Shell-Script Bash
I have this loop in the shell script, but it seems I don't fully understand what it does:
especially gawk -v
for ((k=$1; $k<3 + 3 ; k=$k + $3))
do
echo $k | gawk -v s=4 '{print $1*s}'
done
Assume the arguments we have are 2 10 4
shell-script awk gawk
edited Dec 11 '18 at 15:55
ctrl-alt-delor
11.2k42058
asked Dec 11 '18 at 15:23
fatimafatima
92
|
show 3 more comments
I have this loop in the shell script, but it seems I don't fully understand what it does:
especially gawk -v
for ((k=$1; $k<3 + 3 ; k=$k + $3))
do
echo $k | gawk -v s=4 '{print $1*s}'
done
Assume the arguments we have are 2 10 4
shell-script awk gawk
edited Dec 11 '18 at 15:55
ctrl-alt-delor
11.2k42058
asked Dec 11 '18 at 15:23
fatimafatima
92
2
I'd be happy to answer none school work questions. Could you elaborate on in what context this is used and how it fits into a larger picture? You specify three arguments - but only $1 and $3 is used..
– Claus Andersen
Dec 11 '18 at 15:27
@ClausAndersen gawk -v means that we are assigning the valuse of s to 4 but how about the{print $1*s}
– fatima
Dec 11 '18 at 15:50
This is not a question, just a statement.
– ctrl-alt-delor
Dec 11 '18 at 15:56
What shell is it for?
– ctrl-alt-delor
Dec 11 '18 at 15:57
1
Did you run the script and see what it does?
– RudiC
Dec 11 '18 at 16:32
|
show 3 more comments
I have this loop in the shell script, but it seems I don't fully understand what it does:
especially gawk -v
for ((k=$1; $k<3 + 3 ; k=$k + $3))
do
echo $k | gawk -v s=4 '{print $1*s}'
done
Assume the arguments we have are 2 10 4
shell-script awk gawk
edited Dec 11 '18 at 15:55
ctrl-alt-delor
11.2k42058
asked Dec 11 '18 at 15:23
fatimafatima
92
I have this loop in the shell script, but it seems I don't fully understand what it does:
especially gawk -v
for ((k=$1; $k<3 + 3 ; k=$k + $3))
do
echo $k | gawk -v s=4 '{print $1*s}'
done
Assume the arguments we have are 2 10 4
shell-script awk gawk
shell-script awk gawk
edited Dec 11 '18 at 15:55
ctrl-alt-delor
11.2k42058
asked Dec 11 '18 at 15:23
fatimafatima
92
edited Dec 11 '18 at 15:55
ctrl-alt-delor
11.2k42058
asked Dec 11 '18 at 15:23
fatimafatima
92
edited Dec 11 '18 at 15:55
ctrl-alt-delor
11.2k42058
edited Dec 11 '18 at 15:55
ctrl-alt-delor
11.2k42058
edited Dec 11 '18 at 15:55
ctrl-alt-delor
11.2k42058
11.2k42058
asked Dec 11 '18 at 15:23
fatimafatima
92
asked Dec 11 '18 at 15:23
fatimafatima
92
asked Dec 11 '18 at 15:23
fatimafatima
92
92
2
I'd be happy to answer none school work questions. Could you elaborate on in what context this is used and how it fits into a larger picture? You specify three arguments - but only $1 and $3 is used..
– Claus Andersen
Dec 11 '18 at 15:27
@ClausAndersen gawk -v means that we are assigning the valuse of s to 4 but how about the{print $1*s}
– fatima
Dec 11 '18 at 15:50
This is not a question, just a statement.
– ctrl-alt-delor
Dec 11 '18 at 15:56
What shell is it for?
– ctrl-alt-delor
Dec 11 '18 at 15:57
1
Did you run the script and see what it does?
– RudiC
Dec 11 '18 at 16:32
|
show 3 more comments
2
I'd be happy to answer none school work questions. Could you elaborate on in what context this is used and how it fits into a larger picture? You specify three arguments - but only $1 and $3 is used..
