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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.