如何编写antlr4访问者 [英] How to write a antlr4 visitor
问题描述
我正在尝试为一个简单的antlr4语法写一个访问者-我从书中的以下示例改编而成:
I am trying to write a visitor for a simple antlr4 grammar - I am adapting from the following example from the book:
* directory tour
* example: LabeledExpr.g4, EvalVisitor.java, Calc.java
基于Java代码，我编写了以下go代码:
Based on the java code, I have written the following go code:
package main
import (
"os"
"./parser"
"github.com/antlr/antlr4/runtime/Go/antlr"
)
type evalVisitor struct {
*parser.BaseLabeledExprVisitor
}
func (v *evalVisitor) VisitAddSub(c *parser.AddSubContext) int {
left := v.Visit(c.Expr(0))
right := v.Visit(c.Expr(1))
if(c.GetOp().GetTokenType() == parser.LabeledExprParserADD) {
return left + right //error: invalid operation: left + right (operator + not defined on interface)
} else {
return left - right
}
}
func main() {
input, _ := antlr.NewFileStream(os.Args[1])
lexer := parser.NewLabeledExprLexer(input)
stream := antlr.NewCommonTokenStream(lexer, antlr.TokenDefaultChannel)
p := parser.NewLabeledExprParser(stream)
tree := p.Prog()
var visitor evalVisitor
visitor.Visit(tree)
}
我在上面显示一位访客，其他访客的写法也与此类似.我收到一些编译错误，如上面的注释所示.如何解决此错误?
I am showing one of the visitors above, the other visitors would be written similarly. I am getting some compile errors as shown in comments above. How to fix this error?
在访问者的顶层调用中似乎也有一个错误，因为当我注释掉"left + right"时，行，我遇到了SIGSEGV故障.
There also seems to be an error in the top level calling of the visitor, because when I commented out the "left+right" line, I got a SIGSEGV fault.
供您参考，我在原始的Java代码下面显示:
For your reference, I am showing below the original java code:
public Integer visitAddSub(LabeledExprParser.AddSubContext ctx) {
int left = visit(ctx.expr(0)); // get value of left subexpression
int right = visit(ctx.expr(1)); // get value of right subexpression
if ( ctx.op.getType() == LabeledExprParser.ADD ) return left + right;
return left - right; // must be SUB
}
此外，语法是这样的:
grammar LabeledExpr;
prog: stat+ ;
stat: expr NEWLINE # printExpr
| ID '=' expr NEWLINE # assign
| NEWLINE # blank
;
expr: expr op=('*'|'/') expr # MulDiv
| expr op=('+'|'-') expr # AddSub
| INT # int
| ID # id
| '(' expr ')' # parens
;
MUL : '*' ; // assigns token name to '*' used above in grammar
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
注意:我搜索了一个示例访问者代码，但是在 54992660 上遇到了一些负面评论，并且该评论也已发布在antlr问题上.该问题的答案不完整，无法编译.那么，访客在antlr4的Go目标中是否工作?并为此提供了示例代码吗?
NOTE: I searched around for a sample visitor code, but I hit some negative comments at 54992660, and that is also posted on antlr issues. That question has an answer that is incomplete and does not compile. So, do visitors at all work in the Go target of antlr4? And is there a sample code available for that?
