C++多字符串字符统计时后续字符串首字符未计数问题排查

我编写了一段C++代码，用于统计多个字符串中的元音、辅音及空格数量，但仅第一个字符串的统计结果正确，后续第2、3个及更多字符串的首字符未被计数，请问这是什么原因？

代码如下：

#include <iostream>
using namespace std;

int main()
{
    string sentence;
    int countv = 0, countc = 0, countspace = 0, number, s = 1;
    cout << "How many sentence would you like to check? - ";
    cin >> number;
    
    while(s <= number)
    {
        cout << "\nSentence " << s << ":";
        cin.ignore();
        getline(cin, sentence);
        
        for(int i = 0; i < sentence.length(); i++)
        {
            if(sentence[i] == 'a' || sentence[i] == 'e' || sentence[i] == 'i' || sentence[i] == 'o' || sentence[i] == 'u')
                countv++;
            else if(isspace(sentence[i]))
                countspace++;
            else
                countc++;
        }
        cout << "\nSentence " << s << " result:";
        cout << "\nThere are " << countv << " vowels in the sentence.";
        cout << "\nThere are " << countc << " consonants in the sentence.";
        cout << "\nThere are " << countspace << " whitespace in the sentence.";
        countc = 0, countv = 0, countspace = 0;
        s++;
        cout << "\n";
    }
}

问题原因 #

问题出在cin.ignore()的重复调用上：

• 当用cin >> number读取数字后，输入缓冲区会残留一个换行符（按下回车确认数字时产生）。第一次循环时，cin.ignore()会跳过这个换行符，让getline能正确读取第一个字符串。
• 从第二次循环开始，输入缓冲区里没有残留的换行符了，此时cin.ignore()会直接跳过你输入的当前字符串的首字符，导致后续字符串的首字符未被统计。

解决方案 #

只需要在读取数字后调用一次cin.ignore()，清除缓冲区里的换行符，不要在每次循环中重复调用：

修改后的代码：

#include <iostream>
#include <limits> // 包含头文件以使用numeric_limits
using namespace std;

int main()
{
    string sentence;
    int countv = 0, countc = 0, countspace = 0, number, s = 1;
    cout << "How many sentence would you like to check? - ";
    cin >> number;
    // 一次性清除缓冲区中直到换行符的所有残留字符
    cin.ignore(numeric_limits<streamsize>::max(), '\n');
    
    while(s <= number)
    {
        cout << "\nSentence " << s << ":";
        getline(cin, sentence); // 不再重复调用ignore
        
        for(int i = 0; i < sentence.length(); i++)
        {
            if(sentence[i] == 'a' || sentence[i] == 'e' || sentence[i] == 'i' || sentence[i] == 'o' || sentence[i] == 'u')
                countv++;
            else if(isspace(sentence[i]))
                countspace++;
            else
                countc++;
        }
        cout << "\nSentence " << s << " result:";
        cout << "\nThere are " << countv << " vowels in the sentence.";
        cout << "\nThere are " << countc << " consonants in the sentence.";
        cout << "\nThere are " << countspace << " whitespace in the sentence.";
        countc = 0, countv = 0, countspace = 0;
        s++;
        cout << "\n";
    }
}

额外说明 #

• 使用cin.ignore(numeric_limits<streamsize>::max(), '\n')比单纯的cin.ignore()更可靠，它会清除换行符之前的所有残留字符，避免因输入数字后多输入其他字符导致的问题。
• 如果只需要统计空格（而非所有空白字符如制表符），可以把isspace(sentence[i])改为sentence[i] == ' '。

内容的提问来源于stack exchange，提问作者Khairul Amri