实验5

task1_1.c

#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");
}

 相较于原始数组，发生数组元素值交换是2次

task1_2.c

#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");
}

 相较于原始数组，发生数组元素值交换的是6次

 

task2.c

#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 (item==x[mid]) 
                return mid;
            else if (item<x[mid]) 
                high = mid - 1; 
            else
                low = mid + 1;
    }
    return -1;
}

 

 

task3_2.c  

#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 ; i++) {
        k = i;
        for (j = i + 1; j < n; j++)
            if (strcmp(str[j] ,str[k])<0)      
        {
            strcpy(temp, str[j]);
            strcpy(str[j], str[i]);
            strcpy(str[i], temp);
        }
    }
}

 

task4.c

#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0x135f970 变量n里存放的数是42

② 指针变量pn的地址是0x135f964 变量pn里存放的是0x135f970 ③ 通过 *pn 间接访问的是42   task5.c

#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;
}

 

① 通过 a[i] 和 *(p+i) 都可以访问到数组元素a[i] ② 通过 &a[i] 和 p+i 都可以获得元素a[i]的地址  

 

标签：int,++,实验,printf,output,include,pn
来源： https://www.cnblogs.com/lu0525/p/15680263.html