So nice to know this that you can compare your own code using this
methodology. I was completely unaware of this. Thank you so much.
Fredrik Lundh wrote:
> David Harvey wrote:
>
> > Suppose I write
> >
> > if x in ("abc", "def", "xyz"):
> > doStuff()
> >
> > elif x in ("pqr", "tuv", "123"):
> > doOtherStuff()
> >
> > elif ...
> > When is python building the tuples? Does it need to build the tuple
> > every time it comes through this code? Or does it somehow recognise
> > that they are constant and cache them?
>
> when in doubt, ask the compiler:
>
> def code(x):
> if x in ("abc", "def", "xyz"):
> doStuff()
> elif x in ("pqr", "tuv", "123"):
> doOtherStuff()
>
> import dis
> dis.dis(code)
>
> prints:
>
> 2 0 LOAD_FAST 0 (x)
> 3 LOAD_CONST 7 (('abc', 'def', 'xyz'))
> 6 COMPARE_OP 6 (in)
> 9 JUMP_IF_FALSE 11 (to 23)
> 12 POP_TOP
>
> 3 13 LOAD_GLOBAL 1 (doStuff)
> 16 CALL_FUNCTION 0
> 19 POP_TOP
> 20 JUMP_FORWARD 25 (to 48)
> >> 23 POP_TOP
>
> 4 24 LOAD_FAST 0 (x)
> 27 LOAD_CONST 8 (('pqr', 'tuv', '123'))
> 30 COMPARE_OP 6 (in)
> 33 JUMP_IF_FALSE 11 (to 47)
> 36 POP_TOP
>
> 5 37 LOAD_GLOBAL 2 (doOtherStuff)
> 40 CALL_FUNCTION 0
> 43 POP_TOP
> 44 JUMP_FORWARD 1 (to 48)
> >> 47 POP_TOP
> >> 48 LOAD_CONST 0 (None)
> 51 RETURN_VALUE
>
> so the answer is "yes, in this specific case".
>
> > (The tuples I have in mind are of course much longer than three
> > elements)
>
> in that case, you should probably use pre-calculated sets instead of tuples.
>
> </F>
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