为什么我的二进制搜索程序中没有输出？ [英] Why there is no output in my Binary Search program?

问题描述

//这里是代码，没有输出。

//我是入门级学生，所以请帮帮我。

//在这里，我试图计算二进制搜索程序中的迭代次数。

//here is the code, there is no output.
//I am an entry level student to this, so please help me out.
//here I am trying to count the number of iterations in a binary search program.

#include<stdio.h>
int main()
{ int array[10]={2,10,44,55,66,78,87,89,90},m,left,right,key=90,counter;
  left=0;
  right=8;
  while(left<=right)
  { m=(left+right)/2;
    if(key==array[m])
        {counter++; return 1;}
    else if(key>array[m])
        {counter++; left=m+1;}
    else
        {counter++; right=m-1;}
  }
printf("the number of iterations are %d",counter);
return 0;

}

推荐答案

因为您的实施不正确。尝试：

Because your implementation is not correct. Try:

#include<stdio.h>
int main()
{
  int array[]={2,10,44,55,66,78,87,89,90},m,left,right,key=90,counter=0;
  left=0;
  right=sizeof(array)/sizeof(array[0]) -1;

  while(left <= right)
  {
    m=(left+right)/2;
    counter++;
    if(key == array[m])
      break;
    else if(key>array[m])
      left = m + 1;
    else
      right=m-1;
  }
  printf("the number of iterations are %d\n",counter);
  if ( key == array[m])
    printf("the item was found at index %d\n", m);

  return 0;

}

你在开玩笑;-)

程序结束于返回1;

你可以使用break语句。





提示：



1.在一行上的每个语句。甚至是int

的声明2.在一行上的每个括号。

3.使用调试器单步模式



原因：当代码变得更多时代码保持清洁和可读。

you are kidding ;-)
the program is ending by "return 1;
you can use the break statement.


tips:

1. every statement on ONE line. even a declarations of int
2. every bracket on ONE line.
3. use the debugger in single step mode

Reason: the code stays cleaner and readble when it gets more.

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