fill array, show arrary , reverse_array
#include <iostream> using namespace std; int Fill_array(double arr[], int n) { int count = 0; cout << "Enter your price: "<< endl; for(int i =0; i < n; i++) { cin >> arr[i]; if((i+1) == n) cout <<"Enough! You are a rich man!"<<endl; else if(!cin) break; count++; } return count; } void Show_array(const double arr[], int n) { cout << "Your price list below: " << endl; for (int i = 0; i < n; i++) cout << "Price #" << i+1 << ": "<< arr[i] << endl; } void Reverse_array(double arr[], int n) { cout << "OK! Your price is: " << endl; while(n--) { cout <<"$"<< arr[n] << endl; } } const int LEN = 5; int main() { double name[LEN]; int ct = Fill_array(name, LEN); cout << "There are " << ct << " cases of dirty money!"<<endl; Show_array(name, ct); Reverse_array(name, ct); }
sizeof(arr) / sizeof(double) 得到数组元素个数??
arr[LEN] == `\0`;
然后在for里用 i+1== LEN 来判断更容易理解,如果为true说明数组已经填满。
一会用指针做一遍。
#include <iostream> using namespace std; double *fill_array(double *ps, const double *pe) { while(ps!=pe) { if(!(cin >> *ps)) { cin.clear(); while (cin.get() != '\n') continue; cout << "Bad input!" << endl; } else if (*ps < 0) break; else ps++; } return ps; } void show_array(double *ps, const double *pe) { while (ps != pe) { cout << *ps << endl; ps++; } cout << "Good!" << endl; } double *revalue_array(double *ps, const double *pe, float n) { for(;ps != pe;ps++) *ps *= n; return ps; } const int ARSIZE = 5; int main() { double arr[ARSIZE]; double *pt = fill_array(arr, arr + ARSIZE); cout << "Let's see what you have: " << endl; show_array(arr, pt); cout << "Let's change the value: " << endl; float factor = 1.5; double *pf = revalue_array(arr, pt, factor); show_array(arr, pf); cout << "OK! You are rich now! " << endl; return 0; }
指针还是方便,不需要复制数据,返回指针就可以了