Using multiple "*"-style arguments

[Dave Opstad]

File this one under "enhancement request" I guess, but it would sure be
nice. I wrote this line the other day, thinking it would just work:

x = struct.pack(">HhhhhhhHHHL", i, *bounds, *b[0:2], *b[6:9], len(s))

Unfortunately, it doesn't. The only valid place for an "*"-style
argument is at the end of the arguments list. But this line seems so
natural! Since struct.pack expects all its arguments separately (rather
than gathered into a list or tuple), I have to be able to accommodate
its needs, and since I had several arguments already in lists, I thought
I could get away with this notation.

Certainly I can get around this limitation by either spelling out the
arguments (i.e. b[0], b[1], b[6]...), or by writing multiple lines of
code to gather the arguments into a single list which could then have
the "*" applied. But I'd love a simpler, more compact notation, not
unlike what I've written above. Does anyone know of a current Python
idiom for unravelling list arguments in calls?

Dave

[Andrew Koenig]

"Dave Opstad" <da*********@agfamonotype.com> wrote in message
news:da*******************************@reader0901. news.uu.net...

File this one under "enhancement request" I guess, but it would sure be
nice. I wrote this line the other day, thinking it would just work:

x = struct.pack(">HhhhhhhHHHL", i, *bounds, *b[0:2], *b[6:9], len(s))

Unfortunately, it doesn't. The only valid place for an "*"-style
argument is at the end of the arguments list. But this line seems so
natural! Since struct.pack expects all its arguments separately (rather
than gathered into a list or tuple), I have to be able to accommodate
its needs, and since I had several arguments already in lists, I thought
I could get away with this notation.

Does this do what you want?

x = struct.pack((">HhhhhhhHHHL", i, *(bounds + b[0:2] + b[6:9] +
[len(s)]))

[Dave Opstad]

In article <C2***************@bgtnsc05-news.ops.worldnet.att.net>,
"Andrew Koenig" <ar*@acm.org> wrote:

Does this do what you want?

x = struct.pack((">HhhhhhhHHHL", i, *(bounds + b[0:2] + b[6:9] +
[len(s)]))

No, unfortunately, because the arguments in the bounds and b lists are
numeric, so adding them isn't the right thing.

But thanks for the suggestion!
Dave

[Dave Opstad]

In article <C2***************@bgtnsc05-news.ops.worldnet.att.net>,
"Andrew Koenig" <ar*@acm.org> wrote:

Does this do what you want?

x = struct.pack((">HhhhhhhHHHL", i, *(bounds + b[0:2] + b[6:9] +
[len(s)]))

Wait, ignore my previous reply. Yes, I think this does work, and it fits
within the Python idiom I was seeking.

Thanks!
Dave

[Andrew Koenig]

> > Does this do what you want?

x = struct.pack((">HhhhhhhHHHL", i, *(bounds + b[0:2] + b[6:9] +
[len(s)]))

No, unfortunately, because the arguments in the bounds and b lists are
numeric, so adding them isn't the right thing.

Have you tried it? b[0:2] is a list, so + means concatenation, not
addition.

[John Roth]

"Dave Opstad" <da*********@agfamonotype.com> wrote in message
news:da*******************************@reader0901. news.uu.net...

File this one under "enhancement request" I guess, but it would sure be
nice. I wrote this line the other day, thinking it would just work:

x = struct.pack(">HhhhhhhHHHL", i, *bounds, *b[0:2], *b[6:9], len(s))

Unfortunately, it doesn't. The only valid place for an "*"-style
argument is at the end of the arguments list. But this line seems so
natural! Since struct.pack expects all its arguments separately (rather
than gathered into a list or tuple), I have to be able to accommodate
its needs, and since I had several arguments already in lists, I thought
I could get away with this notation.

Certainly I can get around this limitation by either spelling out the
arguments (i.e. b[0], b[1], b[6]...), or by writing multiple lines of
code to gather the arguments into a single list which could then have
the "*" applied. But I'd love a simpler, more compact notation, not
unlike what I've written above. Does anyone know of a current Python
idiom for unravelling list arguments in calls?
While I haven't tested it, I suspect that this might work:

x = struct.pack("....", *[i] + bounds + b[0:2] + b[6:9] + [len(s)])

John Roth
Dave