Problem using xargs --max-args --replace with default delimiter
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
This works as expected:
$ echo a b c | xargs --replace="" echo x "" y
x a b c y
This does too:
$ echo a b c | xargs --max-args=1 echo x
x a
x b
x c
But this doesn't work as expected:
$ echo a b c | xargs --max-args=1 --replace="" echo x "" y
x a b c y
And neither does this:
$ echo a b c | xargs --delimiter=' ' --max-args=1 --replace="" echo x "" y
x a y
x b y
x c
y
I expected this output:
x a y
x b y
x c y
As a workaround, I am using printf and two xargs, but that is ugly:
$ echo a b c | xargs printf '%s' |
> xargs --null --max-args=1 --replace="" echo x "" y
x a y
x b y
x c y
Any idea why this is happening?
linux xargs gnu
add a comment |Â
up vote
3
down vote
favorite
This works as expected:
$ echo a b c | xargs --replace="" echo x "" y
x a b c y
This does too:
$ echo a b c | xargs --max-args=1 echo x
x a
x b
x c
But this doesn't work as expected:
$ echo a b c | xargs --max-args=1 --replace="" echo x "" y
x a b c y
And neither does this:
$ echo a b c | xargs --delimiter=' ' --max-args=1 --replace="" echo x "" y
x a y
x b y
x c
y
I expected this output:
x a y
x b y
x c y
As a workaround, I am using printf and two xargs, but that is ugly:
$ echo a b c | xargs printf '%s' |
> xargs --null --max-args=1 --replace="" echo x "" y
x a y
x b y
x c y
Any idea why this is happening?
linux xargs gnu
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
This works as expected:
$ echo a b c | xargs --replace="" echo x "" y
x a b c y
This does too:
$ echo a b c | xargs --max-args=1 echo x
x a
x b
x c
But this doesn't work as expected:
$ echo a b c | xargs --max-args=1 --replace="" echo x "" y
x a b c y
And neither does this:
$ echo a b c | xargs --delimiter=' ' --max-args=1 --replace="" echo x "" y
x a y
x b y
x c
y
I expected this output:
x a y
x b y
x c y
As a workaround, I am using printf and two xargs, but that is ugly:
$ echo a b c | xargs printf '%s' |
> xargs --null --max-args=1 --replace="" echo x "" y
x a y
x b y
x c y
Any idea why this is happening?
linux xargs gnu
This works as expected:
$ echo a b c | xargs --replace="" echo x "" y
x a b c y
This does too:
$ echo a b c | xargs --max-args=1 echo x
x a
x b
x c
But this doesn't work as expected:
$ echo a b c | xargs --max-args=1 --replace="" echo x "" y
x a b c y
And neither does this:
$ echo a b c | xargs --delimiter=' ' --max-args=1 --replace="" echo x "" y
x a y
x b y
x c
y
I expected this output:
x a y
x b y
x c y
As a workaround, I am using printf and two xargs, but that is ugly:
$ echo a b c | xargs printf '%s' |
> xargs --null --max-args=1 --replace="" echo x "" y
x a y
x b y
x c y
Any idea why this is happening?
linux xargs gnu
edited Mar 5 at 17:02
asked Mar 5 at 16:04
Pedro Gimeno
1304
1304
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
According to the POSIX documentation, xargs
should run the given utility with arguments delimited by either spaces or newlines, and this is what happens in the two first examples of yours.
However, when --replace
(or -I
) is used, only newlines will delimit arguments. The remedy is to give xargs
arguments on separate lines:
$ printf '%sn' a b c | xargs --max-args=1 --replace="" echo x "" y
x a y
x b y
x c y
Using POSIX options:
printf '%sn' a b c | xargs -n 1 -I "" echo x "" y
Here, I give xargs
not one line but three. It takes one line (at most) and executes the utility with that as the argument.
Note also that -n 1
(or --max-args=1
) in the above is not needed as it's the number of replacements made by -I
that determines the number of arguments used:
$ printf '%sn' a b c | xargs -I "" echo x "" y
x a y
x b y
x c y
In fact, the Rationale section of the POSIX spec on xargs
says (my emphasis)
The
-I
,-L
, and-n
options are mutually-exclusive. Some implementations use the last one specified if more than one is given on a command line; other implementations treat combinations of the options in different ways.
