AWK: Why $* works inside bash function but not under pipe?
@terdon in this post answered the related question of mine, but I missed one more question in that post.
Plz refer to the following commands:
calc(){ awk "BEGIN{ print $* }" ;}; calc "((3+(2^3)) * 34^2 / 9)-75.89"
The above commands work fine with calculated result '1337'.
echo '((3+(2^3)) * 34^2 / 9)-75.89' | awk "BEGIN{ print $* }"
But the above commands don't give any result while @terdon explained well about why.
Could you advise what made the first example work with $*?
awk
asked Jan 2 at 3:11
user58029user58029
615
add a comment |
@terdon in this post answered the related question of mine, but I missed one more question in that post.
Plz refer to the following commands:
calc(){ awk "BEGIN{ print $* }" ;}; calc "((3+(2^3)) * 34^2 / 9)-75.89"
The above commands work fine with calculated result '1337'.
echo '((3+(2^3)) * 34^2 / 9)-75.89' | awk "BEGIN{ print $* }"
But the above commands don't give any result while @terdon explained well about why.
Could you advise what made the first example work with $*?
awk
asked Jan 2 at 3:11
user58029user58029
615
add a comment |
@terdon in this post answered the related question of mine, but I missed one more question in that post.
Plz refer to the following commands:
calc(){ awk "BEGIN{ print $* }" ;}; calc "((3+(2^3)) * 34^2 / 9)-75.89"
The above commands work fine with calculated result '1337'.
echo '((3+(2^3)) * 34^2 / 9)-75.89' | awk "BEGIN{ print $* }"
But the above commands don't give any result while @terdon explained well about why.
Could you advise what made the first example work with $*?
awk
asked Jan 2 at 3:11
user58029user58029
615
@terdon in this post answered the related question of mine, but I missed one more question in that post.
Plz refer to the following commands:
calc(){ awk "BEGIN{ print $* }" ;}; calc "((3+(2^3)) * 34^2 / 9)-75.89"
The above commands work fine with calculated result '1337'.
echo '((3+(2^3)) * 34^2 / 9)-75.89' | awk "BEGIN{ print $* }"
But the above commands don't give any result while @terdon explained well about why.
Could you advise what made the first example work with $*?
awk
awk
asked Jan 2 at 3:11
user58029user58029
615
asked Jan 2 at 3:11
user58029user58029
615
asked Jan 2 at 3:11
user58029user58029
615
asked Jan 2 at 3:11
user58029user58029
615
asked Jan 2 at 3:11
user58029user58029
615
615
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$* refers to positional parameters - those variables which are referenced by $1 and $2 and so on, and are provided as arguments to scripts and functions. That's the key to your question.
When you have interactive shell , there's no positional parameters set by default, so $* is empty. You can make it work if you set those via set "((3+(2^3)) * 34^2 / 9)-75.89" command, which will make $1 equal to that string.
The difference with calc(){ awk "BEGIN{ print $* }" ;}; is that it's a function and functions can process positional parameters (theirs, not the shell's). When you call calc "((3+(2^3)) * 34^2 / 9)-75.89" you're calling a function with positional parameter "((3+(2^3)) * 34^2 / 9)-75.89". There $* won't be empty:
$ calc(){ echo "Params: '$*'"; awk "BEGIN{ print $* }" ;}; calc "((3+(2^3)) * 34^2 / 9)-75.89"
Params: '((3+(2^3)) * 34^2 / 9)-75.89'
1337
edited Jan 2 at 9:32
terdon♦
65.3k12138218
answered Jan 2 at 3:34
Sergiy KolodyazhnyySergiy Kolodyazhnyy
70.9k9147310
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$* refers to positional parameters - those variables which are referenced by $1 and $2 and so on, and are provided as arguments to scripts and functions. That's the key to your question.
When you have interactive shell , there's no positional parameters set by default, so $* is empty. You can make it work if you set those via set "((3+(2^3)) * 34^2 / 9)-75.89" command, which will make $1 equal to that string.
