如何解决如何解决Bison可选else表达式移位/减少错误？
我正在尝试采用受Scala启发的语法来实现玩具语言。我使用了Scala Syntax Specification的一部分：
Expr1 : ‘if’ ‘(’ Expr ‘)’ {nl} Expr [[semi] ‘else’ Expr]
| ...
在这种玩具语言中，所有内容都是一个表达式（if-else，for和while具有返回值），并用;或\n分隔。
这里是parser.y：
%code top {
#include <cstdio>
}
%union {
int n;
Ast *ast;
}
%code requires {
class Ast;
int yylex(void);
void yyerror(const char *msg);
}
%token<n> NUM
%token<n> PLUS '+'
%token<n> MINUS '-'
%token<n> TIMES '*'
%token<n> DIVIDE '/'
%token<n> SEMICOLON ';'
%token<n> NEWLINE '\n'
%token<n> IF "if"
%token<n> ELSE "else"
%token<n> LPAREN '('
%token<n> RPAREN ')'
%type<ast> prog expr primaryExpr optionalElse semi optionalSemi optionalNewlines newlines
%left PLUS MINUS
%left TIMES DIVIDE
%start prog
%%
prog : expr
;
expr : "if" '(' primaryExpr ')' optionalNewlines expr optionalElse { $$ = nullptr; }
| primaryExpr
;
optionalElse : optionalSemi "else" expr { $$ = nullptr; }
| %empty { $$ = nullptr; }
;
primaryExpr : NUM { $$ = nullptr; }
| primaryExpr '+' NUM { $$ = nullptr; }
| primaryExpr '-' NUM { $$ = nullptr; }
| primaryExpr '*' NUM { $$ = nullptr; }
| primaryExpr '/' NUM { $$ = nullptr; }
;
semi : ';' { $$ = nullptr; }
| '\n' { $$ = nullptr; }
;
optionalSemi : semi { $$ = nullptr; }
| %empty { $$ = nullptr; }
;
optionalNewlines : newlines { $$ = nullptr; }
| %empty { $$ = nullptr; }
;
newlines : '\n' { $$ = nullptr; }
| newlines '\n' { $$ = nullptr; }
;
%%
void yyerror(const char *msg) {
fprintf(stderr,"%s\n",msg);
}
这是我的语法，编译为：bison --debug --verbose -Wcounterexamples -o grammar.tab.cpp --defines=grammar.tab.h grammar.y
它给我输出：
Terminals unused in grammar
PLUS
MINUS
TIMES
DIVIDE
SEMICOLON
NEWLINE
LPAREN
RPAREN
Rules useless in parser due to conflicts
14 optionalSemi: %empty
State 21 conflicts: 2 shift/reduce,1 reduce/reduce
Grammar
0 $accept: prog $end
1 prog: expr
2 expr: "if" '(' primaryExpr ')' optionalNewlines expr optionalElse
3 | primaryExpr
4 optionalElse: optionalSemi "else" expr
5 | %empty
6 primaryExpr: NUM
7 | primaryExpr '+' NUM
8 | primaryExpr '-' NUM
9 | primaryExpr '*' NUM
10 | primaryExpr '/' NUM
11 semi: ';'
12 | '\n'
13 optionalSemi: semi
14 | %empty
15 optionalNewlines: newlines
16 | %empty
17 newlines: '\n'
18 | newlines '\n'
Terminals,with rules where they appear
$end (0) 0
'\n' <n> (10) 12 17 18
'(' <n> (40) 2
')' <n> (41) 2
'*' <n> (42) 9
'+' <n> (43) 7
'-' <n> (45) 8
'/' <n> (47) 10
';' <n> (59) 11
error (256)
NUM <n> (258) 6 7 8 9 10
PLUS <n> (259)
MINUS <n> (260)
TIMES <n> (261)
DIVIDE <n> (262)
SEMICOLON <n> (263)
NEWLINE <n> (264)
"if" <n> (265) 2
"else" <n> (266) 4
LPAREN <n> (267)
RPAREN <n> (268)
Nonterminals,with rules where they appear
$accept (22)
on left: 0
prog <ast> (23)
on left: 1
on right: 0
expr <ast> (24)
on left: 2 3
on right: 1 2 4
optionalElse <ast> (25)
on left: 4 5
on right: 2
primaryExpr <ast> (26)
on left: 6 7 8 9 10
on right: 2 3 7 8 9 10
semi <ast> (27)
on left: 11 12
on right: 13
optionalSemi <ast> (28)
on left: 13 14
on right: 4
optionalNewlines <ast> (29)
on left: 15 16
on right: 2
newlines <ast> (30)