While testing this, I noticed that OpenBSD's version of xargs
will do the the following if -n
and -I
are used together:
$ echo a b c | xargs -n 1 -I "" echo x "" y
x a y
x b y
x c y
This is different from what GNU coreutils' xargs
does (which produces x a b c y
). This is due to the implementation accepting spaces as argument delimiter with -n
, even though -I
is used. So, don't use -I
and -n
together (it's not needed anyway).
1
That behaviour (that newline should be the only delimiter with -I) is documented in the GNUxargs
documentation and the POSIX specification (under XSI, so it's not surprising that OpenBSD doesn't comply there). In any case, using-n
with-I
makes little sense asis to replace one argument anyway.
â Stéphane Chazelas
Mar 5 at 16:35
@StéphaneChazelas Ah. I was looking too closely at-n
and not at-I
. Will update.
â Kusalananda
Mar 5 at 16:41
@StéphaneChazelas It seems that OpenBSD'sxargs
does the correct thing if-I
and-n
is not mixed. I rewrote that part of the answer.
â Kusalananda
Mar 5 at 16:46
Thanks for the pointer to the POSIXxargs
. The way I read it, it's nearly useless, as it says that-I
,-L
and-n
are mutually exclusive. Glad we have the GNU version. // Actuallyecho a b c
was a placeholder for any command that generates multiple space-separated parameters, hence why I used| xargs printf
instead of usingprintf
directly. // I'm still digesting your post. You've written that OpenBSDxargs
will do the wrong thing, is that a typo?
â Pedro Gimeno
Mar 5 at 17:16
1
@PedroGimeno No typo. It's the wrong output since it uses the three letters as separate arguments even though-I
is used. It does, however, do the right thing if-n
is not used. As you say,-I
and-n
are mutually exclusive, and an implementation may choose to handle this situation in any way it wants, so "wrong" is maybe the wrong word...
â Kusalananda
Mar 5 at 17:20
 |Â
show 4 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
According to the POSIX documentation, xargs
should run the given utility with arguments delimited by either spaces or newlines, and this is what happens in the two first examples of yours.
However, when --replace
(or -I
) is used, only newlines will delimit arguments. The remedy is to give xargs
arguments on separate lines:
$ printf '%sn' a b c | xargs --max-args=1 --replace="" echo x "" y
x a y
x b y
x c y
Using POSIX options:
printf '%sn' a b c | xargs -n 1 -I "" echo x "" y
Here, I give xargs
not one line but three. It takes one line (at most) and executes the utility with that as the argument.
Note also that -n 1
(or --max-args=1
) in the above is not needed as it's the number of replacements made by -I
that determines the number of arguments used:
$ printf '%sn' a b c | xargs -I "" echo x "" y
x a y
x b y
x c y
In fact, the Rationale section of the POSIX spec on xargs
says (my emphasis)
The
-I
,-L
, and-n
options are mutually-exclusive. Some implementations use the last one specified if more than one is given on a command line; other implementations treat combinations of the options in different ways.
While testing this, I noticed that OpenBSD's version of xargs
will do the the following if -n
and -I
are used together:
$ echo a b c | xargs -n 1 -I "" echo x "" y
x a y
x b y
x c y
This is different from what GNU coreutils' xargs
does (which produces x a b c y
). This is due to the implementation accepting spaces as argument delimiter with -n
, even though -I
is used. So, don't use -I
and -n
together (it's not needed anyway).
1
That behaviour (that newline should be the only delimiter with -I) is documented in the GNUxargs
documentation and the POSIX specification (under XSI, so it's not surprising that OpenBSD doesn't comply there). In any case, using-n
with-I
makes little sense asis to replace one argument anyway.
â Stéphane Chazelas
Mar 5 at 16:35
@StéphaneChazelas Ah. I was looking too closely at-n
and not at-I
. Will update.
â Kusalananda
Mar 5 at 16:41
@StéphaneChazelas It seems that OpenBSD'sxargs
does the correct thing if-I
and-n
is not mixed. I rewrote that part of the answer.
â Kusalananda
Mar 5 at 16:46
Thanks for the pointer to the POSIXxargs
. The way I read it, it's nearly useless, as it says that-I
,-L
and-n
are mutually exclusive. Glad we have the GNU version. // Actuallyecho a b c
was a placeholder for any command that generates multiple space-separated parameters, hence why I used| xargs printf
instead of usingprintf
directly. // I'm still digesting your post. You've written that OpenBSDxargs
will do the wrong thing, is that a typo?
