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[C] Pointer method Enter 10 numbers, swap the smallest number with the first number, and the largest number with the last number.

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Posted on 12/15/2015 2:19:52 PM | | |
  1. #include "stdafx.h"
  2. #include "stdio.h"
  3. int main(int argc, char* argv[])
  4. {void input(int *number);
  5. void maxmin(int *number);
  6. void output(int *number);
  7. int number[10];
  8. input(number);
  9. maxmin(number);
  10. output(number);
  11. return 0;
  12. }
  13. void input(int *number)
  14. {printf("please input 10 integer numbers:");
  15. int i;
  16.     for(i=0;i<10;i++)
  17.     scanf("%d",&number<i>);
  18. }
  19. void maxmin(int *number) //交换函数
  20. {int *max, *min,*p,temp;  //定义最大最小，存储变量
  21. max=min=number;   //开始是最大最小值指向第一个数
  22. for(p=number+1;p<number+10;p++)//循环他的第一个数
  23. if(*p>*max)max=p;        //若p指向的数字大于max指向的数，就使max指向p指向的大数字
  24. else if(*p<*min)min=p;   //若p指向的数字小于min指向的数，就使min指向p指向的小数字
  25. temp=number[0];number[0]=*min;*min=temp; //将最小数与第一个数number【0】交换
  26. if(max==number) max=min; //如果max和number相等，,表示第一个数是最大数，则使max指向当前的最大数
  27. temp=number[9];number[9]=*max;*max=temp;//将最大数与最后一个数交换
  28. }
  29. void output(int *number)
  30. {int *p;
  31. printf("now,they are:   ");
  32. for(p=number;p<number+10;p++)
  33. printf("%d,",*p);
  34. printf("\n");
  35. }</i>
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This function is still relatively complex, it uses three functions, and uses the pointer method to solve it






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