Is there a better way to check Python code for a syntax errors without external modules?

Question

As the title says, is there a better way to check a Python source for syntax errors without the use of external modules?

I mean, in sense of a more Pythonic style or a more performant way.

def CheckSyntax(source, raw = False):
    lines = source.count("\n")
    source += "\nThis is a SyntaxError"  # add a syntax error to source, it shouldn't be executed at all
    try:
        exec source in {}, {}
    except SyntaxError, e:
        if e.lineno != lines + 2:
            if raw:
                return e
            else:
                return e.lineno, e.offset, e.text

EDIT: Ideally it would be performant enough for real-time syntax checking.

Well, it's an external module, but could be an option. Thanks — Niklas R, Jul 29 '11 at 12:53
Do you want to verify if your programs are syntactically correct before running them or are you writing some kind of validation tool? — brandizzi, Jul 29 '11 at 13:11
It's some sort of development environment which should have realtime syntax-checking implemented. — Niklas R, Jul 29 '11 at 13:17

score 6 · Accepted Answer · answered Jul 29 '11 at 12:36

6

exec doesn't sound like a particularly good idea. What if the script in question has side effects (e.g. creates or modifies files) or takes a long time to run?

Take a look at the compile function and the parser module.

answered Jul 29 '11 at 12:36

NPE

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In the function, I intentionally add a SyntaxError to the source so only thy parser will run over the source when executing, and if there are no SyntaxErrors in the original source, it will raise one at the end for sure. – Niklas R Jul 29 '11 at 12:40
Using the compile - function is not as fast as the alternative way above. (Just some milliseconds ..) – Niklas R Jul 29 '11 at 12:53

Is there a better way to check Python code for a syntax errors without external modules?

1 Answers1