â Pedro Gimeno
Mar 5 at 17:16
1
@PedroGimeno No typo. It's the wrong output since it uses the three letters as separate arguments even though-I
is used. It does, however, do the right thing if-n
is not used. As you say,-I
and-n
are mutually exclusive, and an implementation may choose to handle this situation in any way it wants, so "wrong" is maybe the wrong word...
â Kusalananda
Mar 5 at 17:20
 |Â
show 4 more comments
up vote
2
down vote
accepted
According to the POSIX documentation, xargs
should run the given utility with arguments delimited by either spaces or newlines, and this is what happens in the two first examples of yours.
However, when --replace
(or -I
) is used, only newlines will delimit arguments. The remedy is to give xargs
arguments on separate lines:
$ printf '%sn' a b c | xargs --max-args=1 --replace="" echo x "" y
x a y
x b y
x c y
Using POSIX options:
printf '%sn' a b c | xargs -n 1 -I "" echo x "" y
Here, I give xargs
not one line but three. It takes one line (at most) and executes the utility with that as the argument.
Note also that -n 1
(or --max-args=1
) in the above is not needed as it's the number of replacements made by -I
that determines the number of arguments used:
$ printf '%sn' a b c | xargs -I "" echo x "" y
x a y
x b y
x c y
In fact, the Rationale section of the POSIX spec on xargs
says (my emphasis)
The
-I
,-L
, and-n
options are mutually-exclusive. Some implementations use the last one specified if more than one is given on a command line; other implementations treat combinations of the options in different ways.
While testing this, I noticed that OpenBSD's version of xargs
will do the the following if -n
and -I
are used together:
$ echo a b c | xargs -n 1 -I "" echo x "" y
x a y
x b y
x c y
This is different from what GNU coreutils' xargs
does (which produces x a b c y
). This is due to the implementation accepting spaces as argument delimiter with -n
, even though -I
is used. So, don't use -I
and -n
together (it's not needed anyway).
1
That behaviour (that newline should be the only delimiter with -I) is documented in the GNUxargs
documentation and the POSIX specification (under XSI, so it's not surprising that OpenBSD doesn't comply there). In any case, using-n
with-I
makes little sense asis to replace one argument anyway.
â Stéphane Chazelas
Mar 5 at 16:35
@StéphaneChazelas Ah. I was looking too closely at-n
and not at-I
. Will update.
â Kusalananda
Mar 5 at 16:41
@StéphaneChazelas It seems that OpenBSD'sxargs
does the correct thing if-I
and-n
is not mixed. I rewrote that part of the answer.
â Kusalananda
Mar 5 at 16:46
Thanks for the pointer to the POSIXxargs
. The way I read it, it's nearly useless, as it says that-I
,-L
and-n
are mutually exclusive. Glad we have the GNU version. // Actuallyecho a b c
was a placeholder for any command that generates multiple space-separated parameters, hence why I used| xargs printf
instead of usingprintf
directly. // I'm still digesting your post. You've written that OpenBSDxargs
will do the wrong thing, is that a typo?
â Pedro Gimeno
Mar 5 at 17:16
1
@PedroGimeno No typo. It's the wrong output since it uses the three letters as separate arguments even though-I
is used. It does, however, do the right thing if-n
is not used. As you say,-I
and-n
are mutually exclusive, and an implementation may choose to handle this situation in any way it wants, so "wrong" is maybe the wrong word...
â Kusalananda
Mar 5 at 17:20
 |Â
show 4 more comments
up vote
2
down vote
accepted
up vote
2
down vote
accepted
According to the POSIX documentation, xargs
should run the given utility with arguments delimited by either spaces or newlines, and this is what happens in the two first examples of yours.
However, when --replace
(or -I
) is used, only newlines will delimit arguments. The remedy is to give xargs
arguments on separate lines:
$ printf '%sn' a b c | xargs --max-args=1 --replace="" echo x "" y
x a y
x b y
x c y
Using POSIX options:
printf '%sn' a b c | xargs -n 1 -I "" echo x "" y
Here, I give xargs
not one line but three. It takes one line (at most) and executes the utility with that as the argument.
