#include<stdio.h>//二分查找法void binary_search(int a[],int start,int mid,int end);int main(){    int iLength,istars,i,iTimes,iNumber,n;    int a[100];    printf("please enter the length of the array:\n ");    scanf("%d",&iLength);    printf("please enter the number:\n");    for(i=0;i<iLength;i++)    {        scanf("%d",&a[i]);    }    printf("please enter the number that you want:\n");    scanf("%d",&istars);    /*if(n==istars)    {            }    else    {        printf("sorry ,without this number\n");    }*/     binary_search(a,0,istars,iLength);    }void  binary_search(int a[],int start,int iPut,int end){    int i,j,k,n;    i=start;    j=end;    k=(i+j)/2;    n=0;    while(i<j)    //while(i<=k && j>=k) 这个写的本身就有漏洞，就是对整个程序的不理解    {        n++;        if(iPut>a[k])        {            i=k+1;        }        if(iPut<a[k])        {            j=k-1;        }        if(iPut==a[k])        {            printf("the number need %d times to find......a[%d]=%d\n",n,k,iPut);                    break;        }            }

if(n==0)        printf("failed\n");    }

 这是开始参考书目，自己编写的程序，但是程序本身能准确运行范围内的数字，一旦输入范围外的数字则出现卡壳现象。因此，要在源代码中改进。

#include<stdio.h>//二分查找法void binary_search(int a[],int start,int iPut,int end);int main(){    int iLength,istars,i,iTimes,iNumber,n;    int a[100];    printf("please enter the length of the array:\n ");    scanf("%d",&iLength);    printf("please enter the number:\n");    for(i=0;i<iLength;i++)    {        scanf("%d",&a[i]);    }    printf("please enter the number that you want:\n");    scanf("%d",&istars);    /*if(n==istars)    {            }    else    {        printf("sorry ,without this number\n");    }*/     binary_search(a,0,istars,iLength);    }void binary_search(int a[],int start,int iPut,int end){    int i,j,k,n,m;    m=0;    i=start;    j=end;    n=0;    while(i<j)    //while(i<=k && j>=k) 这个写的本身就有漏洞，就是对整个程序的不理解    {        n++;        k=(i+j)/2;//这个为关键！！！！！        if(iPut>a[k])        {            i=k+1;        }        if(iPut<a[k])        {            j=k-1;        }        if(iPut==a[k])        {            printf("the number need %d times to find......a[%d]=%d\n",n,k,iPut);            m++;            break;        }            }    if(m==0)        printf("failed\n");    }