[Python-Dev] AST Optimization: Branch Elimination in Generator Functions (original) (raw)
Jeremy Hylton jeremy at alum.mit.edu
Tue May 6 17:41:09 CEST 2008
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I'm not sure I understand the problem exactly. If we have one pass converting the concrete syntax to the AST, we can mark functions as generators as part of that pass. In a later pass, you can remove the unreachable code.
Jeremy
On Sat, May 3, 2008 at 12:51 AM, Thomas Lee <tom at vector-seven.com> wrote:
The next problem that cropped up during the implementation of the AST code optimizer is related to branch elimination and the elimination of any code after a return.
Within a FunctionDef node, we would (ideally) like to blow away If nodes with a constant - but false - test expression. e.g.: def foo(): if False: # ... stuff ... For most functions, this will cause no problems and the code will behave as expected. However, if the eliminated branch contains a "yield" expression, the function is actually a generator function - even if the yield expression can never be reached: def foo(): if False: yield 5 In addition to this, the following should also be treated as a generator even though we'd like to be able to get rid of all the code following the "return" statement: def foo(): return yield 5 Again, blowing away the yield results in a normal function instead of a generator. Not what we want: we need to preserve the generator semantics. Upon revisiting this, it's actually made me reconsider the use of a Const node for the earlier problem relating to arbitrary constants. We may be better off with annotations after all ... then we could mark FunctionDef nodes as being generators at the AST level to force the compiler to produce code for a generator, but eliminate the branches anyway. The other alternative I can think of is injecting a yield node somewhere unreachable and ensuring it doesn't get optimized away, but this seems pretty hacky in comparison. Any other ideas? Cheers, Tom
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