Note also that -n 1
(or --max-args=1
) in the above is not needed as it's the number of replacements made by -I
that determines the number of arguments used:
$ printf '%sn' a b c | xargs -I "" echo x "" y
x a y
x b y
x c y
In fact, the Rationale section of the POSIX spec on xargs
says (my emphasis)
The
-I
,-L
, and-n
options are mutually-exclusive. Some implementations use the last one specified if more than one is given on a command line; other implementations treat combinations of the options in different ways.
While testing this, I noticed that OpenBSD's version of xargs
will do the the following if -n
and -I
are used together:
$ echo a b c | xargs -n 1 -I "" echo x "" y
x a y
x b y
x c y
This is different from what GNU coreutils' xargs
does (which produces x a b c y
). This is due to the implementation accepting spaces as argument delimiter with -n
, even though -I
is used. So, don't use -I
and -n
together (it's not needed anyway).
According to the POSIX documentation, xargs
should run the given utility with arguments delimited by either spaces or newlines, and this is what happens in the two first examples of yours.
However, when --replace
(or -I
) is used, only newlines will delimit arguments. The remedy is to give xargs
arguments on separate lines:
$ printf '%sn' a b c | xargs --max-args=1 --replace="" echo x "" y
x a y
x b y
x c y
Using POSIX options:
printf '%sn' a b c | xargs -n 1 -I "" echo x "" y
Here, I give xargs
not one line but three. It takes one line (at most) and executes the utility with that as the argument.
Note also that -n 1
(or --max-args=1
) in the above is not needed as it's the number of replacements made by -I
that determines the number of arguments used:
$ printf '%sn' a b c | xargs -I "" echo x "" y
x a y
x b y
x c y
In fact, the Rationale section of the POSIX spec on xargs
says (my emphasis)
The
-I
,-L
, and-n
options are mutually-exclusive. Some implementations use the last one specified if more than one is given on a command line; other implementations treat combinations of the options in different ways.
While testing this, I noticed that OpenBSD's version of xargs
will do the the following if -n
and -I
are used together:
$ echo a b c | xargs -n 1 -I "" echo x "" y
x a y
x b y
x c y
This is different from what GNU coreutils' xargs
does (which produces x a b c y
). This is due to the implementation accepting spaces as argument delimiter with -n
, even though -I
is used. So, don't use -I
and -n
together (it's not needed anyway).
edited Mar 5 at 17:23
answered Mar 5 at 16:09
Kusalananda
103k13202318
103k13202318
1
That behaviour (that newline should be the only delimiter with -I) is documented in the GNUxargs
documentation and the POSIX specification (under XSI, so it's not surprising that OpenBSD doesn't comply there). In any case, using-n
with-I
makes little sense asis to replace one argument anyway.
â Stéphane Chazelas
Mar 5 at 16:35
@StéphaneChazelas Ah. I was looking too closely at-n
and not at-I
. Will update.
â Kusalananda
Mar 5 at 16:41
@StéphaneChazelas It seems that OpenBSD'sxargs
does the correct thing if-I
and-n
is not mixed. I rewrote that part of the answer.
â Kusalananda
Mar 5 at 16:46
Thanks for the pointer to the POSIXxargs
. The way I read it, it's nearly useless, as it says that-I
,-L
and-n
are mutually exclusive. Glad we have the GNU version. // Actuallyecho a b c
was a placeholder for any command that generates multiple space-separated parameters, hence why I used| xargs printf
instead of usingprintf
directly. // I'm still digesting your post. You've written that OpenBSDxargs
will do the wrong thing, is that a typo?
â Pedro Gimeno
Mar 5 at 17:16
1
@PedroGimeno No typo. It's the wrong output since it uses the three letters as separate arguments even though-I
is used. It does, however, do the right thing if-n
is not used. As you say,-I
and-n
are mutually exclusive, and an implementation may choose to handle this situation in any way it wants, so "wrong" is maybe the wrong word...
â Kusalananda
Mar 5 at 17:20
 |Â
show 4 more comments
1
That behaviour (that newline should be the only delimiter with -I) is documented in the GNUxargs
documentation and the POSIX specification (under XSI, so it's not surprising that OpenBSD doesn't comply there). In any case, using-n
with-I
makes little sense asis to replace one argument anyway.
