How to properly use cin.peek()

Question

This function is supposed to read a fraction and place it in an array. If the user enters '0' the function is supposed to exit. I am trying to do this using the cin.peek() function but execution always goes into the if statement and doesn't allow the user to exit.

How should I properly code this (I am open to not using peek(), I thought it was the simplest way of doing it.)

Thanks!

void enterFrac(Fraction* fracs[], int& index)
    {
        int n, d;
        char c, slash;
        cout << "Enter fractions (end by entering a 0): ";
        c = cin.peek();

        if ( c != '0')
        {
            cin >> n >> slash >> d;
            Fraction* f = new Fraction();
            f->num = n;
            f->den = d;
            fracs[index] = f;
            index++;
        }
    }

This test of peek() works however:

#include <iostream>
using namespace std;

int main () {
  char c;
  int n;
  char str[256];

  cout << "Enter a number or a word: ";
  c=cin.peek();

  if ( (c >= '0') && (c <= '9') )
  {
    cin >> n;
    cout << "You have entered number " << n << endl;
  }
  else
  {
    cin >> str;
    cout << " You have entered word " << str << endl;
  }

  return 0;
}

Answer 1

您使用以下两个问题std::istream::peek():

1. 这个函数访问下一个字符,也不会跳过前导空白.您可能希望在确定下一个字符之前跳过前导空格,例如,使用操纵器std::ws:(std::cin >> std::ws).peek().
2. 结果std::istream::peek()不是char.相反,它是一个std::char_traits<char>::int_type(这是一个奇特的拼写int).结果可能是,例如,std::char_traits<char>::eof()并且如果值'0'恰好为负(我不知道它的任何平台;但是,例如,我名字中的有趣字符'ü'在char签名的平台上是负值)你也不会得到正确的结果.也就是说,您通常将结果std::istream::peek()与结果进行比较std::char_traits<char>::to_int_type(),即您使用以下内容:std::cin.peek() == std::char_traits<char>::to_int_type('0')

也就是说,你的程序不会检查它是否能成功读取以斜线分隔的分母和分母.你总是想验证阅读是否成功,例如,使用类似的东西

if ((std::cin >> nominator >> slash >> denominator) && slash == '/') {
    ...
}

只是为了娱乐,你可以创建一个操纵器来测试角色是斜线,的确如下:

std::istream& slash(std::istream& in) {
    if ((in >> std::ws).peek() != std::char_traits<char>::to_int_type('/')) {
        in.setstate(std::ios_base::failbit);
    }
    return in;
}

这样,您就可以封装斜杠测试.如果你需要在多个地方使用它,这是非常方便的.