Calling a function is faster than not calling it?

[Peter Otten]

Steven D'Aprano wrote:

> On Sun, 10 May 2015 08:43 pm, Peter Otten wrote:
> 
>> A significant part of the extra time is apparently spent on stack
>> inspection:
> 
> I don't know what you mean by "stack inspection", or how you come to that
> conclusion.
> 
> 
>> $ python3 -m timeit -s 'f = (lambda: 42); code = f.__code__; ns = {}'
>> 'f()' 10000000 loops, best of 3: 0.179 usec per loop
>> 
>> $ python3 -m timeit -s 'f = (lambda: 42); code = f.__code__; ns = {}'
>> 'eval(code)'
>> 1000000 loops, best of 3: 0.852 usec per loop
>> 
>> $ python3 -m timeit -s 'f = (lambda: 42); code = f.__code__; ns = {}'
>> 'eval(code, ns)'
>> 1000000 loops, best of 3: 0.433 usec per loop
> 
> Curious.
> 
> If I'm reading that correctly, supplying an explicit namespace cuts the
> time by a factor of two. That surprises me, as your function doesn't do
> any name lookups in the body of the function, so I would have thought that
> would be irrelevant.

I had a quick look at the implementation in bltinmodule.c, and the only 
piece of code that I could prevent from being run was:

    if (globals == Py_None) {
        globals = PyEval_GetGlobals();
        if (locals == Py_None) {
            locals = PyEval_GetLocals();
            if (locals == NULL)
                return NULL;
        }
    }

When there was an actual speed-up I also had a look at 
PyEval_GetGlobals/Locals() which in turn call 

PyEval_GetFrame() 

and

PyEvalPyFrame_FastToLocalsWithError()

whatever these do. (The first function reminded me of sys._getframe() hence 
the mention of stack inspection)

>> $ python3 -m timeit -s 'f = (lambda: 42); code = f.__code__; ns = {}'
>> 'eval; ns; f()'
>> 1000000 loops, best of 3: 0.263 usec per loop
> 
> So, roughly speaking, calling eval(code, ns) takes about 1.6 times as long
> as calling the function; calling eval(code) without providing a namespace
> takes about 3.2 times as long. That's roughly consistent with the results
> I'm getting.