Question

我写了一个程序，一次读取一个单词，直到一个'q'进入程序然后报告以元音开头的单词数量，以辅音开头的数字，以及不适合这些类别的数字。

#include <iostream>
#include <cstdlib>

int main()
{
char ch;
bool cont = true; //for controlling the loop
bool space = false; //see if there is a space in the input
    int i = 0; //checking if the input is the first word
int consta, vowel, others;
consta = vowel = others = 0;
std::cout<<"Enter words (q to quit)\n";

while (cont && std::cin>>ch) //continue while cont is true and the input succeded
{
    if (i == 0) //check if this is the first word
    {
        if (isalpha(ch))
            if ((ch == 'a' ||ch == 'e' ||ch== 'i' ||ch== 'o' ||ch== 'u') || (ch == 'A' ||ch== 'E' ||ch== 'I' ||ch== 'O' ||ch== 'U'))
                ++vowel;
            else
                ++consta;
        else
            ++others;
        ++i; //add 1 to i so this if statement wont run again
    }


    if (space == true) //check if the last input was a space
    {
        if (!isspace(ch)) //check if the current input is not a space
        {
         if (ch != 'q') //and ch is not 'q'
         {
        if (isalpha(ch))
            if ((ch == 'a' ||ch == 'e' ||ch== 'i' ||ch== 'o' ||ch==   'u') || (ch == 'A' ||ch== 'E' ||ch== 'I' ||ch== 'O' ||ch== 'U'))
                ++vowel;
            else
                ++consta;
        else
            ++others;

        space = false;
        }

        }
        else
            cont = false;
    }
    if (isspace(ch)) //check if ch is a space
        space = true;
}

std::cout<<"\n"<<consta<<" words beginnig with constants\n";
std::cout<<vowel<<" words beginnig with vowels\n";
std::cout<<others<<" words beginning with others\n";

system("pause");
return 0;
}

但是当我输入空格和'q'时它不会终止。但是如果我放'^ Z'它会终止但是constantans总是1而其他的总是0。

Answer 1

问题是std::cin默认忽略所有标签和空格。所以在你的代码中

if (isspace(ch)) //check if ch is a space
    space = true;

永远不会将space设置为true。避免此问题的一种方法是使用std::cin.get(ch)代替std::cin>>ch

Answer 2

希望这有助于你

#include<iostream>
#include<cstdlib>
#include<string>
int main()
{
    char ch[50];
    int consta, vowel, others;
    consta = vowel = others = 0;
    bool flag = false;
    std::cout<<"Enter words (q to quit)\n";
    while(1)
    {
        if(flag == true)
        {
            break;
        }
        gets(ch);
        char* pch;
        pch = strtok(ch," ");
        if(strcmp("q",pch)==0)
        {
            break;
        }
        while (pch != NULL)
        {
            if(strcmp("q",pch)==0)
            {
                flag = true;
                break;
            }
            if(isalpha(pch[0]))
            {
                if ((pch[0] == 'a' ||pch[0] == 'e' ||pch[0]== 'i' ||pch[0]== 'o' ||pch[0]== 'u') || (pch[0] == 'A' ||pch[0]== 'E' ||pch[0]== 'I' ||pch[0]== 'O' ||pch[0]== 'U'))
                    ++vowel;
                else
                    ++consta;
            }
            else
                ++others;
            pch = strtok (NULL, " ");
        }
    }
    std::cout<<"\n"<<consta<<" words beginnig with constants\n";
    std::cout<<vowel<<" words beginnig with vowels\n";
    std::cout<<others<<" words beginning with others\n";
    std::cin.ignore();
    return 0;
}

请参阅this以获取 strtok 。这将对所有单词进行计数，无论是用空格分隔还是用单独的行分隔。

Answer 3

假设您要使用相同的代码结构，请在while循环中使用

while (cont && std::cin>> std::noskipws >>ch)

std :: noskipws：这会将cin流的行为更改为不忽略空格。它还在流的末尾添加了一个空格。小心末端空间。幸运的是，您的代码处理空格，因此您不会看到它的影响。

当您在代码中处理空格时，这将起作用。此外，默认情况下，cin的行为是忽略空格。

但是我认为你可以降低代码的复杂性，你可能不需要这个声明

需要帮助调试字符输入

3 个答案: