c语言入门-bite-3-5-test.rar

#include<stdio.h> int main() { int aa[2][5] = { 10,9,8,7,6,5,4,3,2,1 }; int* ptr1 = (int*)(&aa + 1); int* ptr2 = (int*)(*(aa + 1)); printf("%d,%d", *(ptr1 - 1), *(ptr2 - 1)); return 0; } void bullbe_sort(int str[], int sz) { int i = 0; for (i = 0; i < sz; i++) { int j = 0; int flag = 0; for (j = 0; j < sz - 1 - i; j++) { if (str[j] < str[j + 1]) { flag = 1; int tmp = str[j]; str[j] = str[j + 1]; str[j + 1] = tmp; } } if (flag == 0) { break; } } } void print(int arr[], int sz) { int a = 0; for (a = 0; a < sz; a++) { printf("%d ", arr[a]); } } ////实现对整形数组的冒泡排序 int main() { int arr[10] = { 2,3,5,34,57,423,634,35,45,78}; int sz = sizeof(arr) / sizeof(arr[0]); bullbe_sort(arr,sz); //print(arr[10], sz); int i = 0; for (i = 0; i < sz; i++) { printf("%d ", arr[i]); } return 0; } void print(int* str, int sz) { int i = 0; for (i = 0; i < sz; i++) { //printf("%d ", *(str + i)); //printf("%d ", str[i]);//注意实际上编译器理解为 *(str+i)==str[i}==i[str} printf("%d ", i[str]); } } ////使用指针打印数组 int main() { int arr[10] = { 1,2,3,4,5,6,7,8,9,10 }; int sz = sizeof(arr) / sizeof(arr[0]); print(arr, sz); return 0; } //一道面试题 #include <stdio.h> int main() { char str1[] = "hello bit."; char str2[] = "hello bit."; char* str3 = "hello bit.";//str3和str4存放的是同一个无法被修改的常量字符串的首字符的地址 char* str4 = "hello bit.";//地址是唯一确定的所以str3=str4 if (str1 == str2) printf("str1 and str2 are same\n"); else printf("str1 and str2 are not same\n"); if (str3 == str4) printf("str3 and str4 are same\n"); else printf("str3 and str4 are not same\n"); return 0; } int main() { char ch[11]= "zhuwenyang"; char* p1 = "zhuwenyang";//p1中存放的是常量字符串的首元素'z'的地址在代码块中 char* p2 = ch; //数组名ch也表示数组ch首元素'z'的地址 printf("%s\n", p1); // 但是p1!=p2 p2==ch printf("%s\n", p2); return 0; }