â Stéphane Chazelas
Mar 5 at 16:35
@StéphaneChazelas Ah. I was looking too closely at-n
and not at-I
. Will update.
â Kusalananda
Mar 5 at 16:41
@StéphaneChazelas It seems that OpenBSD'sxargs
does the correct thing if-I
and-n
is not mixed. I rewrote that part of the answer.
â Kusalananda
Mar 5 at 16:46
Thanks for the pointer to the POSIXxargs
. The way I read it, it's nearly useless, as it says that-I
,-L
and-n
are mutually exclusive. Glad we have the GNU version. // Actuallyecho a b c
was a placeholder for any command that generates multiple space-separated parameters, hence why I used| xargs printf
instead of usingprintf
directly. // I'm still digesting your post. You've written that OpenBSDxargs
will do the wrong thing, is that a typo?
â Pedro Gimeno
Mar 5 at 17:16
1
@PedroGimeno No typo. It's the wrong output since it uses the three letters as separate arguments even though-I
is used. It does, however, do the right thing if-n
is not used. As you say,-I
and-n
are mutually exclusive, and an implementation may choose to handle this situation in any way it wants, so "wrong" is maybe the wrong word...
â Kusalananda
Mar 5 at 17:20
1
1
That behaviour (that newline should be the only delimiter with -I) is documented in the GNU
xargs
documentation and the POSIX specification (under XSI, so it's not surprising that OpenBSD doesn't comply there). In any case, using -n
with -I
makes little sense as
is to replace one argument anyway.â Stéphane Chazelas
Mar 5 at 16:35
That behaviour (that newline should be the only delimiter with -I) is documented in the GNU
xargs
documentation and the POSIX specification (under XSI, so it's not surprising that OpenBSD doesn't comply there). In any case, using -n
with -I
makes little sense as
is to replace one argument anyway.â Stéphane Chazelas
Mar 5 at 16:35
@StéphaneChazelas Ah. I was looking too closely at
-n
and not at -I
. Will update.â Kusalananda
Mar 5 at 16:41
@StéphaneChazelas Ah. I was looking too closely at
-n
and not at -I
. Will update.â Kusalananda
Mar 5 at 16:41
@StéphaneChazelas It seems that OpenBSD's
xargs
does the correct thing if -I
and -n
is not mixed. I rewrote that part of the answer.â Kusalananda
Mar 5 at 16:46
@StéphaneChazelas It seems that OpenBSD's
xargs
does the correct thing if -I
and -n
is not mixed. I rewrote that part of the answer.â Kusalananda
Mar 5 at 16:46
Thanks for the pointer to the POSIX
xargs
. The way I read it, it's nearly useless, as it says that -I
, -L
and -n
are mutually exclusive. Glad we have the GNU version. // Actually echo a b c
was a placeholder for any command that generates multiple space-separated parameters, hence why I used | xargs printf
instead of using printf
directly. // I'm still digesting your post. You've written that OpenBSD xargs
will do the wrong thing, is that a typo?â Pedro Gimeno
Mar 5 at 17:16
Thanks for the pointer to the POSIX
xargs
. The way I read it, it's nearly useless, as it says that -I
, -L
and -n
are mutually exclusive. Glad we have the GNU version. // Actually echo a b c
was a placeholder for any command that generates multiple space-separated parameters, hence why I used | xargs printf
instead of using printf
directly. // I'm still digesting your post. You've written that OpenBSD xargs
will do the wrong thing, is that a typo?â Pedro Gimeno
Mar 5 at 17:16
1
1
@PedroGimeno No typo. It's the wrong output since it uses the three letters as separate arguments even though
-I
is used. It does, however, do the right thing if -n
is not used. As you say, -I
and -n
are mutually exclusive, and an implementation may choose to handle this situation in any way it wants, so "wrong" is maybe the wrong word...â Kusalananda
Mar 5 at 17:20
@PedroGimeno No typo. It's the wrong output since it uses the three letters as separate arguments even though
-I
is used. It does, however, do the right thing if -n
is not used. As you say, -I
and -n
are mutually exclusive, and an implementation may choose to handle this situation in any way it wants, so "wrong" is maybe the wrong word...â Kusalananda
Mar 5 at 17:20
 |Â
show 4 more comments
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