Optimized bytecode in exec statement
Terry Reedy
tjreedy at udel.edu
Mon Feb 1 20:02:01 EST 2010
On 2/1/2010 6:05 PM, Hermann Lauer wrote:
> Dear All,
>
> while trying to optimize some unpack operations with self compiling
> code I wondered howto invoke the optimization in the exec statement.
> Even starting with -O or -OO yields in Python 2.5.2 the same dis.dis code, which
> is appended below.
Currently, as far as I know, -0 just removes asserts.
>
> My interest would be to reduce the LOAD_FAST ops of "s" with DUP_TOP on the stack
> (are there hash lookups behind those ?)
Load_fast is a locals array lookup. load_fast 0 (s) means "put
locals[0] on the top of the stack". The '(s)' is added for the human reader.
> and of course to optimize the integer
> arithmetics following below (removing +-0 etc).
Something + 0 is no necessarily something. It depends on the type of
something. So the interpreter will not optimize it away. What concretely
happens with int+0 is a different matter. You can remove it though, if
you want to get into byte-code hacking, but why write '+0' if you do not
want it?
> Thanks for any ideas,
> greetings
> Hermann
>
>
>>>> import dis
>>>> g=r.rra.LAST3.getter(ds=0,row=0)
>
> class fa(struct.Struct):
> def __init__(s):
> super(fa,s).__init__('d')
>
> def __call__(s,):
> return s.unpack_from(s.buf,5392+(((s.lastrow()-0)%2880)*3+0)*8)[0]
>
>>>> dis.dis(g.__call__)
> 7 0 LOAD_FAST 0 (s)
> 3 LOAD_ATTR 0 (unpack_from)
> 6 LOAD_FAST 0 (s)
> 9 LOAD_ATTR 1 (buf)
> 12 LOAD_CONST 1 (5392)
> 15 LOAD_FAST 0 (s)
> 18 LOAD_ATTR 2 (lastrow)
> 21 CALL_FUNCTION 0
> 24 LOAD_CONST 2 (0)
> 27 BINARY_SUBTRACT
> 28 LOAD_CONST 3 (2880)
> 31 BINARY_MODULO
> 32 LOAD_CONST 4 (3)
> 35 BINARY_MULTIPLY
> 36 LOAD_CONST 2 (0)
> 39 BINARY_ADD
> 40 LOAD_CONST 5 (8)
> 43 BINARY_MULTIPLY
> 44 BINARY_ADD
> 45 CALL_FUNCTION 2
> 48 LOAD_CONST 2 (0)
> 51 BINARY_SUBSCR
> 52 RETURN_VALUE
Terry Jan Reedy
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