– Claus Andersen
Dec 11 '18 at 15:27
@ClausAndersen gawk -v means that we are assigning the valuse of s to 4 but how about the{print $1*s}
– fatima
Dec 11 '18 at 15:50
This is not a question, just a statement.
– ctrl-alt-delor
Dec 11 '18 at 15:56
What shell is it for?
– ctrl-alt-delor
Dec 11 '18 at 15:57
1
Did you run the script and see what it does?
– RudiC
Dec 11 '18 at 16:32
2
2
I'd be happy to answer none school work questions. Could you elaborate on in what context this is used and how it fits into a larger picture? You specify three arguments - but only $1 and $3 is used..
– Claus Andersen
Dec 11 '18 at 15:27
I'd be happy to answer none school work questions. Could you elaborate on in what context this is used and how it fits into a larger picture? You specify three arguments - but only $1 and $3 is used..
– Claus Andersen
Dec 11 '18 at 15:27
@ClausAndersen gawk -v means that we are assigning the valuse of s to 4 but how about the
{print $1*s}– fatima
Dec 11 '18 at 15:50
@ClausAndersen gawk -v means that we are assigning the valuse of s to 4 but how about the
{print $1*s}– fatima
Dec 11 '18 at 15:50
This is not a question, just a statement.
– ctrl-alt-delor
Dec 11 '18 at 15:56
This is not a question, just a statement.
– ctrl-alt-delor
Dec 11 '18 at 15:56
What shell is it for?
– ctrl-alt-delor
Dec 11 '18 at 15:57
What shell is it for?
– ctrl-alt-delor
Dec 11 '18 at 15:57
1
1
Did you run the script and see what it does?
– RudiC
Dec 11 '18 at 16:32
Did you run the script and see what it does?
– RudiC
Dec 11 '18 at 16:32
|
show 3 more comments
2 Answers
2
active
oldest
votes
I can only imagine this is a homework question.
Let's break this down. Firstly you have a loop which starts at $1 and increments by $3 and will stop at 6. So if you pass in 2 10 4, then the loop will start at 2, and increment by 4. It will immediately stop because 2 + 4 = 6.
So the following just print's 2. With the arguments 2 10 4.
for ((k=$1; $k<3 + 3 ; k=$k + $3))
do
echo $k
done
Each time round the loop, you process the output using gawk -v s=4 '{print $1*s}'. This is a very small "awk" program. It sets a variable s=4. Then prints $1*s (ie: it calculates 2 * 4 and just prints 8.
answered Dec 11 '18 at 16:54
coulingcouling
494311
add a comment |
The awk bit just prints the current value of $k (this is $1 in the awk code as it is read from the input) times 4 (this is the value of the awk variable s, as set on the command line).
It would be shorter to do
printf '%dn' "$(( 4*k ))"
The loop goes from whatever the first argument is to 5 in steps of the third argument. The second argument does not make any difference.
Therefore, the whole thing could be simplified down to
seq "$(( 4*$1 ))" "$(( 4*$3 ))" 20
The three arguments to GNU seq are "start, increment, and end". This is for the output, and the output will always be four times the current value of the loop variable. The loop starts at $1, so the output starts at four times that. The loop increments by $3, so we increment by four times that. The loop ends with $k at a maximum of 5 (one less than 3+3), so the output ends at 4*5.
Or, if you want to do that seq call as a bash loop instead:
for (( k = 4*$1; k <= 20; k += 4*$3 )); do
printf '%dn' "$k"
done
And, as you see,
for (( k = $1; k <= 5; k += $3 )); do
printf '%dn' "$(( 4*k ))"
done
is not far from that.
answered Dec 11 '18 at 17:23
KusalanandaKusalananda
127k16240395
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
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active
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votes
I can only imagine this is a homework question.
Let's break this down. Firstly you have a loop which starts at $1 and increments by $3 and will stop at 6. So if you pass in 2 10 4, then the loop will start at 2, and increment by 4. It will immediately stop because 2 + 4 = 6.
