我必须编写一个程序来计算旅行期间汽车的平均速度,我必须提示用户输入他们之间旅行的两个城市的名字。该程序有效,但我们必须编写的下一个程序是对最后一个程序的补充,其中提示用户程序应该运行多少次。我为用户需要程序运行的次数声明了一个新变量。但是当程序进入循环时,它会跳过第一个getline()方法(询问原始城市)并直接跳到第二个getline()(请求目标城市的方法)。我已经尝试清除缓冲区并以不同的方式声明循环,但无论我做什么,字符串仍然作为空字符串读入程序。只是想知道我是否在某些事情上犯了错误,或者在这种情况下我是否不能使用getline()。
在GNU编译器中使用C ++和Codeblocks IDE(我也尝试过其他编译器)
无论如何这里是代码。
#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
//Declare Variables
string orgcity = "";
string destcity = "";
int hours = 0;
double minutes = 0.0;
int hoursmin = 0;
int minutesmin = 0;
int minutesmax = 60;
double dist = 0.0;
double totalhours = 0.0;
double avespeed = 0.0;
double num1 = 0;
cout << "How many times would you like to calculate the average speed: ";
cin >> num1;
for(num1; num1 > 0; --num1)
{
//Collect Data
cout << "Enter the city of origin: ";
getline(cin, orgcity);
cout << "Enter the destination city: ";
getline(cin, destcity);
//If Statements and Loops...Start here
do {
cout << "Enter the number of hours spent in travel: ";
cin >> hours;
if (hours < hoursmin)
cout << " Invalid number of hours - must be >= 0" << endl;
} while (hours < hoursmin);
do {
cout << "Enter the number of minutes spent in travel: ";
cin >> minutes;
if (minutes >= minutesmax){
cout <<" Invalid number of minutes - must be in range 0..59" << endl;
}
if (minutes <= minutesmin) {
cout << " Invalid number of minutes - must be in range 0..59" << endl;
}
} while (minutes >= minutesmax);
//End Here
cout << "Enter the distance (in miles) between the two cities: ";
cin >> dist;
cout << endl;
//Formula and Final Prompt
totalhours = (hours + (minutes / 60.0));
avespeed = dist / totalhours;
cout << "The average speed of the vehicle traveling" << endl;
cout << "between " << orgcity << " and " << destcity << " is " << fixed << setprecision(2) << avespeed << " miles per hour." << endl;
cout << "-------------------------------------------------------------------------------" << endl;
}
return 0;
}
答案 0 :(得分:0)
当您读取循环运行的次数时,输入操作符会读取该数字,但会将新行留在缓冲区中。这意味着第一次调用getline
会读取单个换行符,而不再是。{/ p>
为确保您在该号码之后跳过任何内容,包括换行符,您可以使用例如。
std::cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');