i=0;i<2;i++)
{if (i==0)
  aa[i][i+1]=*p+1;
else
  ++p;
printf("%d",*p+1);}
}
请大家解释一下这道题的解题思路，谢谢了。

结果是34,

过程如下：

for (i=0;i<2;i++) {if (i==0) aa[i][i+1]=*p+1; i==0时：*p=a[0][0],a[0][1]=a[0][0]+1=2+1=3 else ++p; p=&a[0][1],*p=3 printf("%d",*p+1);}

所以结果为：34

int aa[3][3]={{2,0,0},{4,0,0},{6,0,0}};
int main(void)
{
        int i,*p=&amp;aa[0][0];
        for (i=0;i&lt;2;i++)
                {
                if (i==0)
                        {
                        aa[i][i+1]=*p+1;
                        }
                else{
                        ++p;
                    }
                printf("%d\n",*p+1);
                }
        getch();
        return 0;
}