#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[];

    printf("please enter the length of the array:\n ");

    scanf("%d",&iLength);

    printf("please enter the number:\n");

    for(i=;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,,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)/;

    n=;

    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+;

        }

        if(iPut<a[k])

        {

            j=k-;

        }

        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[];

    printf("please enter the length of the array:\n ");

    scanf("%d",&iLength);

    printf("please enter the number:\n");

    for(i=;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,,istars,iLength);

}

void binary_search(int a[],int start,int iPut,int end)

{

    int i,j,k,n,m;

    m=;

    i=start;

    j=end;

    n=;

    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)/;//这个为关键！！！！！

        if(iPut>a[k])

        {

            i=k+;

        }

        if(iPut<a[k])

        {

            j=k-;

        }

        if(iPut==a[k])

        {

            printf("the number need %d times to find......a[%d]=%d\n",n,k,iPut);

            m++;

            break;

        }

    }

    if(m==)

        printf("failed\n");

}