/*
[The "BSD licence"]
Copyright (c) 2004 Terence Parr and Loring Craymer
All rights reserved.
Redistribution and use in source and binary forms, with or without
modification, are permitted provided that the following conditions
are met:
1. Redistributions of source code must retain the above copyright
notice, this list of conditions and the following disclaimer.
2. Redistributions in binary form must reproduce the above copyright
notice, this list of conditions and the following disclaimer in the
documentation and/or other materials provided with the distribution.
3. The name of the author may not be used to endorse or promote products
derived from this software without specific prior written permission.
THIS SOFTWARE IS PROVIDED BY THE AUTHOR ``AS IS'' AND ANY EXPRESS OR
IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE IMPLIED WARRANTIES
OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE ARE DISCLAIMED.
IN NO EVENT SHALL THE AUTHOR BE LIABLE FOR ANY DIRECT, INDIRECT,
INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL DAMAGES (INCLUDING, BUT
NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES; LOSS OF USE,
DATA, OR PROFITS; OR BUSINESS INTERRUPTION) HOWEVER CAUSED AND ON ANY
THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT
(INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT OF THE USE OF
THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF SUCH DAMAGE.
*/
import org.antlr.runtime.*;
import java.util.*;
/** Python does not explicitly provide begin and end nesting signals.
Rather, the indentation level indicates when you begin and end.
This is an interesting lexical problem because multiple DEDENT
tokens should be sent to the parser sometimes without a corresponding
input symbol! Consider the following example:
a=1
if a>1:
print a
b=3
Here the "b" token on the left edge signals that a DEDENT is needed
after the "print a \n" and before the "b". The sequence should be
... 1 COLON NEWLINE INDENT PRINT a NEWLINE DEDENT b ASSIGN 3 ...
For more examples, see the big comment at the bottom of this file.
This TokenStream normally just passes tokens through to the parser.
Upon NEWLINE token from the lexer, however, an INDENT or DEDENT token
may need to be sent to the parser. The NEWLINE is the trigger for
this class to do it's job. NEWLINE is saved and then the first token
of the next line is examined. If non-leading-whitespace token,
then check against stack for indent vs dedent. If LEADING_WS, then
the column of the next non-whitespace token will dictate indent vs
dedent. The column of the next real token is number of spaces
in the LEADING_WS token + 1 (to move past the whitespace). The
lexer grammar must set the text of the LEADING_WS token to be
the proper number of spaces (and do tab conversion etc...).
A stack of column numbers is tracked and used to detect changes
in indent level from one token to the next.
A queue of tokens is built up to hold multiple DEDENT tokens that
are generated. Before asking the lexer for another token via
nextToken(), the queue is flushed first one token at a time.
Terence Parr and Loring Craymer
February 2004
*/
public class PythonTokenSource implements TokenSource {
public static final int MAX_INDENTS = 100;
public static final int FIRST_CHAR_POSITION = 0;
/** The stack of indent levels (column numbers) */
int[] indentStack = new int[MAX_INDENTS];
/** stack pointer */
int sp=-1; // grow upwards
/** The queue of tokens */
Vector tokens = new Vector();
/** We pull real tokens from this lexer */
CommonTokenStream stream;
int lastTokenAddedIndex = -1;
public PythonTokenSource(PythonLexer lexer) {
}
public PythonTokenSource(CommonTokenStream stream) {
this.stream = stream;
// "state" of indent level is FIRST_CHAR_POSITION
push(FIRST_CHAR_POSITION);
}
public String getSourceName() {
return stream.getSourceName();
}
/** From http://www.python.org/doc/2.2.3/ref/indentation.html
"Before the first line of the file is read, a single zero is
pushed on the stack; this will never be popped off again. The
numbers pushed on the stack will always be strictly increasing
from bottom to top. At the beginning of each logical line, the
line's indentation level is compared to the top of the
stack. If it is equal, nothing happens. If it is larger, it is
pushed on the stack, and one INDENT token is generated. If it
is smaller, it must be one of the numbers occurring on the
stack; all numbers on the stack that are larger are popped
off, and for each number popped off a DEDENT token is
generated. At the end of the file, a DEDENT token is generated
for each number remaining on the stack that is larger than
zero."
I use char position in line 0..n-1 instead.
The DEDENTS possibly needed at EOF are gracefully handled by forcing
EOF to have char pos 0 even though with UNIX it's hard to get EOF
at a non left edge.
