AWK: why does $(cat) work for stdin, but $* doesn't?
Clash Royale CLAN TAG#URR8PPP
echo '((3+(2^3)) * 34^2 / 9)-75.89' | awk "BEGIN print $(cat) "
The above syntax works fine with the calculated result '1337'.
echo '((3+(2^3)) * 34^2 / 9)-75.89' | awk "BEGIN print $* "
But the above syntax doesn't work, though there's no error.
Plz advise.
bash awk
add a comment |
echo '((3+(2^3)) * 34^2 / 9)-75.89' | awk "BEGIN print $(cat) "
The above syntax works fine with the calculated result '1337'.
echo '((3+(2^3)) * 34^2 / 9)-75.89' | awk "BEGIN print $* "
But the above syntax doesn't work, though there's no error.
Plz advise.
bash awk
add a comment |
echo '((3+(2^3)) * 34^2 / 9)-75.89' | awk "BEGIN print $(cat) "
The above syntax works fine with the calculated result '1337'.
echo '((3+(2^3)) * 34^2 / 9)-75.89' | awk "BEGIN print $* "
But the above syntax doesn't work, though there's no error.
Plz advise.
bash awk
echo '((3+(2^3)) * 34^2 / 9)-75.89' | awk "BEGIN print $(cat) "
The above syntax works fine with the calculated result '1337'.
echo '((3+(2^3)) * 34^2 / 9)-75.89' | awk "BEGIN print $* "
But the above syntax doesn't work, though there's no error.
Plz advise.
bash awk
bash awk
edited Jan 2 at 9:46
heemayl
66.2k8138211
66.2k8138211
asked Jan 2 at 1:31
user58029user58029
615
615
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
The $(command)
syntax will return the output of command
. Here, you are using the very simple cat
program whose only job is to copy everything from standard input (stdin) into standard output (stdout). Since you are running the awk
script inside double quotes, the $(cat)
is expanded by the shell before the awk
script is run, so it reads the echo
output into its stdin and duly copies it to its stdout. This is then passed to the awk
script. You can see this in action with set -x
:
$ set -x
$ echo '((3+(2^3)) * 34^2 / 9)-75.89' | awk "BEGIN print $(cat) "
+ echo '((3+(2^3)) * 34^2 / 9)-75.89'
++ cat
+ awk 'BEGIN print ((3+(2^3)) * 34^2 / 9)-75.89 '
1337
So, awk
is actually running BEGIN print ((3+(2^3)) * 34^2 / 9)-75.89 '
which returns 1337.
Now, the $*
is a special shell variable that expands to all the positional parameters given to a shell script (see man bash
):
* Expands to the positional parameters, starting from one. When the expan‐
sion is not within double quotes, each positional parameter expands to a
separate word. In contexts where it is performed, those words are sub‐
ject to further word splitting and pathname expansion. When the expan‐
sion occurs within double quotes, it expands to a single word with the
value of each parameter separated by the first character of the IFS spe‐
cial variable. That is, "$*" is equivalent to "$1c$2c...", where c is
the first character of the value of the IFS variable. If IFS is unset,
the parameters are separated by spaces. If IFS is null, the parameters
are joined without intervening separators.
However, this variable is empty here. Therefore, the awk
script becomes:
$ echo '((3+(2^3)) * 34^2 / 9)-75.89' | awk "BEGIN print $* "
+ awk 'BEGIN print '
+ echo '((3+(2^3)) * 34^2 / 9)-75.89'
The $*
expands to an empty string, and awk
is told to print an empty string, and this is why you get no output.
You might want to just use bc
instead:
$ echo '((3+(2^3)) * 34^2 / 9)-75.89' | bc
1336.11
Should probably usebc -l
, otherwise you'll get the discrepancy you've posted above (where the result of the division has been truncated).
– cmbuckley
Jan 2 at 15:25
@cmbuckley I was trying to get it to print 1337 by playing around withscale=
(I assume the OP wants to play with leetspeak) but I couldn't find a way.bc -l
returns1336.99888888888888888888
on my system.
– terdon♦
Jan 2 at 15:40
add a comment |
Your Answer
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1 Answer
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1 Answer
1
active
oldest
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oldest
votes
active
oldest
votes
The $(command)
syntax will return the output of command
. Here, you are using the very simple cat
program whose only job is to copy everything from standard input (stdin) into standard output (stdout). Since you are running the awk
script inside double quotes, the $(cat)
is expanded by the shell before the awk
script is run, so it reads the echo
output into its stdin and duly copies it to its stdout. This is then passed to the awk
script. You can see this in action with set -x
:
$ set -x
$ echo '((3+(2^3)) * 34^2 / 9)-75.89' | awk "BEGIN print $(cat) "
+ echo '((3+(2^3)) * 34^2 / 9)-75.89'
++ cat
+ awk 'BEGIN print ((3+(2^3)) * 34^2 / 9)-75.89 '
1337
So, awk
is actually running BEGIN print ((3+(2^3)) * 34^2 / 9)-75.89 '
which returns 1337.
