如何处理 LL(1) 解析器中的运算符优先级
How to handle operator precedence in an LL(1) parser
我正在为表达式语法编写 LL(1) 解析器。我有以下语法:
E -> E + E
E -> E - E
E -> E * E
E -> E / E
E -> INT
然而,这是左递归的,我用以下语法删除了左递归:
E -> INT E'
E' -> + INT E'
E' -> - INT E'
E' -> * INT E'
E' -> / INT E'
E' -> ε
如果我要使用表达式 1 + 2 * 3
,解析器如何知道在加法之前计算乘法?
试试这个:
; an expression is an addition
E -> ADD
; an addition is a multiplication that is optionally followed
; by +- and another addition
ADD -> MUL T
T -> PM ADD
T -> ε
PM -> +
PM -> -
; a multiplication is an integer that is optionally followed
; by */ and another multiplication
MUL -> INT G
G -> MD MUL
G -> ε
MD -> *
MD -> /
; an integer is a digit that is optionally followed by an integer
INT -> DIGIT J
J -> INT
J -> ε
; digits
DIGIT -> 0
DIGIT -> 1
DIGIT -> 2
DIGIT -> 3
DIGIT -> 4
DIGIT -> 5
DIGIT -> 6
DIGIT -> 7
DIGIT -> 8
DIGIT -> 9
我正在为表达式语法编写 LL(1) 解析器。我有以下语法:
E -> E + E
E -> E - E
E -> E * E
E -> E / E
E -> INT
然而,这是左递归的,我用以下语法删除了左递归:
E -> INT E'
E' -> + INT E'
E' -> - INT E'
E' -> * INT E'
E' -> / INT E'
E' -> ε
如果我要使用表达式 1 + 2 * 3
,解析器如何知道在加法之前计算乘法?
试试这个:
; an expression is an addition
E -> ADD
; an addition is a multiplication that is optionally followed
; by +- and another addition
ADD -> MUL T
T -> PM ADD
T -> ε
PM -> +
PM -> -
; a multiplication is an integer that is optionally followed
; by */ and another multiplication
MUL -> INT G
G -> MD MUL
G -> ε
MD -> *
MD -> /
; an integer is a digit that is optionally followed by an integer
INT -> DIGIT J
J -> INT
J -> ε
; digits
DIGIT -> 0
DIGIT -> 1
DIGIT -> 2
DIGIT -> 3
DIGIT -> 4
DIGIT -> 5
DIGIT -> 6
DIGIT -> 7
DIGIT -> 8
DIGIT -> 9