break outside of loop not caught

[pyflakes-bot]

Original report by asmeurer (@asmeurer?) on Launchpad:

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$cat test.py
def test():
    break
$pyflakes test.py
$python test.py
  File "test.py", line 2
    break
SyntaxError: 'break' outside loop

[pyflakes-bot]

Original comment by asmeurer (@asmeurer?) on Launchpad:

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Same with continue

[pyflakes-bot]

Original comment by asmeurer (@asmeurer?) on Launchpad:

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Same with return or yield outside a function.

It seems the problem is that ast.parse (or compile with ast.PyCF_ONLY_AST) ignores these errors. Apparently the fact that continue and break have to be inside a loop or that return and yield have to be inside a function are not specified in the grammar (would that make it context-sensitive?). This is clear if you look at the grammar definition at https://docs.python.org/2/library/ast.html#abstract-grammar or even the official one at http://hg.python.org/cpython/file/96ea15ee8525/Grammar/Grammar.

Is there a function that can be used to try to compile an ast object further, in order to get these syntax errors, or is the only fix to also call compile without ast.PyCF_ONLY_AST on the source code?

[pyflakes-bot]

Original comment by asmeurer (@asmeurer?) on Launchpad:

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I have a fix at #8. The problem is that these SyntaxErrors are not encoded in the ast grammar, so they are not detected when just compiling to ast. You must compile down to bytecode to catch them. The advantage here is that we can still check for other kinds of errors in this case, because the ast is still valid.

[pyflakes-bot]

Original comment by asmeurer (@asmeurer?) on Launchpad:

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#20 was merged.