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Parsing polynomials with the Pratt algorithm

· 4 min read


Using a Pratt parser I aim to parse expressions like this x^2+y^2-1, x=1, y=1 and xy, x=2, y=3. Parsing mathematical expressions it's not hard but it already contains some interesting behaviour such as associativity between operators x+y*z is equal to x+(y*z) and not (x+y)*z. For no particular reason I decided to put the variable assignments after the polynomial. The expression x^2+y^2-1, x=1, y=1 is to be interpreted as you would with this pseudo code

print (x*2+y^2+1)

For the full source code see here

The Pratt algorithm allows to create a "general purpose parser" in which is possible to plug in all the parser rules needed for your language or grammar. In my implementation I decided to build a parsed with fixed parsing rules by adding them into the constructors. But as you can see in the Java example (see references) it's easy to create a generic parser and plug the rules by subclassing it.

The important stuff happens in the ParseExpression method. The first time it is called with precedence = 0 and let's see how it works with the polynomial x+y*z:

  1. advance to the next token, which is x
  2. retrieve the corresponding prefix denotation VarDenotation. The prefix denotation cannot be null if the polynomial expressions is correct since we are at the beginning of the expression so the only valid tokens are exactly the ones associated with a prefix denotation, for example a polynomial cannot start with a '*'.
  3. use the denotation to parse the token, VarDenotation return a VariableExpr. Remember that a denotation can call ParseExpression, so here we have a kind of recursion.
  4. we check if the next token + as an higher precedence with respect of the current precedence parameters. Since the precedence of the BinaryOperatorDenotation of + is 1 we enter the while
  5. advance to +
  6. get the infix denotation
  7. use the denotation to parse the token. The parse method on an infix denotation needs the previous expression, the token and the parser to call ParseExpression again. It's easy to see the in order to parse an AddExpr we need the expression before the + and the expression after.
  8. the BinaryOperatorDenotation will call the ParseExpression method with precedence=1
  9. advance to y
  10. retrieve VarDenotation
  11. parse y to a VariableExpr
  12. PeekPrecedence() will return the precedence related to the BinaryOperatorDenotation of '*' which is 2 so we enter again the while again. This reflects the fact that * has a higher precedence of + and entering the while means that the VariableExpr of y will be a parameter of the * instead of + as one would expect.
  13. advance to *
  14. get the infix denotation
  15. the parse of the * call parse with precedence=2.
  16. (we're almost over) advance to z
  17. parse the VarDenotation
  18. PeekPrecedence() returns 0 since we are at the end of the expression
  19. we skip the while and returns the z parselect
  20. we're back in the binary operator related to * which returns the ProductExpr representing y*z
  21. we're back in the ParseExpression with precedence=1
  22. we're back in the binary operator related to + which returns the final AddExpr representing x+(y*z)
  23. we're back in the ParseExpression with precedence=0 and we return the result.
public IExpr ParseExpression(int precedence = 0)
var token = Advance();
var prefix = GetPrefixDenotation(token.Type);

if (prefix is null) throw new ParseError($"Could not parse \"{token.Lexeme}\" at column {token.Column}");

var left = prefix.Parse(this, token);

while (precedence < PeekPrecedence())
token = Advance();

var infix = GetInfixDenotation(token.Type);
left = infix.Parse(this, left, token);

return left;


I came across the Pratt parsing algorithm while reading, which is a great resource on interpreters and compilers. From the same author, a java implementation of a Pratt parser

More formal resources are the paper from Pratt in which he explains the algorithm for the first time and a thesis which is more approachable