英文:
Using antlr grammar rules to parse expression in Go
问题
基本上是从https://stackoverflow.com/questions/64842665/复制的所有内容。
我有以下语法文件:
grammar Expr;
@parser::header {
import (
"os"
)
}
@parser::members {
func eval(left int, op antlr.Token, right int) int {
if (op.GetText() == "*") {
return left * right
} else if (op.GetText() == "+") {
return left + right
} else if (op.GetText() == "-") {
return left - right
} else {
return 0
}
}
}
stat: e NEWLINE
| NEWLINE
;
e returns [int v]
: a=e op=('+'|'-') b=e {
$v = eval($a.v, $op, $b.v)
fmt.Fprintf(os.Stdout, "got args=%d %d\n", $a.v, $b.v)
}
| INT {
$v = $INT.int
fmt.Fprintf(os.Stdout, "got number=%d\n", $v)
}
;
MUL : '*';
DIV : '/';
ADD : '+';
SUB : '-';
ID : [a-zA-Z]+ ; // match identifiers
INT : [0-9]+ ; // match integers
NEWLINE:'\r'? '\n' ; // return newlines to parser (is end-statement signal)
WS : [ \t]+ -> skip ; // toss out whitespace
这是测试代码:
package main
import (
"os"
"./parser"
"github.com/antlr/antlr4/runtime/Go/antlr"
)
func calc(inputfile string) {
input, _ := antlr.NewFileStream(inputfile)// Setup the input
lexer := parser.NewExprLexer(input)// Create the Lexer
stream := antlr.NewCommonTokenStream(lexer, antlr.TokenDefaultChannel)
p := parser.NewExprParser(stream)// Create the Parser
p.Stat()
}
func main() {
calc(os.Args[1])
}
这些是我运行的命令:
java org.antlr.v4.Tool -Dlanguage=Go -o parser -no-listener Expr.g4
go build expr_t.go
上述两个程序都可以正常工作。然而,如何打印最终输出?我尝试了以下代码,但它不起作用:
stat: e NEWLINE {
$v = $e.int
fmt.Printf("=%d\n", $v)
}
| NEWLINE
;
我阅读了整个ANTLR Mega教程的部分,但没有找到有关此方面的详细信息。
英文:
Basically duplicating everything from https://stackoverflow.com/questions/64842665/
I have the following grammar file:
grammar Expr;
@parser::header {
import (
"os"
)
}
@parser::members {
func eval(left int, op antlr.Token, right int) int {
if (op.GetText() == "*") {
return left * right
} else if (op.GetText() == "+") {
return left + right
} else if (op.GetText() == "-") {
return left - right
} else {
return 0
}
}
}
stat: e NEWLINE
| NEWLINE
;
e returns [int v]
: a=e op=('+'|'-') b=e {
$v = eval($a.v, $op, $b.v)
fmt.Fprintf(os.Stdout, "got args=%d %d\n", $a.v, $b.v)
}
| INT {
$v = $INT.int
fmt.Fprintf(os.Stdout, "got number=%d\n", $v)
}
;
MUL : '*' ;
DIV : '/' ;
ADD : '+' ;
SUB : '-' ;
ID : [a-zA-Z]+ ; // match identifiers
INT : [0-9]+ ; // match integers
NEWLINE:'\r'? '\n' ; // return newlines to parser (is end-statement signal)
WS : [ \t]+ -> skip ; // toss out whitespace
And this is the test code:
package main
import (
"os"
"./parser"
"github.com/antlr/antlr4/runtime/Go/antlr"
)
func calc(inputfile string) {
input, _ := antlr.NewFileStream(inputfile)// Setup the input
lexer := parser.NewExprLexer(input)// Create the Lexer
stream := antlr.NewCommonTokenStream(lexer, antlr.TokenDefaultChannel)
p := parser.NewExprParser(stream)// Create the Parser
p.Stat()
}
func main() {
calc(os.Args[1])
}
These are the commands that I have run:
java org.antlr.v4.Tool -Dlanguage=Go -o parser -no-listener Expr.g4
go build expr_t.go
The above 2 programs work correctly. However, how to print the final output? I tried with this and it is not working:
stat: e NEWLINE {
$v = $e.int
fmt.Printf("=%d\n", $v)
}
| NEWLINE
;
I went through the whole section of The ANTLR Mega Tutorial, but wasn't able to find the details on the aspect.
答案1
得分: 1
e
规则返回一个名为v
的值:
e returns [int v]
: ...
| ...
;
所以,不要使用$e.int
,而应该使用$e.v
:
stat
: e NEWLINE { fmt.Printf("=%d\n", $e.v) }
| NEWLINE
;
但是,你也可以让入口规则stat
返回一个值:
stat returns [int v]
: a=e NEWLINE {$v = $a.v}
| NEWLINE {$v = 0}
;
然后在你的代码中,可以这样做:
func calc(inputfile string) {
input, _ := antlr.NewFileStream(inputfile)
lexer := parser.NewExprLexer(input)
stream := antlr.NewCommonTokenStream(lexer, antlr.TokenDefaultChannel)
p := parser.NewExprParser(stream)
result := p.Stat()
fmt.Fprintf(os.Stderr, "%s", result.GetText())
fmt.Printf("%d\n", result.GetV())
}
代码测试通过。
英文:
The e
rule returns a value called v
:
e returns [int v]
: ...
| ...
;
So, instead of doing $e.int
you should do $e.v
:
stat
: e NEWLINE { fmt.Printf("=%d\n", $e.v) }
| NEWLINE
;
But, you could also let the entry rule stat
return a value:
stat returns [int v]
: a=e NEWLINE {$v = $a.v}
| NEWLINE {$v = 0}
;
and then in your code, do something like this:
func calc(inputfile string) {
input, _ := antlr.NewFileStream(inputfile)
lexer := parser.NewExprLexer(input)
stream := antlr.NewCommonTokenStream(lexer, antlr.TokenDefaultChannel)
p := parser.NewExprParser(stream)
result := p.Stat()
fmt.Fprintf(os.Stderr, "%s", result.GetText())
fmt.Printf("%d\n", result.GetV())
}
Code tested fine.
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