[pypy-dev] call_function

Ben.Young at risk.sungard.com Ben.Young at risk.sungard.com
Fri Dec 16 15:23:56 CET 2005

Carl Friedrich Bolz <cfbolz at gmx.de> wrote on 16/12/2005 14:08:38:

> Hi Ben!
> > I was having a quick look at how function calling is implemented in 
> > and I can't seem to understand why the CALL_FUNCTION opcode and the 
> > call_function method seem to do duplicate things. 
> > 
> > Whay doesn't CALL_FUNCTION just pop oparg items off the stack and call 
> > space call_function? It doesn't seem like the functionality would be 
> > different as the varargs cases are handled differently anyway?
> Well, the way we handle function arguments is definitively a bit
> convoluted. But for this case what you plan would not suffice: the oparg
> argument to the CALL_FUNCTION bytecode is not just the number of
> arguments of that function. Quoting from the CPython docs
> (http://python.org/doc/2.4.2/lib/bytecodes.html):
>      Calls a function. The low byte of argc indicates the number of
> positional parameters, the high byte the number of keyword parameters.
> On the stack, the opcode finds the keyword parameters first. For each
> keyword argument, the value is on top of the key. Below the keyword
> parameters, the positional parameters are on the stack, with the
> right-most parameter on top. Below the parameters, the function object
> to call is on the stack.
> That is why the fast paths in our CALL_FUNCTION implementation work: the
> test oparg == 1 tests that the function has exactly one positional and
> no keyword argument.

Well, you always check the high byte is zero and then call down. It just 
seems like then fact that there are 4 fastcall "hacks" doesn't need to be 
known at this level (as it is already known at the level below (i.e in the 

Thanks for the reply.


> Cheers,
> Carl Friedrich

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