Hi Richard, On Sat, Aug 28, 2004 at 10:22:31PM +0100, Richard Emslie wrote:
Also BTW you can't overload c++ return types - unless you know something I don't.
That's right, but it's not necessary. The C version (genc.h) has a type code for return type too (e.g. OP_ADD_iii) but that's only so that typer.py can know what the return type will be, and convert it if needed. In general I don't think we need to overload on the return type. If we define only: Int op_add(const Int&, const Int&) Ref op_add(const Ref&, const Ref&) and then do: someref = op_add(someint, someint) then the first signature will be choosen and the return type will be automatically converted from Int to Ref, which is fine. Armin
Hi Armin, On Mon, 30 Aug 2004, Armin Rigo wrote:
I get it (partially). :-). Either we only support ops which explicitly match the signature and convert the return type via a temporary or not-so-obviously let c++ create a temporary when passing in a const c++ reference (c++ : why??? if it wasn't reference then fair enuf... but *sigh*). """ Int i; Ref r; r = op_add(i, i); // calls Int op_add(const Int&, const Int&) // converts return type :-) r = op_add(i, r); // i converted to Ref as temporary :-( """ but I assume the latter is not what you mean. Actually, to make things more explicit - I would prefer to only have an assignment operator for Ref: class Ref... { ... Ref & operator= (const Int &ref) { ... return *this; } } Cheers, Richard
Hello Richard, On Mon, Aug 30, 2004 at 05:36:50PM +0100, Richard Emslie wrote:
No: it *is* exactly what I mean. There is no other way to do that with the C API of CPython. To add an int i and a PyObject* r you have no choice but do: tmp = PyInt_FromLong(i); res = PyNumber_Add(tmp, r); Py_DECREF(tmp); This is exactly what the C++ compiler will generate with automatic conversions. Armin
Hi Armin, On Mon, 30 Aug 2004, Armin Rigo wrote:
I get it (partially). :-). Either we only support ops which explicitly match the signature and convert the return type via a temporary or not-so-obviously let c++ create a temporary when passing in a const c++ reference (c++ : why??? if it wasn't reference then fair enuf... but *sigh*). """ Int i; Ref r; r = op_add(i, i); // calls Int op_add(const Int&, const Int&) // converts return type :-) r = op_add(i, r); // i converted to Ref as temporary :-( """ but I assume the latter is not what you mean. Actually, to make things more explicit - I would prefer to only have an assignment operator for Ref: class Ref... { ... Ref & operator= (const Int &ref) { ... return *this; } } Cheers, Richard
Hello Richard, On Mon, Aug 30, 2004 at 05:36:50PM +0100, Richard Emslie wrote:
No: it *is* exactly what I mean. There is no other way to do that with the C API of CPython. To add an int i and a PyObject* r you have no choice but do: tmp = PyInt_FromLong(i); res = PyNumber_Add(tmp, r); Py_DECREF(tmp); This is exactly what the C++ compiler will generate with automatic conversions. Armin
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Armin Rigo -
Richard Emslie