*/
public Token nextToken() {
// if something in queue, just remove and return it
if ( tokens.size()>0 ) {
Token t = (Token)tokens.firstElement();
tokens.removeElementAt(0);
// System.out.println(t);
return t;
}
insertImaginaryIndentDedentTokens();
return nextToken();
}
protected void insertImaginaryIndentDedentTokens()
{
Token t = stream.LT(1);
stream.consume();
// if not a NEWLINE, doesn't signal indent/dedent work; just enqueue
if ( t.getType()!=PythonLexer.NEWLINE ) {
List hiddenTokens = stream.getTokens(lastTokenAddedIndex+1,t.getTokenIndex()-1);
if ( hiddenTokens!=null ) {
tokens.addAll(hiddenTokens);
}
lastTokenAddedIndex = t.getTokenIndex();
tokens.addElement(t);
return;
}
// save NEWLINE in the queue
//System.out.println("found newline: "+t+" stack is "+stackString());
List hiddenTokens = stream.getTokens(lastTokenAddedIndex+1,t.getTokenIndex()-1);
if ( hiddenTokens!=null ) {
tokens.addAll(hiddenTokens);
}
lastTokenAddedIndex = t.getTokenIndex();
tokens.addElement(t);
// grab first token of next line
t = stream.LT(1);
stream.consume();
hiddenTokens = stream.getTokens(lastTokenAddedIndex+1,t.getTokenIndex()-1);
if ( hiddenTokens!=null ) {
tokens.addAll(hiddenTokens);
}
lastTokenAddedIndex = t.getTokenIndex();
// compute cpos as the char pos of next non-WS token in line
int cpos = t.getCharPositionInLine(); // column dictates indent/dedent
if ( t.getType()==Token.EOF ) {
cpos = -1; // pretend EOF always happens at left edge
}
else if ( t.getType()==PythonLexer.LEADING_WS ) {
cpos = t.getText().length();
}
//System.out.println("next token is: "+t);
// compare to last indent level
int lastIndent = peek();
//System.out.println("cpos, lastIndent = "+cpos+", "+lastIndent);
if ( cpos > lastIndent ) { // they indented; track and gen INDENT
push(cpos);
//System.out.println("push("+cpos+"): "+stackString());
Token indent = new ClassicToken(PythonParser.INDENT,"");
indent.setCharPositionInLine(t.getCharPositionInLine());
indent.setLine(t.getLine());
tokens.addElement(indent);
}
else if ( cpos < lastIndent ) { // they dedented
// how far back did we dedent?
int prevIndex = findPreviousIndent(cpos);
//System.out.println("dedented; prevIndex of cpos="+cpos+" is "+prevIndex);
// generate DEDENTs for each indent level we backed up over
for (int d=sp-1; d>=prevIndex; d--) {
Token dedent = new ClassicToken(PythonParser.DEDENT,"");
dedent.setCharPositionInLine(t.getCharPositionInLine());
dedent.setLine(t.getLine());
tokens.addElement(dedent);
}
sp = prevIndex; // pop those off indent level
}
if ( t.getType()!=PythonLexer.LEADING_WS ) { // discard WS
tokens.addElement(t);
}
}
// T O K E N S T A C K M E T H O D S
protected void push(int i) {
if (sp>=MAX_INDENTS) {
throw new IllegalStateException("stack overflow");
}
sp++;
indentStack[sp] = i;
}
protected int pop() {
if (sp<0) {
throw new IllegalStateException("stack underflow");
}
int top = indentStack[sp];
sp--;
return top;
}
protected int peek() {
return indentStack[sp];
}
/** Return the index on stack of previous indent level == i else -1 */
protected int findPreviousIndent(int i) {
for (int j=sp-1; j>=0; j--) {
if ( indentStack[j]==i ) {
return j;
}
}
return FIRST_CHAR_POSITION;
}
public String stackString() {
StringBuffer buf = new StringBuffer();
for (int j=sp; j>=0; j--) {
buf.append(" ");
buf.append(indentStack[j]);
}
return buf.toString();
}
}
/* More example input / output pairs with code simplified to single chars
------- t1 -------
a a
b b
c
d
a a \n INDENT b b \n c \n DEDENT d \n EOF
------- t2 -------
a c
b
c
a c \n INDENT b \n DEDENT c \n EOF
------- t3 -------
a
b
c
d
a \n INDENT b \n INDENT c \n DEDENT DEDENT d \n EOF
------- t4 -------
a
c
d
e
f
g
h
i
j
k
a \n INDENT c \n INDENT d \n DEDENT e \n f \n INDENT g \n h \n i \n INDENT j \n DEDENT DEDENT k \n DEDENT EOF
------- t5 -------
a
b
c
d
e
a \n INDENT b \n c \n INDENT d \n e \n DEDENT DEDENT EOF
*/