Shell-Script Bash
up vote
1
down vote
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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 yesterday
ctrl-alt-delor
10.4k41955
asked yesterday
fatima
62
|
show 3 more comments
up vote
1
down vote
favorite
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 yesterday
ctrl-alt-delor
10.4k41955
asked yesterday
fatima
62
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
yesterday
@ClausAndersen gawk -v means that we are assigning the valuse of s to 4 but how about the{print $1*s}
– fatima
yesterday
This is not a question, just a statement.
– ctrl-alt-delor
yesterday
What shell is it for?
– ctrl-alt-delor
yesterday
1
Did you run the script and see what it does?
– RudiC
yesterday
|
show 3 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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 yesterday
ctrl-alt-delor
10.4k41955
asked yesterday
fatima
62
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 yesterday
ctrl-alt-delor
10.4k41955
asked yesterday
fatima
62
edited yesterday
ctrl-alt-delor
10.4k41955
asked yesterday
fatima
62
edited yesterday
ctrl-alt-delor
10.4k41955
edited yesterday
ctrl-alt-delor
10.4k41955
edited yesterday
ctrl-alt-delor
10.4k41955
10.4k41955
asked yesterday
fatima
62
asked yesterday
fatima
62
asked yesterday
fatima
62
62
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
yesterday
@ClausAndersen gawk -v means that we are assigning the valuse of s to 4 but how about the{print $1*s}
– fatima
yesterday
This is not a question, just a statement.
– ctrl-alt-delor
yesterday
What shell is it for?
– ctrl-alt-delor
yesterday
1
Did you run the script and see what it does?
– RudiC
yesterday
|
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
yesterday
@ClausAndersen gawk -v means that we are assigning the valuse of s to 4 but how about the{print $1*s}
– fatima
yesterday
This is not a question, just a statement.
– ctrl-alt-delor
yesterday
What shell is it for?
– ctrl-alt-delor
yesterday
1
Did you run the script and see what it does?
– RudiC
yesterday
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
yesterday
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
yesterday
@ClausAndersen gawk -v means that we are assigning the valuse of s to 4 but how about the
{print $1*s}– fatima
yesterday
@ClausAndersen gawk -v means that we are assigning the valuse of s to 4 but how about the
{print $1*s}– fatima
yesterday
This is not a question, just a statement.
– ctrl-alt-delor
yesterday
This is not a question, just a statement.
– ctrl-alt-delor
yesterday
What shell is it for?
– ctrl-alt-delor
yesterday
What shell is it for?
– ctrl-alt-delor
yesterday
1
1
Did you run the script and see what it does?
– RudiC
yesterday
Did you run the script and see what it does?
– RudiC
yesterday
|
show 3 more comments
2 Answers
2
active
oldest
votes
up vote
3
down vote
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 yesterday
couling
263210
add a comment |
up vote
1
down vote
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 yesterday
Kusalananda
119k16225367
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
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 yesterday
couling
263210
add a comment |
up vote
3
down vote
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 yesterday
couling
263210
add a comment |
up vote
3
down vote
up vote
3
down vote
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 yesterday
couling
263210
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 yesterday
couling
263210
edited 19 hours ago
answered yesterday
couling
263210
answered yesterday
couling
263210
answered yesterday
couling
263210
263210
add a comment |
add a comment |
up vote
1
down vote
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 yesterday
Kusalananda
119k16225367
add a comment |
up vote
1
down vote
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 yesterday
Kusalananda
119k16225367
add a comment |
up vote
1
down vote
up vote
1
down vote
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 yesterday
Kusalananda
119k16225367
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 yesterday
Kusalananda
119k16225367
edited yesterday
answered yesterday
Kusalananda
119k16225367
answered yesterday
Kusalananda
119k16225367
answered yesterday
Kusalananda
119k16225367
119k16225367
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
yesterday
@ClausAndersen gawk -v means that we are assigning the valuse of s to 4 but how about the
{print $1*s}– fatima
yesterday
This is not a question, just a statement.
– ctrl-alt-delor
yesterday
What shell is it for?
– ctrl-alt-delor
yesterday
1
Did you run the script and see what it does?
– RudiC
yesterday