on left: 17 18
on right: 15 18
State 0
0 $accept: • prog $end
NUM shift,and go to state 1
"if" shift,and go to state 2
prog go to state 3
expr go to state 4
primaryExpr go to state 5
State 1
6 primaryExpr: NUM •
$default reduce using rule 6 (primaryExpr)
State 2
2 expr: "if" • '(' primaryExpr ')' optionalNewlines expr optionalElse
'(' shift,and go to state 6
State 3
0 $accept: prog • $end
$end shift,and go to state 7
State 4
1 prog: expr •
$default reduce using rule 1 (prog)
State 5
3 expr: primaryExpr •
7 primaryExpr: primaryExpr • '+' NUM
8 | primaryExpr • '-' NUM
9 | primaryExpr • '*' NUM
10 | primaryExpr • '/' NUM
'+' shift,and go to state 8
'-' shift,and go to state 9
'*' shift,and go to state 10
'/' shift,and go to state 11
$default reduce using rule 3 (expr)
State 6
2 expr: "if" '(' • primaryExpr ')' optionalNewlines expr optionalElse
NUM shift,and go to state 1
primaryExpr go to state 12
State 7
0 $accept: prog $end •
$default accept
State 8
7 primaryExpr: primaryExpr '+' • NUM
NUM shift,and go to state 13
State 9
8 primaryExpr: primaryExpr '-' • NUM
NUM shift,and go to state 14
State 10
9 primaryExpr: primaryExpr '*' • NUM
NUM shift,and go to state 15
State 11
10 primaryExpr: primaryExpr '/' • NUM
NUM shift,and go to state 16
State 12
2 expr: "if" '(' primaryExpr • ')' optionalNewlines expr optionalElse
7 primaryExpr: primaryExpr • '+' NUM
8 | primaryExpr • '-' NUM
9 | primaryExpr • '*' NUM
10 | primaryExpr • '/' NUM
'+' shift,and go to state 11
')' shift,and go to state 17
State 13
7 primaryExpr: primaryExpr '+' NUM •
$default reduce using rule 7 (primaryExpr)
State 14
8 primaryExpr: primaryExpr '-' NUM •
$default reduce using rule 8 (primaryExpr)
State 15
9 primaryExpr: primaryExpr '*' NUM •
$default reduce using rule 9 (primaryExpr)
State 16
10 primaryExpr: primaryExpr '/' NUM •
$default reduce using rule 10 (primaryExpr)
State 17
2 expr: "if" '(' primaryExpr ')' • optionalNewlines expr optionalElse
'\n' shift,and go to state 18
$default reduce using rule 16 (optionalNewlines)
optionalNewlines go to state 19
newlines go to state 20
State 18
17 newlines: '\n' •
$default reduce using rule 17 (newlines)
State 19
2 expr: "if" '(' primaryExpr ')' optionalNewlines • expr optionalElse
NUM shift,and go to state 2
expr go to state 21
primaryExpr go to state 5
State 20
15 optionalNewlines: newlines •
18 newlines: newlines • '\n'
'\n' shift,and go to state 22
$default reduce using rule 15 (optionalNewlines)
State 21
2 expr: "if" '(' primaryExpr ')' optionalNewlines expr • optionalElse
';' shift,and go to state 23
'\n' shift,and go to state 24
';' [reduce using rule 5 (optionalElse)]
'\n' [reduce using rule 5 (optionalElse)]
"else" reduce using rule 5 (optionalElse)
"else" [reduce using rule 14 (optionalSemi)]
$default reduce using rule 5 (optionalElse)
optionalElse go to state 25
semi go to state 26
optionalSemi go to state 27
shift/reduce conflict on token ';':
5 optionalElse: • %empty
11 semi: • ';'
Example: "if" '(' primaryExpr ')' optionalNewlines "if" '(' primaryExpr ')' optionalNewlines expr • ';' "else" expr