So the following just print's 2. With the arguments 2 10 4.
for ((k=$1; $k<3 + 3 ; k=$k + $3))
do
echo $k
done
Each time round the loop, you process the output using gawk -v s=4 '{print $1*s}'. This is a very small "awk" program. It sets a variable s=4. Then prints $1*s (ie: it calculates 2 * 4 and just prints 8.
answered Dec 11 '18 at 16:54
coulingcouling
494311
add a comment |
I can only imagine this is a homework question.
Let's break this down. Firstly you have a loop which starts at $1 and increments by $3 and will stop at 6. So if you pass in 2 10 4, then the loop will start at 2, and increment by 4. It will immediately stop because 2 + 4 = 6.
So the following just print's 2. With the arguments 2 10 4.
for ((k=$1; $k<3 + 3 ; k=$k + $3))
do
echo $k
done
Each time round the loop, you process the output using gawk -v s=4 '{print $1*s}'. This is a very small "awk" program. It sets a variable s=4. Then prints $1*s (ie: it calculates 2 * 4 and just prints 8.
answered Dec 11 '18 at 16:54
coulingcouling
494311
add a comment |
I can only imagine this is a homework question.
Let's break this down. Firstly you have a loop which starts at $1 and increments by $3 and will stop at 6. So if you pass in 2 10 4, then the loop will start at 2, and increment by 4. It will immediately stop because 2 + 4 = 6.
So the following just print's 2. With the arguments 2 10 4.
for ((k=$1; $k<3 + 3 ; k=$k + $3))
do
echo $k
done
Each time round the loop, you process the output using gawk -v s=4 '{print $1*s}'. This is a very small "awk" program. It sets a variable s=4. Then prints $1*s (ie: it calculates 2 * 4 and just prints 8.
answered Dec 11 '18 at 16:54
coulingcouling
494311
I can only imagine this is a homework question.
Let's break this down. Firstly you have a loop which starts at $1 and increments by $3 and will stop at 6. So if you pass in 2 10 4, then the loop will start at 2, and increment by 4. It will immediately stop because 2 + 4 = 6.
So the following just print's 2. With the arguments 2 10 4.
for ((k=$1; $k<3 + 3 ; k=$k + $3))
do
echo $k
done
Each time round the loop, you process the output using gawk -v s=4 '{print $1*s}'. This is a very small "awk" program. It sets a variable s=4. Then prints $1*s (ie: it calculates 2 * 4 and just prints 8.
answered Dec 11 '18 at 16:54
coulingcouling
494311
edited Dec 12 '18 at 9:30
answered Dec 11 '18 at 16:54
coulingcouling
494311
answered Dec 11 '18 at 16:54
coulingcouling
494311
answered Dec 11 '18 at 16:54
coulingcouling
494311
494311
add a comment |
add a comment |
The awk bit just prints the current value of $k (this is $1 in the awk code as it is read from the input) times 4 (this is the value of the awk variable s, as set on the command line).
It would be shorter to do
printf '%dn' "$(( 4*k ))"
The loop goes from whatever the first argument is to 5 in steps of the third argument. The second argument does not make any difference.
Therefore, the whole thing could be simplified down to
seq "$(( 4*$1 ))" "$(( 4*$3 ))" 20
The three arguments to GNU seq are "start, increment, and end". This is for the output, and the output will always be four times the current value of the loop variable. The loop starts at $1, so the output starts at four times that. The loop increments by $3, so we increment by four times that. The loop ends with $k at a maximum of 5 (one less than 3+3), so the output ends at 4*5.
Or, if you want to do that seq call as a bash loop instead:
for (( k = 4*$1; k <= 20; k += 4*$3 )); do
printf '%dn' "$k"
done
And, as you see,
for (( k = $1; k <= 5; k += $3 )); do
printf '%dn' "$(( 4*k ))"
done
is not far from that.
answered Dec 11 '18 at 17:23
KusalanandaKusalananda
127k16240395
add a comment |
The awk bit just prints the current value of $k (this is $1 in the awk code as it is read from the input) times 4 (this is the value of the awk variable s, as set on the command line).
It would be shorter to do
printf '%dn' "$(( 4*k ))"
The loop goes from whatever the first argument is to 5 in steps of the third argument. The second argument does not make any difference.
