实验5

task1

#include<stdio.h>
#define N 5

void output(int x[],int n);

int main()
{
    int x[N] = {9, 55, 30, 27, 22};
    int i;
    int k;
    int t;
    
    printf("original array:\n");
    output(x, N);
    
    k = 0;
    for(i=1; i<N; ++i)
        if(x[i]>x[k])
            k = i;
            
    if(k != N-1)
    {
        t = x[N-1];
        x[N-1] = x[k];
        x[k] = t;
    }
    
    printf("after swapped:\n");
    output(x, N);
    
    return 0;
}

void output(int x[], int n)
{
    int i;
    
    for(i = 0; i<n; ++i)
        printf("%d", x[i]);
        
    printf("\n");
} 
 

 

#include <stdio.h>
#define N 5

void output(int x[], int n);

int main()
{
    int x[N]={9, 55, 30, 27, 22};
    int i;
    int t;
    
    printf("original array:\n");
    output(x, N);
    
    for(i=0; i<N-1; ++i)
        if(x[i] > x[i+1])
        {
            t = x[i];
            x[i] = x[i+1];
            x[i+1] = t;
        }
    
    
    printf("after swapped:\n");
    output(x, N);
    
    return 0;
}

void output(int x[], int n)
{
    int i;
    
    for(i=0; i<n; ++i)
        printf("%d", x[i]);
        
    printf("\n");
}

 

第一个交换了一次

第二个交换了三次

第一个是把最大元素向后放，而第二个是比较两者间的较小值往前放

task2

#include<stdio.h>

#define N 5

int binarySearch(int x[], int n, int item);

int main()
{
    int a[N]={2, 7, 19, 45, 66};
    int i, index, key;
    
    printf("数组a中的数据:\n");
    for(i = 0;i < N; ++i)
        printf("%d", a[i]);
    printf("\n");
    
    printf("输入待查找的数据项：");
    scanf("%d", &key);
    
    index = binarySearch(a, N, key);
    
    
    
    if(index >= 0)
        printf("%d 在数组中，下标为%d\n", key, index);
    else
        printf("%d 不在数组中\n", key);

    return 0;
}

int binarySearch(int x[], int n, int item)
{
    int low,high,mid;
    
    low = 0;
    high = n-1;
    
    while(low<= high)
    {
        mid = (low + high)/2;
        
        if(x[mid] == item)
            return mid;
        else if(x[mid] > item)
            high = mid-1;
        else
            low = mid +1;
    }
    
    return -1;
}

 

 

 

task3

#include <stdio.h>
#include <string.h>
#define N 5

void selectSort(char str[][20], int n);

int main()
{
    char name[][20] = {"Bob","Bill","Joseph","Taylor","George"};
    int i;
    
    printf("输出初始名单:\n");
    for(i=0;i<N;i++)
        printf("%s\n",name[i]);
        
    selectSort(name,N);
    
    printf("按字典序输出名单:\n");
    for(i=0;i<N;i++)
        printf("%s\n",name[i]);
        
    return 0;
}

void selectSort(char str[][20], int n)
{
    int i,j,k;
    char temp[20];
    
    for(i=0;i<n-1;i++)
    {
        k=i;
        
        for(j=i+1;j<n;j++)
            if(strcmp(str[j],str[k])<0)
                k=j;
                
        if(k!=i)
        {
            strcpy(temp,str[i]);
            strcpy(str[i],str[k]);
            strcpy(str[k],temp);
        }
    }
}

 

task4

#include <stdio.h>

int main()
{
   int n;
   int *pn;
   
   n = 42;
   pn = &n;
   
   printf("&n = %#x, n = %d\n", &n, n);
   printf("&pn = %#x, pn = %#x\n", &pn, pn);
   printf("*pn = %d\n", *pn);
   
   return 0;
}

 

n的地址是0x62felc 存放的是42

pn的地址是0x62fe10 存放的是n的地址0x62felc

通过*pn间接访问的是n

task5

#include <stdio.h>

#define N 5

int main()
{
    int a[N]= {1, 9, 2, 0, 7};
    int i;
    int *p;
    
    for(i=0; i<N; ++i)
       printf("&a[%d] = %#x, a[%d] = %d\n", i, &a[i], i, a[i]);
       
    printf("\n");
    
    for(i=0; i<N; ++i)
       printf("a+%d = %#x, *(a+%d) = %d\n", i, a+i, i, *(a+i));
       
    printf("\n");
    
    p = a;
    for(i=0; i<N; ++i)
       printf("p+%d = %#x, *(p+%d) = %d\n", i, p+i, i, *(p+i));
       
    return 0;
}

 可以

可以