From `compile`'s doc:
    "Compile the source into a code or AST object."

The docs don't say how to compile into an AST object with `compile`, though. As it says later:
    "If you want to parse Python code into its AST representation, see ast.parse()."

I checked 3.4-3.2, 3.0, 2.7, and 2.6. Versions before 3.4, and version 2.6, are missing the `ast.parse` line, but still have the first line.

What you're looking for is in the 2nd paragraph of the ast docs:

    An abstract syntax tree can be generated by passing
    ast.PyCF_ONLY_AST as a flag to the compile() built-in
    function, or using the parse() helper provided in this
    module. The result will be a tree of objects whose
    classes all inherit from ast.AST. An abstract syntax
    tree can be compiled into a Python code object using
    the built-in compile() function.

For example:

    >>> mod = compile('42', '', 'exec', ast.PyCF_ONLY_AST)
    >>> mod
    <_ast.Module object at 0x7f0e45b15be0
    >>> ast.dump(mod)
    'Module(body=[Expr(value=Num(n=42))])'

In the discussion of `flags`, I think the compile docs should explicitly list ast.PyCF_ONLY_AST and the CO_FUTURE_* flags in a table.

> What you're looking for is in the 2nd paragraph of the ast docs:

Oh. I considered that, but then compile's docs say:

    The optional arguments flags and dont_inherit
    control which future statements (see PEP 236)
    affect the compilation of source.

We've added a reference to the compiler flags into the compile(), see issue 40484 for details.

resolution: duplicate
stage: resolved

versions: + Python 3.7, Python 3.8, Python 3.9, - Python 3.2, Python 3.3, Python 3.4