[issue28858] Fastcall uses more C stack

[STINNER Victor]

STINNER Victor added the comment:

When I wrote the _PyObject_CallArg1(), it looks as a cool hack:

#define _PyObject_CallArg1(func, arg) \
    _PyObject_FastCall((func), (PyObject **)&(arg), 1)

It hacks the declaration of an explicit "stack" like:

   PyObject *stack[1];
   stack[0] = arg;
   res = _PyObject_FastCall(func, stack, 1);

And I expected that the C compiler magically computes the memory address of the argument. But it seems like requesting the memory address of an argument allocates something on the C stack.

On x86_64, first function arguments are passed with CPU registers. Maybe requesting the memory address of an argument requires to allocate a local variable, copy the register into the variable, to get the address of the local variable?

So, I suggest to *remove* the _PyObject_CallArg1() macro, and use existing functions like PyObject_CallFunctionObjArgs().

What do you think Serhiy?

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Python tracker <report at bugs.python.org>
<http://bugs.python.org/issue28858>
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