Now, the $*
is a special shell variable that expands to all the positional parameters given to a shell script (see man bash
):
* Expands to the positional parameters, starting from one. When the expan‐
sion is not within double quotes, each positional parameter expands to a
separate word. In contexts where it is performed, those words are sub‐
ject to further word splitting and pathname expansion. When the expan‐
sion occurs within double quotes, it expands to a single word with the
value of each parameter separated by the first character of the IFS spe‐
cial variable. That is, "$*" is equivalent to "$1c$2c...", where c is
the first character of the value of the IFS variable. If IFS is unset,
the parameters are separated by spaces. If IFS is null, the parameters
are joined without intervening separators.
However, this variable is empty here. Therefore, the awk
script becomes:
$ echo '((3+(2^3)) * 34^2 / 9)-75.89' | awk "BEGIN print $* "
+ awk 'BEGIN print '
+ echo '((3+(2^3)) * 34^2 / 9)-75.89'
The $*
expands to an empty string, and awk
is told to print an empty string, and this is why you get no output.
You might want to just use bc
instead:
$ echo '((3+(2^3)) * 34^2 / 9)-75.89' | bc
1336.11
Should probably usebc -l
, otherwise you'll get the discrepancy you've posted above (where the result of the division has been truncated).
– cmbuckley
Jan 2 at 15:25
@cmbuckley I was trying to get it to print 1337 by playing around withscale=
(I assume the OP wants to play with leetspeak) but I couldn't find a way.bc -l
returns1336.99888888888888888888
on my system.
– terdon♦
Jan 2 at 15:40
add a comment |
The $(command)
syntax will return the output of command
. Here, you are using the very simple cat
program whose only job is to copy everything from standard input (stdin) into standard output (stdout). Since you are running the awk
script inside double quotes, the $(cat)
is expanded by the shell before the awk
script is run, so it reads the echo
output into its stdin and duly copies it to its stdout. This is then passed to the awk
script. You can see this in action with set -x
:
$ set -x
$ echo '((3+(2^3)) * 34^2 / 9)-75.89' | awk "BEGIN print $(cat) "
+ echo '((3+(2^3)) * 34^2 / 9)-75.89'
++ cat
+ awk 'BEGIN print ((3+(2^3)) * 34^2 / 9)-75.89 '
1337
So, awk
is actually running BEGIN print ((3+(2^3)) * 34^2 / 9)-75.89 '
which returns 1337.
Now, the $*
is a special shell variable that expands to all the positional parameters given to a shell script (see man bash
):
* Expands to the positional parameters, starting from one. When the expan‐
sion is not within double quotes, each positional parameter expands to a
separate word. In contexts where it is performed, those words are sub‐
ject to further word splitting and pathname expansion. When the expan‐
sion occurs within double quotes, it expands to a single word with the
value of each parameter separated by the first character of the IFS spe‐
cial variable. That is, "$*" is equivalent to "$1c$2c...", where c is
the first character of the value of the IFS variable. If IFS is unset,
the parameters are separated by spaces. If IFS is null, the parameters
are joined without intervening separators.
However, this variable is empty here. Therefore, the awk
script becomes:
$ echo '((3+(2^3)) * 34^2 / 9)-75.89' | awk "BEGIN print $* "
+ awk 'BEGIN print '
+ echo '((3+(2^3)) * 34^2 / 9)-75.89'
The $*
expands to an empty string, and awk
is told to print an empty string, and this is why you get no output.
You might want to just use bc
instead:
$ echo '((3+(2^3)) * 34^2 / 9)-75.89' | bc
1336.11
Should probably usebc -l
, otherwise you'll get the discrepancy you've posted above (where the result of the division has been truncated).
– cmbuckley
Jan 2 at 15:25
@cmbuckley I was trying to get it to print 1337 by playing around withscale=
(I assume the OP wants to play with leetspeak) but I couldn't find a way.bc -l
returns1336.99888888888888888888
on my system.
– terdon♦
Jan 2 at 15:40
add a comment |
The $(command)
syntax will return the output of command
. Here, you are using the very simple cat
program whose only job is to copy everything from standard input (stdin) into standard output (stdout). Since you are running the awk
script inside double quotes, the $(cat)
is expanded by the shell before the awk
script is run, so it reads the echo
output into its stdin and duly copies it to its stdout. This is then passed to the awk
script. You can see this in action with set -x
:
$ set -x
$ echo '((3+(2^3)) * 34^2 / 9)-75.89' | awk "BEGIN print $(cat) "
+ echo '((3+(2^3)) * 34^2 / 9)-75.89'
++ cat
+ awk 'BEGIN print ((3+(2^3)) * 34^2 / 9)-75.89 '
1337
So, awk
is actually running BEGIN print ((3+(2^3)) * 34^2 / 9)-75.89 '
which returns 1337.