Shift derivation
expr
↳ "if" '(' primaryExpr ')' optionalNewlines expr optionalElse
↳ "if" '(' primaryExpr ')' optionalNewlines expr optionalElse ↳ ε
↳ optionalSemi "else" expr
↳ semi
↳ • ';'
Reduce derivation
expr
↳ "if" '(' primaryExpr ')' optionalNewlines expr optionalElse
↳ "if" '(' primaryExpr ')' optionalNewlines expr optionalElse ↳ optionalSemi "else" expr
↳ • ↳ semi
↳ ';'
shift/reduce conflict on token '\n':
5 optionalElse: • %empty
12 semi: • '\n'
Example: "if" '(' primaryExpr ')' optionalNewlines "if" '(' primaryExpr ')' optionalNewlines expr • '\n' "else" expr
Shift derivation
expr
↳ "if" '(' primaryExpr ')' optionalNewlines expr optionalElse
↳ "if" '(' primaryExpr ')' optionalNewlines expr optionalElse ↳ ε
↳ optionalSemi "else" expr
↳ semi
↳ • '\n'
Reduce derivation
expr
↳ "if" '(' primaryExpr ')' optionalNewlines expr optionalElse
↳ "if" '(' primaryExpr ')' optionalNewlines expr optionalElse ↳ optionalSemi "else" expr
↳ • ↳ semi
↳ '\n'
reduce/reduce conflict on token "else":
5 optionalElse: • %empty
14 optionalSemi: • %empty
Example: •
First reduce derivation
optionalElse
↳ •
Second reduce derivation
optionalSemi
↳ •
State 22
18 newlines: newlines '\n' •
$default reduce using rule 18 (newlines)
State 23
11 semi: ';' •
$default reduce using rule 11 (semi)
State 24
12 semi: '\n' •
$default reduce using rule 12 (semi)
State 25
2 expr: "if" '(' primaryExpr ')' optionalNewlines expr optionalElse •
$default reduce using rule 2 (expr)
State 26
13 optionalSemi: semi •
$default reduce using rule 13 (optionalSemi)
State 27
4 optionalElse: optionalSemi • "else" expr
"else" shift,and go to state 28
State 28
4 optionalElse: optionalSemi "else" • expr
NUM shift,and go to state 2
expr go to state 29
primaryExpr go to state 5
State 29
4 optionalElse: optionalSemi "else" expr •
$default reduce using rule 4 (optionalElse)
我应该如何修复State 21 conflicts: 2 shift/reduce,1 reduce/reduce？
解决方法
目前，我使用一种棘手的方法来解决此if-else移位/减少错误。
这个想法来自两个答案：
- https://stackoverflow.com/a/12734499/4438921-使用
%prec来修复if-else移位/减少错误。 - https://stackoverflow.com/a/63000285/4438921-将2个标记合并为一个标记，以简化野牛语法。
首先，在弹性SEMICOLON_ELSE中添加令牌token.l：
%option noyywrap
%option bison-bridge
%{
#include "parser.tab.hh"
%}
%%
[\n]+ { return NEWLINE; }
";" { return SEMICOLON; }
"else"{ return ELSE; }
[\n]+[^ \t\v\f\r]*"else" { return SEMICOLON_ELSE; }
";"[^ \t\v\f\r]*"else" { return SEMICOLON_ELSE; }
%%
假设; / \n和else之间没有评论。使它们成为一个令牌。
您可以使用更复杂的正则表达式来允许它们之间的注释。
然后放入野牛parser.y：
%token<n> SEMICOLON_ELSE "semicolon_else"
%token<n> ELSE "else"
%nonassoc "then"
%nonassoc "else"
%nonassoc "semicolon_else"
%%
expr : "if" '(' primaryExpr ')' optionalNewlines expr %prec "then" { $$ = nullptr; }
| "if" '(' primaryExpr ')' optionalNewlines expr "semicolon_else" expr %prec "semicolon_else" { $$ = nullptr; }
| "if" '(' primaryExpr ')' optionalNewlines expr "else" expr %prec "else" { $$ = nullptr; }
| primaryExpr
;
%%
在这种情况下，%prec将适用于解析器！
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