2015-04-28 16:51:06 +0100, Pádraig Brady:
[...]
> > POSIX is already clear that anyone parsing for literal tabs is broken
> > when trying to parse du output. The only safe way to parse du output is
> > to break on all whitespace (the way awk already does). I'm 70-30 in
> > favor of changing to sp
On 28/04/15 15:42, Eric Blake wrote:
> On 04/28/2015 03:59 AM, Pádraig Brady wrote:
>
The output from du shall consist of the amount of space allocated
to a file and the name of the file, in the following format:
"%d %s\n", ,
Instead, GNU du uses "%d\t%s\
On 04/28/2015 03:59 AM, Pádraig Brady wrote:
>>> The output from du shall consist of the amount of space allocated
>>> to a file and the name of the file, in the following format:
>>>
>>> "%d %s\n", ,
>>>
>>> Instead, GNU du uses "%d\t%s\n", i.e., a tab character as delimiter,
>>> eve
On 28/04/15 08:44, Andreas Schwab wrote:
> Bernhard Voelker
> writes:
>
>> Reading POSIX [0] again because of bug#20442 [1], I think that
>> GNU du is violating the spec:
>>
>> The output from du shall consist of the amount of space allocated
>> to a file and the name of the file, in the
On 04/28/2015 09:44 AM, Andreas Schwab wrote:
Bernhard Voelker
writes:
Reading POSIX [0] again because of bug#20442 [1], I think that
GNU du is violating the spec:
The output from du shall consist of the amount of space allocated
to a file and the name of the file, in the following
Bernhard Voelker
writes:
> Reading POSIX [0] again because of bug#20442 [1], I think that
> GNU du is violating the spec:
>
> The output from du shall consist of the amount of space allocated
> to a file and the name of the file, in the following format:
>
> "%d %s\n", ,
>
> Instead,
Reading POSIX [0] again because of bug#20442 [1], I think that
GNU du is violating the spec:
The output from du shall consist of the amount of space allocated
to a file and the name of the file, in the following format:
"%d %s\n", ,
Instead, GNU du uses "%d\t%s\n", i.e., a tab chara
