Using #define with functions

Question

I've been reading through the Python source code for kicks the last couple days, and I came across something I didn't understand. In the "abstract.c" file I found the below code snippet. I thought I understood how it worked, but then I realized I had no clue where the *v and *w come from. Could somebody please explain what is going on in this code, and an example of how you would use it?

#define BINARY_FUNC(func, op, op_name) \
    PyObject * \
    func(PyObject *v, PyObject *w) { \
        return binary_op(v, w, NB_SLOT(op), op_name); \
}

BINARY_FUNC(PyNumber_Or, nb_or, "|")
BINARY_FUNC(PyNumber_Xor, nb_xor, "^")
BINARY_FUNC(PyNumber_And, nb_and, "&")
BINARY_FUNC(PyNumber_Lshift, nb_lshift, "<<")
BINARY_FUNC(PyNumber_Rshift, nb_rshift, ">>")
BINARY_FUNC(PyNumber_Subtract, nb_subtract, "-")
BINARY_FUNC(PyNumber_Divmod, nb_divmod, "divmod()")

They don't come from anywhere, they're just literal parts of the macro expansion. They're the names given to the parameters of the functions being defined. — Barmar
– Barmar, Commented May 15, 2013 at 17:43
It's no different than normally defining a function: int f(int x, int y){return x*y;)}. x and y are only defined for this function. To fully explain what's going on, you'd have to let us know how binary_op and NB_SLOT are defined. — jerry
– jerry, Commented May 15, 2013 at 17:49
Run it through the pre-processor and inspect its output, this might help to enlighten. — alk
– alk, Commented May 15, 2013 at 18:02

Barmar · Accepted Answer · 2013-05-15 17:52:04Z

2

v and w are not macro parameters, they're literal parts of the expansion. The macro expands into a function definition, and the function takes parameters -- they're always named v and w, and they're of type PythonObject *. For example:

BINARY_FUNC(PyNumber_Or, nb_or, "|")

expands into:

PythonObject *
PyNumber_Or(PyObject *v, PyObject *w) {
    return binary_op(v, w, NB_SLOT(nb_or), "|");
}

answered May 15, 2013 at 17:52

Barmar

789k57 gold badges555 silver badges669 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

janovak Over a year ago

Ahh, gotcha. I don't know why I didn't see that before. Thank you!

Michael Dorgan · Accepted Answer · 2013-05-15 17:51:24Z

0

#define BINARY_FUNC(func, op, op_name) \
PyObject * \
func(PyObject *v, PyObject *w) { \
    return binary_op(v, w, NB_SLOT(op), op_name); \
}

BINARY_FUNC(PyNumber_Or, nb_or, "|")

The BINARY_FUNC defination I included above expands to:

PyObject *PyNumber_Or(PyObject *v, PyObject *w) { 
    return binary_op(v, w, NB_SLOT(nb_or), "|");
}

An entire function where the u and v pointers are part of the function. Of course, there is no parenthesis safety so it would be pretty darn easy to break when passing in "bad" values...

answered May 15, 2013 at 17:51

Michael Dorgan

12.5k3 gold badges36 silver badges63 bronze badges

Collectives™ on Stack Overflow

Using #define with functions

2 Answers 2

1 Comment

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Related