推荐答案
我用Google搜索了一下，然后一起入侵了以下Go访问者:
I Googled a bit, and hacked the following Go visitor together:
package antlr4demo
import (
"strconv"
"github.com/antlr/antlr4/runtime/Go/antlr"
)
type EvalVisitor struct {
BaseExpressionVisitor
Results map[int]float64
}
func (v *EvalVisitor) Visit(tree antlr.ParseTree) float64 {
switch val := tree.(type) {
case *ParseContext:
return v.VisitParse(val)
case *MultDivExprContext:
return v.VisitMultDivExpr(val)
case *NumberExprContext:
return v.VisitNumberExpr(val)
case *PlusSubExprContext:
return v.VisitPlusSubExpr(val)
case *NestedExprContext:
return v.VisitNestedExpr(val)
case *UnaryExprContext:
return v.VisitUnaryExpr(val)
default:
panic("Unknown context")
}
}
func (v *EvalVisitor) VisitParse(ctx *ParseContext) float64 {
for index, expr := range ctx.expr_list {
v.Results[index] = v.Visit(expr)
}
return v.Results[len(v.Results)-1]
}
func (v *EvalVisitor) VisitMultDivExpr(ctx *MultDivExprContext) float64 {
lhs := v.Visit(ctx.lhs)
rhs := v.Visit(ctx.rhs)
if ctx.op.GetTokenType() == ExpressionLexerMULT {
return lhs * rhs
} else {
return lhs / rhs
}
}
func (v *EvalVisitor) VisitPlusSubExpr(ctx *PlusSubExprContext) float64 {
lhs := v.Visit(ctx.lhs)
rhs := v.Visit(ctx.rhs)
if ctx.op.GetTokenType() == ExpressionLexerPLUS {
return lhs + rhs
} else {
return lhs - rhs
}
}
func (v *EvalVisitor) VisitNumberExpr(ctx *NumberExprContext) float64 {
val, _ := strconv.ParseFloat(ctx.NUMBER().GetText(), 10)
return val
}
func (v *EvalVisitor) VisitNestedExpr(ctx *NestedExprContext) float64 {
return v.Visit(ctx.Expr())
}
func (v *EvalVisitor) VisitUnaryExpr(ctx *UnaryExprContext) float64 {
return -v.Visit(ctx.Expr())
}
文件:./Expression.g4
grammar Expression;
parse
: expr_list+=expr+ EOF
;
expr
: '(' expr ')' #NestedExpr
| SUB expr #UnaryExpr
| lhs=expr op=( MULT | DIV ) rhs=expr #MultDivExpr
| lhs=expr op=( PLUS | SUB ) rhs=expr #PlusSubExpr
| NUMBER #NumberExpr
;
MULT : '*';
DIV : '/';
PLUS : '+';
SUB : '-';
NUMBER
: ( D* '.' )? D+
;
SPACES
: [ \t\r\n] -> skip
;
fragment D : [0-9];
首先下载ANTLR 4.9 JAR，生成解析器和visitor Go文件，然后将它们移至 antlr4demo 文件夹:
First download the ANTLR 4.9 JAR, generate the parser and visitor Go files and move them to the antlr4demo folder:
wget https://www.antlr.org/download/antlr-4.9-complete.jar
java -cp antlr-4.9-complete.jar org.antlr.v4.Tool -Dlanguage=Go -o antlr4demo -package antlr4demo -visitor -no-listener Expression.g4
如果您现在运行以下Go脚本:
If you now run the following Go script:
package main
import (
"fmt"
"./antlr4demo"
"github.com/antlr/antlr4/runtime/Go/antlr"
)
func main() {
expression := "1000 25/5 (1 + 2) * -3.14159265"
input := antlr.NewInputStream(expression)
lexer := antlr4demo.NewExpressionLexer(input)
stream := antlr.NewCommonTokenStream(lexer, 0)
parser := antlr4demo.NewExpressionParser(stream)
parser.BuildParseTrees = true
tree := parser.Parse()
visitor := antlr4demo.EvalVisitor{
Results: make(map[int]float64),
}
var result = visitor.Visit(tree)
fmt.Println(expression, "=", result)
fmt.Println("All results: ", visitor.Results)
}
您将看到输出:
$ go run main.go
1000 25/5 (1 + 2) * -3.14159265 = -9.424777950000001
All results: map[0:1000 1:5 2:-9.424777950000001]
请注意，我从未在Go中编写任何程序:我确定代码是一团糟，但是，嘿，行之有效".
Note that I have never programed anything in Go: I'm sure the code is a mess, but hey, "it works".
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