重新编写解析表达式语法(PEG)，而无需左递归 [英] Re-write Parsing Expression Grammar (PEG) without left recursion

问题描述

使用 https://github.com/JetBrains/Grammar-Kit 如何重写没有左递归的语法?

Using https://github.com/JetBrains/Grammar-Kit how to rewrite grammar without left recursion?

grammar ::= exprs
exprs::= (sum_expr (';')?)*
private sum_expr::= sum_expr_infix | sum_expr_prefix
sum_expr_infix ::= number sum_expr_prefix


left sum_expr_prefix::= op_plus number


private op_plus ::= '+'    
number ::= float | integer
float ::= digit+ '.' digit*
integer ::= digit+
private digit ::=('0'|'1'|'2'|'3'|'4'|'5'|'6'|'7'|'8'|'9')

样本输入:

10+20+30.0;
10+20+30.0

答案应保持解析树属性，其中节点包含2/3个子元素:

Answer shall maintain parse tree property that nodes contain 2/3 children:

推荐答案

此问题朝着正确的方向发展: 解析布尔表达式而无需左手递归

this question lead in the right direction: Parsing boolean expression without left hand recursion

grammar ::= e*
e ::=  math separator?

math ::= add
add ::=
    mul op_plus math
 |  mul op_minus math
 |  mul


mul ::=
    factorial op_mul mul
  | factorial op_div mul
  | factorial

factorial ::= term op_factorial space* | term
op_factorial ::= '!'

term ::= parentheses | space* number space*
parentheses ::= '(' math ')'


op_minus ::= '-'
op_plus ::= '+'
op_div ::= '/'
op_mul ::= '*'
number ::= float | integer
float ::= (digit+'.') digit*
integer ::=digit+
digit ::= '0'|'1'|'2'|'3'|'4'|'5'|'6'|'7'|'8'|'9'
space ::= ' ' | '\t'
separator ::= ';'

测试输入:

1!
3*2+1
3*2+1+3.0!
3*2+1 + 3.0!
1+1+(1+1)!

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