The difference with calc(){ awk "BEGIN{ print $* }" ;}; is that it's a function and functions can process positional parameters (theirs, not the shell's). When you call calc "((3+(2^3)) * 34^2 / 9)-75.89" you're calling a function with positional parameter "((3+(2^3)) * 34^2 / 9)-75.89". There $* won't be empty:
$ calc(){ echo "Params: '$*'"; awk "BEGIN{ print $* }" ;}; calc "((3+(2^3)) * 34^2 / 9)-75.89"
Params: '((3+(2^3)) * 34^2 / 9)-75.89'
1337
edited Jan 2 at 9:32
terdon♦
65.3k12138218
answered Jan 2 at 3:34
Sergiy KolodyazhnyySergiy Kolodyazhnyy
70.9k9147310
add a comment |
$* refers to positional parameters - those variables which are referenced by $1 and $2 and so on, and are provided as arguments to scripts and functions. That's the key to your question.
When you have interactive shell , there's no positional parameters set by default, so $* is empty. You can make it work if you set those via set "((3+(2^3)) * 34^2 / 9)-75.89" command, which will make $1 equal to that string.
The difference with calc(){ awk "BEGIN{ print $* }" ;}; is that it's a function and functions can process positional parameters (theirs, not the shell's). When you call calc "((3+(2^3)) * 34^2 / 9)-75.89" you're calling a function with positional parameter "((3+(2^3)) * 34^2 / 9)-75.89". There $* won't be empty:
$ calc(){ echo "Params: '$*'"; awk "BEGIN{ print $* }" ;}; calc "((3+(2^3)) * 34^2 / 9)-75.89"
Params: '((3+(2^3)) * 34^2 / 9)-75.89'
1337
edited Jan 2 at 9:32
terdon♦
65.3k12138218
answered Jan 2 at 3:34
Sergiy KolodyazhnyySergiy Kolodyazhnyy
70.9k9147310
add a comment |
$* refers to positional parameters - those variables which are referenced by $1 and $2 and so on, and are provided as arguments to scripts and functions. That's the key to your question.
When you have interactive shell , there's no positional parameters set by default, so $* is empty. You can make it work if you set those via set "((3+(2^3)) * 34^2 / 9)-75.89" command, which will make $1 equal to that string.
The difference with calc(){ awk "BEGIN{ print $* }" ;}; is that it's a function and functions can process positional parameters (theirs, not the shell's). When you call calc "((3+(2^3)) * 34^2 / 9)-75.89" you're calling a function with positional parameter "((3+(2^3)) * 34^2 / 9)-75.89". There $* won't be empty:
$ calc(){ echo "Params: '$*'"; awk "BEGIN{ print $* }" ;}; calc "((3+(2^3)) * 34^2 / 9)-75.89"
Params: '((3+(2^3)) * 34^2 / 9)-75.89'
1337
edited Jan 2 at 9:32
terdon♦
65.3k12138218
answered Jan 2 at 3:34
Sergiy KolodyazhnyySergiy Kolodyazhnyy
70.9k9147310
$* refers to positional parameters - those variables which are referenced by $1 and $2 and so on, and are provided as arguments to scripts and functions. That's the key to your question.
When you have interactive shell , there's no positional parameters set by default, so $* is empty. You can make it work if you set those via set "((3+(2^3)) * 34^2 / 9)-75.89" command, which will make $1 equal to that string.
The difference with calc(){ awk "BEGIN{ print $* }" ;}; is that it's a function and functions can process positional parameters (theirs, not the shell's). When you call calc "((3+(2^3)) * 34^2 / 9)-75.89" you're calling a function with positional parameter "((3+(2^3)) * 34^2 / 9)-75.89". There $* won't be empty:
$ calc(){ echo "Params: '$*'"; awk "BEGIN{ print $* }" ;}; calc "((3+(2^3)) * 34^2 / 9)-75.89"
Params: '((3+(2^3)) * 34^2 / 9)-75.89'
1337
edited Jan 2 at 9:32
terdon♦
65.3k12138218
answered Jan 2 at 3:34
Sergiy KolodyazhnyySergiy Kolodyazhnyy
70.9k9147310
edited Jan 2 at 9:32
terdon♦
65.3k12138218
edited Jan 2 at 9:32
terdon♦
65.3k12138218
edited Jan 2 at 9:32
terdon♦
65.3k12138218
65.3k12138218
answered Jan 2 at 3:34
Sergiy KolodyazhnyySergiy Kolodyazhnyy
70.9k9147310
answered Jan 2 at 3:34
Sergiy KolodyazhnyySergiy Kolodyazhnyy
70.9k9147310
answered Jan 2 at 3:34
Sergiy KolodyazhnyySergiy Kolodyazhnyy
70.9k9147310
70.9k9147310
add a comment |
add a comment |
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