如何使用 ANTLR 获取 Python 中每个函数的 ast
How to get ast for every function in Python using ANTLR
我已经有了 Python 语法。我想为 Python 中定义的每个函数生成一个单独的 AST。这是我的代码:
public class TestGrammar extends Python3BaseListener {
public static void main(String[] args) throws IOException {
PlagiarismPercentage obj = new PlagiarismPercentage();
String source = obj
.readFile("C:\Users\Paridhi\quickSort.py");
ANTLRInputStream inputCharStream = new ANTLRInputStream(new StringReader(source));
TokenSource lexer = new Python3Lexer(inputCharStream);
// Python3Lexer lexer = new Python3Lexer(CharStreams.fromString(source));
Python3Parser parser = new Python3Parser(new CommonTokenStream(lexer));
ParseTreeWalker.DEFAULT.walk(new Python3BaseListener() {
AstPrinter astPrinter = new AstPrinter();
@Override
public void enterFuncdef(Python3Parser.FuncdefContext ctx) {
System.out.printf("NAME=%s\n", ctx.NAME().getText());
// System.out.println(ctx.toString());
TestGrammar grammar = new TestGrammar();
StringBuilder str = grammar.explore(ctx, false, 0, new StringBuilder(""));
System.out.println("----------------------------");
System.out.println(str);
System.out.println("----------------------------");
}
}, parser.single_input());
}
private StringBuilder explore(RuleContext ctx, boolean verbose, int indentation, StringBuilder str) {
boolean toBeIgnored = !verbose && ctx.getChildCount() == 1 && ctx.getChild(0) instanceof ParserRuleContext;
if (!toBeIgnored) {
String ruleName = Python3Parser.ruleNames[ctx.getRuleIndex()];
for (int i = 0; i < indentation; i++) {
System.out.print(" ");
}
System.out.println(ruleName + " " + ctx.getText());
if (indentation != 0) {
str.append(ruleName + " ");
}
}
for (int i = 0; i < ctx.getChildCount(); i++) {
ParseTree element = ctx.getChild(i);
if (element instanceof RuleContext) {
explore((RuleContext) element, verbose, indentation + (toBeIgnored ? 0 : 1), str);
}
}
return str;
}
}
我的Python文件快速排序代码如下:
def quickSort(alist):
quickSortHelper(alist,0,len(alist)-1)
def quickSortHelper(alist,first,last):
if first<last:
splitpoint = partition(alist,first,last)
quickSortHelper(alist,first,splitpoint-1)
quickSortHelper(alist,splitpoint+1,last)
def partition(alist,first,last):
pivotvalue = alist[first]
leftmark = first+1
rightmark = last
done = False
while not done:
while leftmark <= rightmark and alist[leftmark] <= pivotvalue:
leftmark = leftmark + 1
while alist[rightmark] >= pivotvalue and rightmark >= leftmark:
rightmark = rightmark -1
if rightmark < leftmark:
done = True
else:
temp = alist[leftmark]
alist[leftmark] = alist[rightmark]
alist[rightmark] = temp
temp = alist[first]
alist[first] = alist[rightmark]
alist[rightmark] = temp
return rightmark
alist = [54,26,93,17,77,31,44,55,20]
quickSort(alist)
print(alist)
现在,树遍历器只是输出第一个快速排序函数的详细信息,
但是,我想要 python 代码中所有函数定义的详细信息。
我可以在 java 中执行类似 getAllMethods() 的操作吗?
您在 main
方法中实例化一个 java.util.List
,然后在 enterFuncdef
方法中向其添加项目。
我已经有了 Python 语法。我想为 Python 中定义的每个函数生成一个单独的 AST。这是我的代码:
public class TestGrammar extends Python3BaseListener {
public static void main(String[] args) throws IOException {
PlagiarismPercentage obj = new PlagiarismPercentage();
String source = obj
.readFile("C:\Users\Paridhi\quickSort.py");
ANTLRInputStream inputCharStream = new ANTLRInputStream(new StringReader(source));
TokenSource lexer = new Python3Lexer(inputCharStream);
// Python3Lexer lexer = new Python3Lexer(CharStreams.fromString(source));
Python3Parser parser = new Python3Parser(new CommonTokenStream(lexer));
ParseTreeWalker.DEFAULT.walk(new Python3BaseListener() {
AstPrinter astPrinter = new AstPrinter();
@Override
public void enterFuncdef(Python3Parser.FuncdefContext ctx) {
System.out.printf("NAME=%s\n", ctx.NAME().getText());
// System.out.println(ctx.toString());
TestGrammar grammar = new TestGrammar();
StringBuilder str = grammar.explore(ctx, false, 0, new StringBuilder(""));
System.out.println("----------------------------");
System.out.println(str);
System.out.println("----------------------------");
}
}, parser.single_input());
}
private StringBuilder explore(RuleContext ctx, boolean verbose, int indentation, StringBuilder str) {
boolean toBeIgnored = !verbose && ctx.getChildCount() == 1 && ctx.getChild(0) instanceof ParserRuleContext;
if (!toBeIgnored) {
String ruleName = Python3Parser.ruleNames[ctx.getRuleIndex()];
for (int i = 0; i < indentation; i++) {
System.out.print(" ");
}
System.out.println(ruleName + " " + ctx.getText());
if (indentation != 0) {
str.append(ruleName + " ");
}
}
for (int i = 0; i < ctx.getChildCount(); i++) {
ParseTree element = ctx.getChild(i);
if (element instanceof RuleContext) {
explore((RuleContext) element, verbose, indentation + (toBeIgnored ? 0 : 1), str);
}
}
return str;
}
}
我的Python文件快速排序代码如下:
def quickSort(alist):
quickSortHelper(alist,0,len(alist)-1)
def quickSortHelper(alist,first,last):
if first<last:
splitpoint = partition(alist,first,last)
quickSortHelper(alist,first,splitpoint-1)
quickSortHelper(alist,splitpoint+1,last)
def partition(alist,first,last):
pivotvalue = alist[first]
leftmark = first+1
rightmark = last
done = False
while not done:
while leftmark <= rightmark and alist[leftmark] <= pivotvalue:
leftmark = leftmark + 1
while alist[rightmark] >= pivotvalue and rightmark >= leftmark:
rightmark = rightmark -1
if rightmark < leftmark:
done = True
else:
temp = alist[leftmark]
alist[leftmark] = alist[rightmark]
alist[rightmark] = temp
temp = alist[first]
alist[first] = alist[rightmark]
alist[rightmark] = temp
return rightmark
alist = [54,26,93,17,77,31,44,55,20]
quickSort(alist)
print(alist)
现在,树遍历器只是输出第一个快速排序函数的详细信息, 但是,我想要 python 代码中所有函数定义的详细信息。 我可以在 java 中执行类似 getAllMethods() 的操作吗?
您在 main
方法中实例化一个 java.util.List
,然后在 enterFuncdef
方法中向其添加项目。