Re: Why no @ or $ operator in C++?

ram@zedat.fu-berlin.de (Stefan Ram) writes:


  When writing a parser for exponentiation, one needs to know
  that exponentiation is right-associative, while division is
  left-associative. I just wrote this tiny parser to show how
  this might be implemented:

#include <iostream>
#include <ostream>
#include <map>
#include <functional>
#include <cmath>
#include <sstream>
#include <cstring>

::std::map< char, ::std::function< double( double, double )>> ex
{ { '^', []( double const x, double const y )->double
  { return ::std::pow( x, y ); }},
  { '*', []( double const x, double const y )->double{ return x * y; }},
  { '/', []( double const x, double const y )->double{ return x / y; }},
  { '+', []( double const x, double const y )->double{ return x + y; }},
  { '-', []( double const x, double const y )->double{ return x - y; }}};

::std::map< char, bool >left_associative
{ { '^', false }, { '*', true }, { '/', true }, { '+', true }, { '-', true }};

template< typename scanner >struct parser
{ scanner s; parser( char const * s ): s{ s } {}
  char check( char const * const op )
  { return ::std::strchr( op, static_cast< char >( s.peek() )) ?
    static_cast< char >( s.get() ): 0; }
  double numeral(){ return s.get() - '0'; }
  double prefix(){ int sign = 1;
    while( '-' == s.peek() ){ s.get(); sign *= -1; } return sign * numeral(); }
  double parse( char const * const op, double( ::parser< scanner >::*next ) () )
  { double result =( this->*next )();
    for( char sym; sym = check( op ); )result = ex[ sym ]
    ( result, left_associative[ sym ]?( this->*next )(): parse( op, next ));
   return result; };
  double power(){ return parse( "^", &::parser< scanner >::prefix ); }
  double product(){ return parse( "*/", &::parser< scanner >::power ); }
  double sum(){ return parse( "+-", &::parser< scanner >::product ); }
  double start(){ return sum(); }};

int main()
{ using p = ::parser< ::std::stringstream >;
  auto test = []( char const * s ){ ::std::cout << p{ s }.start() << '\n'; };
  test( "2^2^3/2/2" ); test( "2+2*2^2^-3" ); }

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