问题
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')
Sample input:
10+20+30.0;
10+20+30.0
Answer shall maintain parse tree property that nodes contain 2/3 children:
回答1:
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 ::= ';'
test input:
1!
3*2+1
3*2+1+3.0!
3*2+1 + 3.0!
1+1+(1+1)!
来源:https://stackoverflow.com/questions/24688484/re-write-parsing-expression-grammar-peg-without-left-recursion