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Groups > comp.lang.python > #69254
| From | Peter Otten <__peter__@web.de> |
|---|---|
| Subject | Re: Python language hack for C-style programmers [DO NOT USE!] :-) |
| Date | 2014-03-28 08:46 +0100 |
| Organization | None |
| References | <20140327110856.14991bb0@bigbox.christie.dr> <CAPTjJmoSEiBEO9fXfwM7mGywty+dT8erBqeu2mpyrxAf12udcw@mail.gmail.com> <lh2jsg$cqu$1@ger.gmane.org> <CAPTjJmoU-Aa_Y2DDHzCyzMNUfNnuV2i2gsHWZP3gKBhzrXjLzw@mail.gmail.com> |
| Newsgroups | comp.lang.python |
| Message-ID | <mailman.8658.1395992809.18130.python-list@python.org> (permalink) |
Chris Angelico wrote:
> On Fri, Mar 28, 2014 at 12:44 PM, Dave Angel <davea@davea.name> wrote:
>>> if (array m = Regexp.split2(some_pattern, some_string))
>>> do_something(m);
>>>
>>
>> I don't know for certain about if, but you can declare (in C++) a
>> new variable in for, which is a superset of if. Scope ends when
>> the for does.
>
> Yeah, but only a for, AFAIK. Not an if or a while.
Why would you guess if you can check? Just fire up the interactive
interpreter^W^W compiler:
$ cat ifdecl.c
#include <stdio.h>
int main()
{
if(int i=42)
printf("%d\n", i);
}
$ gcc ifdecl.c
ifdecl.c: In function ‘main’:
ifdecl.c:5:6: error: expected expression before ‘int’
if(int i=42)
^
ifdecl.c:6:20: error: ‘i’ undeclared (first use in this function)
printf("%d\n", i);
^
ifdecl.c:6:20: note: each undeclared identifier is reported only once for
each function it appears in
$ g++ ifdecl.c
$ ./a.out
42
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Re: Python language hack for C-style programmers [DO NOT USE!] :-) Peter Otten <__peter__@web.de> - 2014-03-28 08:46 +0100
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