Parsing boolean expression without left hand recursion

Question

I'm trying to match this

f(some_thing) == 'something else'

• f(some_thing) is a function call, which is an expression
• == is a boolean operator
• 'something else' is a string, which also is an expression

so the boolean expression should be

expression operator expression

The problem is I can't figure out how to do that without left recursion These are my rules

expression 
  = 
  bool_expression
  / function_call
  / string
  / real_number
  / integer
  / identifier

bool_expression
  = l:expression space* op:bool_operator space* r:expression 
  { return ... }

Using grammar notation, I have

O := ==|<=|>=|<|>|!=  // operators
E := B|....           // expression, many non terminals
B := EOE

Because my grammar is EOE I don't know how to use the left hand algorithm which is

A := Ab|B
transforms into
A := BA'
A':= e|bA

Where e is empty and b is a terminal

Coachwhip

Answer 1

Something like this ought to do it:

expression
 = bool_expression

bool_expression
 = add_expression "==" bool_expression
 / add_expression "!=" bool_expression
 / add_expression

add_expression
 = mult_expression "+" add_expression
 / mult_expression "-" add_expression
 / mult_expression

mult_expression
 = atom "*" mult_expression
 / atom "/" mult_expression
 / atom

atom
 = function_call 
 / string
 / real_number
 / integer
 / identifier

function_call
 = identifier "(" (expression ("," expression)*)? ")"

string
 = "'" [^']* "'"

identifier
 = [a-zA-Z_]+

integer
 = [0-9]+

real_number
 = integer "." integer?
 / "." integer

Letty

Thank you good sir, that works! I did find out before this answer how to do it, and was about to post it, but you did an awesome job :P – Coachwhip