在 C++ 中读取带空格的字符串

Reading string with spaces in c++

提问by prattom

How can I read input line(type string) with whitespace? I tried getline but it goes into infinite loop. Following is my code.

如何使用空格读取输入行（字符串类型）？我试过 getline 但它进入无限循环。以下是我的代码。

#include <iostream>
#include <cstring>

#define MAX 50 //size of array

//Used G++ 4.6.3 compiler
using namespace std;

int main() {

struct Manager {
string name;
int age;
int working_years;
string phone;
int salary;
}info[MAX];

char inp; //To choose options
int array_pos = 0; //Current position in array of Manager structure
string search_name; //Manager name you want to search

cout << "Press 'i' to insert manager information or 's' to search for manager information by name or 'a' to abort: ";
cin >> inp;

while(inp != 'a') {
int search_num = 0; //array position at which search result is found
int found = 0;
if (inp == 'i' || inp == 's') {
    if (inp == 'i') {
        int k = array_pos;
        cout << "Enter the information of the manager no "<<k+1<<" is : "; 

        cout << "Enter the Name : "; 
                     //infinte loop occurs
        getline(info[array_pos].name, '\n');
        //cin >> info[array_pos].name;

        cout<<"Enter manager age : "; 
        cin >> info[array_pos].age;

        cout << "Enter manage working years : ";
        cin >> info[array_pos].working_years;

        cout << "Enter manager phone no. : ";
        cin >> info[array_pos].phone;

        cout << "Enter manager salary : ";
        cin >> info[array_pos].salary;
        array_pos++;
    }
    if (inp == 's') {
        cout << "Enter the manager name you want to search : ";
        cin >> search_name;
        for(int i = 0; i < array_pos; i++) {
            //using str1.compare(str2) to compare manager name
            if(info[i].name.compare(search_name) == 0) { //manager name found in array of structure
                found = 1;                  
                search_num = i;                 
                cout << "Name : " << info[search_num].name << "\n";
                cout << "Age: " << info[search_num].age << "\n";
                cout << "Working Years: " << info[search_num].working_years << "\n";
                cout << "Phone No. : " << info[search_num].phone << "\n";
                cout << "Salary : " << info[search_num].salary << "\n";
            } //end of if loop for comparing string
        } //end of for loop for searching
        if(found == 0)
            cout << "No Manager by this name exist in record" << "\n"; 

    } //end of if loop

} //end of if loop for  searching or insertion
if(inp == 'a')
    break;

cout << "Press 'i' to insert manager information or 's' to search for manager information by name or 'a' to abort: ";
cin >> inp;
} //end of while loop

return 0;
}

采纳答案by LihO

"How can I read input line(type string) with whitespace?"

“如何使用空格读取输入行（字符串类型）？”

std::string line;
if (std::getline(std::cin, line)) {
    ...
}

Note that apart from checking the return value of std:getlinecall, you should also avoid mixing >>operator with std::getlinecalls. Once you decide reading the file line by line, it seems to be cleaner and more reasonable to just make one huge loop and do the additional parsing while using string stream object, e.g.:

请注意，除了检查std:getline调用的返回值外，还应避免将>>运算符与std::getline调用混合使用。一旦您决定逐行读取文件，在使用字符串流对象的同时进行一个巨大的循环并进行额外的解析似乎更清晰、更合理，例如：

std::string line;
while (std::getline(std::cin, line)) {
    if (line.empty()) continue;
    std::istringstream is(line);
    if (is >> ...) {
        ...
    }
    ...
}

回答by Atul Rokade

Simplest way to read string with spaces without bothering about std namespace is as follows

在不关心 std 命名空间的情况下读取带空格的字符串的最简单方法如下

#include <iostream>
#include <string>
using namespace std;
int main(){
    string str;
    getline(cin,str);
    cout<<str;
    return 0;
}

回答by ytj

Solution #1:

解决方案#1：

char c;
cin >> noskipws;    // Stops all further whitespace skipping
while (cin >> c) {  // Reads whitespace chars now.
    count++;
}

Solution #2:

解决方案#2：

char c;
while (cin.get(c)) {  // Always reads whitespace chars.
    count++;
}