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实验5 数组和指针

实验5 数组和指针

1. 实验任务1

#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");
}

 

task1.c实现方式中,相较于原始数组,发生数据元素交换1次

算法是交换下标

task2.c实现方式中,相较于原始数组,发生数据元素交换3次

算法是交换元素

 

 

2. 实验任务2

#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;
}

 

 

 

 

 

 3.实验任务3

#include<stdio.h>
#include<string.h>
#define N 5
void selectsort(char str[][20],int n);

int main(){
    char name[][20]={"Bob","Bill","Joseph","Talor","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,sign;
    char temp[20];
    for(i=0;i<n-1;i++){
        k=i;
        for(j=i+1;j<n;j++){
            sign=strcmp(str[k],str[j]);
            
            if(sign>0)
                k=j;
        }
        if(k!=i){
            strcpy(temp,str[i]);
            strcpy(str[i],str[k]);
            strcpy(str[k],temp);
        }
    }
}

 

 

 

 4.实验任务4

#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;
}

(1)整型变量n的地址是0x62fe1c,存放的数是42

(2)指针变量pn的地址是0x62fe10,存放的是变量n的地址

(3)通过*pn间接访问的是变量n的值

5.实验任务5

#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;
}

 

 通过 a[i] 和 *(p+i) 都可以访问到数组元素a[i]

通过 &a[i] 和 p+i 都可以获得元素a[i]的地址

posted on 2021-12-11 21:22  葛才璟  阅读(63)  评论(0编辑  收藏  举报

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