什么是scanf格式输入的cin analougus？

Question

With scanf there's, usually, a direct way to take formatted input:

1) line with a real number higher than 0, and less than 1. Ending on 'x', eg: 0.32432523x

scanf("0.%[0-9]x", &number);

2) line represents an addition in the format :30+28=fifty-eight

scanf(":%d+%d=%99s", &number1, &number2, &total);

What is the cin solution, using only the standard library?

Answer 1

You can make a simple class to verify input.

struct confirm_input {
    char const *str;

    confirm_input( char const *in ) : str( in ) {}

    friend std::istream &operator >>
        ( std::istream &s, confirm_input const &o ) {

        for ( char const *p = o.str; * p; ++ p ) {
            if ( std::isspace( * p ) ) {
                std::istream::sentry k( s ); // discard whitespace
            } else if ( (c = s.get() ) != * p ) {
                s.setstate( std::ios::failbit ); // stop extracting
            }
        }
        return s;
     }
};

usage:

std::cin >> x >> confirm_input( " = " ) >> y;

Answer 2

Use the >> operator to read from cin.

  int number1, number2;
  std::string text;
  char plus, equals;
  std::cin >> number1 >> plus >> number2 >> equals >> text;
  if (!std::cin.fail() && plus == '+' && equals == '=' && !text.empty())
    std::cout << "matched";

It's not as good as scanf because you'd have to verify any literals that were in the scanf string yourself. Doing it with streams will almost certainly be a lot more lines of code than scanf.

I would use scanf.