Now, the $*
is a special shell variable that expands to all the positional parameters given to a shell script (see man bash
):
* Expands to the positional parameters, starting from one. When the expan‐
sion is not within double quotes, each positional parameter expands to a
separate word. In contexts where it is performed, those words are sub‐
ject to further word splitting and pathname expansion. When the expan‐
sion occurs within double quotes, it expands to a single word with the
value of each parameter separated by the first character of the IFS spe‐
cial variable. That is, "$*" is equivalent to "$1c$2c...", where c is
the first character of the value of the IFS variable. If IFS is unset,
the parameters are separated by spaces. If IFS is null, the parameters
are joined without intervening separators.
However, this variable is empty here. Therefore, the awk
script becomes:
$ echo '((3+(2^3)) * 34^2 / 9)-75.89' | awk "BEGIN print $* "
+ awk 'BEGIN print '
+ echo '((3+(2^3)) * 34^2 / 9)-75.89'
The $*
expands to an empty string, and awk
is told to print an empty string, and this is why you get no output.
You might want to just use bc
instead:
$ echo '((3+(2^3)) * 34^2 / 9)-75.89' | bc
1336.11
The $(command)
syntax will return the output of command
. Here, you are using the very simple cat
program whose only job is to copy everything from standard input (stdin) into standard output (stdout). Since you are running the awk
script inside double quotes, the $(cat)
is expanded by the shell before the awk
script is run, so it reads the echo
output into its stdin and duly copies it to its stdout. This is then passed to the awk
script. You can see this in action with set -x
:
$ set -x
$ echo '((3+(2^3)) * 34^2 / 9)-75.89' | awk "BEGIN print $(cat) "
+ echo '((3+(2^3)) * 34^2 / 9)-75.89'
++ cat
+ awk 'BEGIN print ((3+(2^3)) * 34^2 / 9)-75.89 '
1337
So, awk
is actually running BEGIN print ((3+(2^3)) * 34^2 / 9)-75.89 '
which returns 1337.
Now, the $*
is a special shell variable that expands to all the positional parameters given to a shell script (see man bash
):
* Expands to the positional parameters, starting from one. When the expan‐
sion is not within double quotes, each positional parameter expands to a
separate word. In contexts where it is performed, those words are sub‐
ject to further word splitting and pathname expansion. When the expan‐
sion occurs within double quotes, it expands to a single word with the
value of each parameter separated by the first character of the IFS spe‐
cial variable. That is, "$*" is equivalent to "$1c$2c...", where c is
the first character of the value of the IFS variable. If IFS is unset,
the parameters are separated by spaces. If IFS is null, the parameters
are joined without intervening separators.
However, this variable is empty here. Therefore, the awk
script becomes:
$ echo '((3+(2^3)) * 34^2 / 9)-75.89' | awk "BEGIN print $* "
+ awk 'BEGIN print '
+ echo '((3+(2^3)) * 34^2 / 9)-75.89'
The $*
expands to an empty string, and awk
is told to print an empty string, and this is why you get no output.
You might want to just use bc
instead:
$ echo '((3+(2^3)) * 34^2 / 9)-75.89' | bc
1336.11
edited Jan 2 at 3:05
answered Jan 2 at 1:54
terdon♦terdon
65.2k12138218
65.2k12138218
Should probably usebc -l
, otherwise you'll get the discrepancy you've posted above (where the result of the division has been truncated).
– cmbuckley
Jan 2 at 15:25
@cmbuckley I was trying to get it to print 1337 by playing around withscale=
(I assume the OP wants to play with leetspeak) but I couldn't find a way.bc -l
returns1336.99888888888888888888
on my system.
– terdon♦
Jan 2 at 15:40
add a comment |
Should probably usebc -l
, otherwise you'll get the discrepancy you've posted above (where the result of the division has been truncated).
– cmbuckley
Jan 2 at 15:25
@cmbuckley I was trying to get it to print 1337 by playing around withscale=
(I assume the OP wants to play with leetspeak) but I couldn't find a way.bc -l
returns1336.99888888888888888888
on my system.
– terdon♦
Jan 2 at 15:40
Should probably use
bc -l
, otherwise you'll get the discrepancy you've posted above (where the result of the division has been truncated).– cmbuckley
Jan 2 at 15:25
Should probably use
bc -l
, otherwise you'll get the discrepancy you've posted above (where the result of the division has been truncated).– cmbuckley
Jan 2 at 15:25
@cmbuckley I was trying to get it to print 1337 by playing around with
scale=
(I assume the OP wants to play with leetspeak) but I couldn't find a way. bc -l
returns 1336.99888888888888888888
on my system.– terdon♦
Jan 2 at 15:40
@cmbuckley I was trying to get it to print 1337 by playing around with
scale=
(I assume the OP wants to play with leetspeak) but I couldn't find a way. bc -l
returns 1336.99888888888888888888
on my system.– terdon♦
Jan 2 at 15:40
add a comment |
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