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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<http://bugs.python.org/issue28858>
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