实验5
作者:互联网
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; }
![](https://www.icode9.com/i/l/?n=20&i=blog/2603405/202112/2603405-20211212191047826-2110141693.png)
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