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Re: bin/49177: usr.bin/uuencode is not functioning correctly, there are some problems in the source code.
The following reply was made to PR bin/49177; it has been noted by GNATS.
From: Dave Huang <khym%azeotrope.org@localhost>
Subject: Re: bin/49177: usr.bin/uuencode is not functioning correctly, there
are some problems in the source code.
Date: Sat, 6 Sep 2014 13:48:53 -0500
On Sep 6, 2014, at 10:15, skyblack%live.cn@localhost wrote:
> I've read through the code carefully, and i found the following =
> it says "usage: %s [-m] [inputfile] outputname\n", right?
> despite of the "-m" , it means if i pass two arguments to it, the =
first one will be treated as input and the second one will be treated as =
It's hard to document how a program works in a one-line usage summary.=20=
The man page should have a better explanation of what "outputname"=20
means (although in the case of uuencode, it's still not very clear,=20
unfortunately). However, it is clear that "uuencode reads inputfile (or=20=
by default the standard input) and writes an encoded version to the=20
In other words, uuencode always writes to stdout. The "outputname"=20
parameter just affects the filename in the "begin" line of the encoded=20=
file; uuencode doesn't actually write to that file. I think the purpose=20=
of "outputname" is if you are encoding stdin; you need to provide=20
uuencode a filename to store in the begin line. E.g.,
$ gzip -c /vmunix | uuencode vmunix.gz | mail khym
P.S. I do think the uuencode man page could be improved. It talks about=20=
"outputname" in the paragraph describing "uudecode", and in the=20
documentation of the options, it lists "-o outputname" for uudecode,=20
when uudecode's synopsis has "-o outputfile"
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