忽略空间在C ++中使用getline [英] Ignore Spaces Using getline in C++

问题描述

嘿，我想写一个程序，从人接受新任务，将其添加到堆栈，能够显示任务，能够将该堆栈保存到文本文件，然后阅读文本文件。问题出现时，我试图接受用户的输入，每当你输入一个字符串，它有一个空格，菜单选择做什么只是循环。我需要一种方法来解决这个问题。任何帮助将非常感激。

  //基本文件io操作
 #include< iostream& 
 #include< fstream> 
 #include< stack> 
 #include< string> 
 using namespace std; 
 
 int main（）{
 //声明堆栈
 stack< string>列表; 
 
 //为菜单开始循环
 string inputLine; 
 cout<< 欢迎来到待办事项列表！ << endl; 
 
 //尝试读取文件
 ifstream myfile（to-do.txt）; 
 if（myfile.is_open（））{
 
 //读取待处理列表的每一行并将其添加到堆栈
 while（myfile.good（）） {
 getline（myfile，inputLine）; 
 list.push（inputLine）; 
} 
 myfile.close（）; 
 cout<< 文件读取成功！ << endl; 
} else {
 cout<< 没有文件要加载...创建一个空白堆栈。 << endl; 
} 
 
 int选项; 
 
 //而我们不想退出
 while（true）{
 //显示程序的选项
 cout< endl 你想干什么？ << endl; 
 cout<< 1.查看堆栈上当前的任务。 << endl; 
 cout<< 2.删除堆栈中的顶层任务。 << endl; 
 cout<< 3.向堆栈添加一个新任务。 << endl; 
 cout<< 4.将当前任务保存到文件。 << endl; 
 cout<< 5.退出。 << endl endl; 
 
 //从用户获取输入
 cin>>选项; 
 
 //使用该选项执行必要的任务
 if（option< 6&& option> 0）{
 if（option == 1）{ 
 //创建一个缓冲区列表以显示所有
 stack< string> buff = list; 
 cout< endl; 
 //打印出栈
 while（！buff.empty（））{
 cout< buff.top（）<< endl; 
 buff.pop（）; 
} 
} else if（option == 2）{
 list.pop（）; 
} else if（option == 3）{
 //使一个字符串保存输入
 string task; 
 cout< endl 输入您要添加的任务：< endl; 
 getline（cin，task）; //这是问题发生在
 cin.ignore（）; 
 
 //添加字符串
 list.push（task）; 
 cout< endl; 
} else if（option == 4）{
 //将堆栈写入文件
 stack< string> buff = list; 
 ofstream myfile（to-do.txt）; 
 if（myfile.is_open（））{
 while（！buff.empty（））{
 myfile< buff.top（）; 
 buff.pop（）; 
 if（！buff.empty（））{
 myfile<< endl; 
} 
} 
} 
 myfile.close（）; 
} else {
 cout<< 谢谢，再见！ << endl; 
 break; 
} 
} else {
 cout< 输入正确的数字！ << endl; 
} 
} 
} 
  

解决方案>

您必须在选择选项后添加 cin.ignore（）：

  //从用户获取输入
 cin>>选项; 
 cin.ignore（）; 
  

并且 cin.ignore（）必须在 getline 后：

  getline //这是问题发生在
 //cin.ignore（）; 
  

问题出现在 options 没有调用 cin.ignore（）之后，选项将包含行尾，循环将继续...

我希望这有助于。

Hey, I'm trying to write a program that will accept new tasks from people, add it to a stack, be able to display the task, be able to save that stack to a text file, and then read the text file. The issue comes when I am trying to accept input from the user, whenever you enter a string with a space in it, the menu to choose what to do just loops. I need a way to fix this. Any help would be greatly appreciated.

// basic file io operations
#include <iostream>
#include <fstream>
#include <stack>
#include <string>
using namespace std;

int main () {
    //Declare the stack
    stack<string> list;

    //Begin the loop for the menu
    string inputLine;
    cout << "Welcome to the to-do list!" << endl;

    //Trying to read the file
    ifstream myfile ("to-do.txt");
    if(myfile.is_open()){

        //read every line of the to-do list and add it to the stack
        while(myfile.good()){
            getline(myfile,inputLine);
            list.push(inputLine);
        }
        myfile.close();
        cout << "File read successfully!" << endl;
    } else {
        cout << "There was no file to load... creating a blank stack." << endl;
    }

    int option;

    //while we dont want to quit
    while(true){
        //display the options for the program
        cout << endl << "What would you like to do?" << endl;
        cout << "1. View the current tasks on the stack." << endl;
        cout << "2. Remove the top task in the stack." << endl;
        cout << "3. Add a new task to the stack." << endl;
        cout << "4. Save the current task to a file." << endl;
        cout << "5. Exit." << endl << endl;

        //get the input from the user
        cin >> option;

        //use the option to do the necessary task
        if(option < 6 && option > 0){
            if(option == 1){
                //create a buffer list to display all
                stack<string> buff = list;
                cout << endl;
                //print out the stack
                while(!buff.empty()){
                    cout << buff.top() << endl;
                    buff.pop();
                }
            }else if (option == 2){
                list.pop();
            }else if (option == 3){
                //make a string to hold the input
                string task;
                cout << endl << "Enter the task that you would like to add:" << endl;
                getline(cin, task); // THIS IS WHERE THE ISSUE COMES IN
                cin.ignore();

                //add the string
                list.push(task);
                cout << endl;
            }else if (option == 4){
                //write the stack to the file
                stack<string> buff = list;
                ofstream myfile ("to-do.txt");
                if (myfile.is_open()){
                    while(!buff.empty()){
                        myfile << buff.top();
                        buff.pop();
                        if(!buff.empty()){
                            myfile << endl;
                        }
                    }
                }
                myfile.close();
            }else{
                cout << "Thank you! And Goodbye!" << endl;
                break;
            }
        } else {
            cout << "Enter a proper number!" << endl;
        }
    }
}

解决方案

You have to add cin.ignore() right after options is chosen:

//get the input from the user
cin >> option;
cin.ignore();

And cin.ignore() is not necessary after your getline:

    getline(cin, task); // THIS IS WHERE THE ISSUE COMES IN
        //cin.ignore();

The problem is in options - if you didn't call cin.ignore() after it, options will contain end of line and loop will continue...

I hope this helps.

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