Therefore, the whole thing could be simplified down to
seq "$(( 4*$1 ))" "$(( 4*$3 ))" 20
The three arguments to GNU seq are "start, increment, and end". This is for the output, and the output will always be four times the current value of the loop variable. The loop starts at $1, so the output starts at four times that. The loop increments by $3, so we increment by four times that. The loop ends with $k at a maximum of 5 (one less than 3+3), so the output ends at 4*5.
Or, if you want to do that seq call as a bash loop instead:
for (( k = 4*$1; k <= 20; k += 4*$3 )); do
printf '%dn' "$k"
done
And, as you see,
for (( k = $1; k <= 5; k += $3 )); do
printf '%dn' "$(( 4*k ))"
done
is not far from that.
answered Dec 11 '18 at 17:23
KusalanandaKusalananda
127k16240395
add a comment |
The awk bit just prints the current value of $k (this is $1 in the awk code as it is read from the input) times 4 (this is the value of the awk variable s, as set on the command line).
It would be shorter to do
printf '%dn' "$(( 4*k ))"
The loop goes from whatever the first argument is to 5 in steps of the third argument. The second argument does not make any difference.
Therefore, the whole thing could be simplified down to
seq "$(( 4*$1 ))" "$(( 4*$3 ))" 20
The three arguments to GNU seq are "start, increment, and end". This is for the output, and the output will always be four times the current value of the loop variable. The loop starts at $1, so the output starts at four times that. The loop increments by $3, so we increment by four times that. The loop ends with $k at a maximum of 5 (one less than 3+3), so the output ends at 4*5.
Or, if you want to do that seq call as a bash loop instead:
for (( k = 4*$1; k <= 20; k += 4*$3 )); do
printf '%dn' "$k"
done
And, as you see,
for (( k = $1; k <= 5; k += $3 )); do
printf '%dn' "$(( 4*k ))"
done
is not far from that.
answered Dec 11 '18 at 17:23
KusalanandaKusalananda
127k16240395
The awk bit just prints the current value of $k (this is $1 in the awk code as it is read from the input) times 4 (this is the value of the awk variable s, as set on the command line).
It would be shorter to do
printf '%dn' "$(( 4*k ))"
The loop goes from whatever the first argument is to 5 in steps of the third argument. The second argument does not make any difference.
Therefore, the whole thing could be simplified down to
seq "$(( 4*$1 ))" "$(( 4*$3 ))" 20
The three arguments to GNU seq are "start, increment, and end". This is for the output, and the output will always be four times the current value of the loop variable. The loop starts at $1, so the output starts at four times that. The loop increments by $3, so we increment by four times that. The loop ends with $k at a maximum of 5 (one less than 3+3), so the output ends at 4*5.
Or, if you want to do that seq call as a bash loop instead:
for (( k = 4*$1; k <= 20; k += 4*$3 )); do
printf '%dn' "$k"
done
And, as you see,
for (( k = $1; k <= 5; k += $3 )); do
printf '%dn' "$(( 4*k ))"
done
is not far from that.
answered Dec 11 '18 at 17:23
KusalanandaKusalananda
127k16240395
edited Dec 11 '18 at 17:30
answered Dec 11 '18 at 17:23
KusalanandaKusalananda
127k16240395
answered Dec 11 '18 at 17:23
KusalanandaKusalananda
127k16240395
answered Dec 11 '18 at 17:23
KusalanandaKusalananda
127k16240395
127k16240395
add a comment |
add a comment |
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2
I'd be happy to answer none school work questions. Could you elaborate on in what context this is used and how it fits into a larger picture? You specify three arguments - but only $1 and $3 is used..
– Claus Andersen
Dec 11 '18 at 15:27
@ClausAndersen gawk -v means that we are assigning the valuse of s to 4 but how about the
{print $1*s}– fatima
Dec 11 '18 at 15:50
This is not a question, just a statement.
– ctrl-alt-delor
Dec 11 '18 at 15:56
What shell is it for?
– ctrl-alt-delor
Dec 11 '18 at 15:57
1
Did you run the script and see what it does?
– RudiC
Dec 11 '18 at 16:32