2 number multiplying and adding
M="Wrong"
P="Wrong"
I=$1
while [ "$I" != "" ]
do
case $I in
-m | --multiplying)
shift
M=`expr $1 * $2`
;;
-p | --plus )
shift
P=`expr $1 + $2`
;;
*) echo "Something wrong"
exit 1
;;
esac
shift
I=$1
done
echo "Multiplying=$M, Adding=$P"
I give two number after each -m and -p
So when i run my code i want this to happen
./code.sh -m 2 2 -p 5 5
Multiplying=4 Adding=10
my code simply skip the and -p part and goes straigh to
*) echo "Something wrong"
What am i doing wrong?
linux shell-script
edited Dec 19 '18 at 6:21
Rui F Ribeiro
39k1479130
asked Dec 19 '18 at 5:24
Knap
11
|
show 2 more comments
M="Wrong"
P="Wrong"
I=$1
while [ "$I" != "" ]
do
case $I in
-m | --multiplying)
shift
M=`expr $1 * $2`
;;
-p | --plus )
shift
P=`expr $1 + $2`
;;
*) echo "Something wrong"
exit 1
;;
esac
shift
I=$1
done
echo "Multiplying=$M, Adding=$P"
I give two number after each -m and -p
So when i run my code i want this to happen
./code.sh -m 2 2 -p 5 5
Multiplying=4 Adding=10
my code simply skip the and -p part and goes straigh to
*) echo "Something wrong"
What am i doing wrong?
linux shell-script
edited Dec 19 '18 at 6:21
Rui F Ribeiro
39k1479130
asked Dec 19 '18 at 5:24
Knap
11
(1) Thank you for describing the behavior you want. But, when you ask a "What am I doing wrong?" question, you should also describe the behavior that you are getting. (2) Addset -xto your script to see what is happening.
– G-Man
Dec 19 '18 at 5:30
my code simply skip the and -p part and goes straigh to *) echo "Something wrong"
– Knap
Dec 19 '18 at 5:32
and doesnt display any result
– Knap
Dec 19 '18 at 5:33
(1) Thank you. I forgot to say, "Please do not respond in comments; edit your question to make it clearer and more complete." (2) Theset -xshould illuminate what's happening.
– G-Man
Dec 19 '18 at 5:34
1
@KhushrajRathod Why? This is the correct site for it.
– Kusalananda
Dec 19 '18 at 6:49
|
show 2 more comments
M="Wrong"
P="Wrong"
I=$1
while [ "$I" != "" ]
do
case $I in
-m | --multiplying)
shift
M=`expr $1 * $2`
;;
-p | --plus )
shift
P=`expr $1 + $2`
;;
*) echo "Something wrong"
exit 1
;;
esac
shift
I=$1
done
echo "Multiplying=$M, Adding=$P"
I give two number after each -m and -p
So when i run my code i want this to happen
./code.sh -m 2 2 -p 5 5
Multiplying=4 Adding=10
my code simply skip the and -p part and goes straigh to
*) echo "Something wrong"
What am i doing wrong?
linux shell-script
edited Dec 19 '18 at 6:21
Rui F Ribeiro
39k1479130
asked Dec 19 '18 at 5:24
Knap
11
M="Wrong"
P="Wrong"
I=$1
while [ "$I" != "" ]
do
case $I in
-m | --multiplying)
shift
M=`expr $1 * $2`
;;
-p | --plus )
shift
P=`expr $1 + $2`
;;
*) echo "Something wrong"
exit 1
;;
esac
shift
I=$1
done
echo "Multiplying=$M, Adding=$P"
I give two number after each -m and -p
So when i run my code i want this to happen
./code.sh -m 2 2 -p 5 5
Multiplying=4 Adding=10
my code simply skip the and -p part and goes straigh to
*) echo "Something wrong"
What am i doing wrong?
linux shell-script
linux shell-script
edited Dec 19 '18 at 6:21
Rui F Ribeiro
39k1479130
asked Dec 19 '18 at 5:24
Knap
11
edited Dec 19 '18 at 6:21
Rui F Ribeiro
39k1479130
asked Dec 19 '18 at 5:24
Knap
11
edited Dec 19 '18 at 6:21
Rui F Ribeiro
39k1479130
edited Dec 19 '18 at 6:21
Rui F Ribeiro
39k1479130
edited Dec 19 '18 at 6:21
Rui F Ribeiro
39k1479130
39k1479130
asked Dec 19 '18 at 5:24
Knap
11
asked Dec 19 '18 at 5:24
Knap
11
asked Dec 19 '18 at 5:24
Knap
11
11
(1) Thank you for describing the behavior you want. But, when you ask a "What am I doing wrong?" question, you should also describe the behavior that you are getting. (2) Addset -xto your script to see what is happening.
– G-Man
Dec 19 '18 at 5:30
my code simply skip the and -p part and goes straigh to *) echo "Something wrong"
– Knap
Dec 19 '18 at 5:32
and doesnt display any result
– Knap
Dec 19 '18 at 5:33
(1) Thank you. I forgot to say, "Please do not respond in comments; edit your question to make it clearer and more complete." (2) Theset -xshould illuminate what's happening.
– G-Man
Dec 19 '18 at 5:34
1
@KhushrajRathod Why? This is the correct site for it.
– Kusalananda
Dec 19 '18 at 6:49
|
show 2 more comments
(1) Thank you for describing the behavior you want. But, when you ask a "What am I doing wrong?" question, you should also describe the behavior that you are getting. (2) Addset -xto your script to see what is happening.
– G-Man
Dec 19 '18 at 5:30
my code simply skip the and -p part and goes straigh to *) echo "Something wrong"
– Knap
Dec 19 '18 at 5:32
and doesnt display any result
– Knap
Dec 19 '18 at 5:33
(1) Thank you. I forgot to say, "Please do not respond in comments; edit your question to make it clearer and more complete." (2) Theset -xshould illuminate what's happening.
– G-Man
Dec 19 '18 at 5:34
1
@KhushrajRathod Why? This is the correct site for it.
– Kusalananda
Dec 19 '18 at 6:49
(1) Thank you for describing the behavior you want. But, when you ask a "What am I doing wrong?" question, you should also describe the behavior that you are getting. (2) Add
set -x to your script to see what is happening.– G-Man
Dec 19 '18 at 5:30
(1) Thank you for describing the behavior you want. But, when you ask a "What am I doing wrong?" question, you should also describe the behavior that you are getting. (2) Add
set -x to your script to see what is happening.– G-Man
Dec 19 '18 at 5:30
my code simply skip the and -p part and goes straigh to *) echo "Something wrong"
– Knap
Dec 19 '18 at 5:32
my code simply skip the and -p part and goes straigh to *) echo "Something wrong"
– Knap
Dec 19 '18 at 5:32
and doesnt display any result
– Knap
Dec 19 '18 at 5:33
and doesnt display any result
– Knap
Dec 19 '18 at 5:33
(1) Thank you. I forgot to say, "Please do not respond in comments; edit your question to make it clearer and more complete." (2) The
set -x should illuminate what's happening.– G-Man
Dec 19 '18 at 5:34
(1) Thank you. I forgot to say, "Please do not respond in comments; edit your question to make it clearer and more complete." (2) The
set -x should illuminate what's happening.– G-Man
Dec 19 '18 at 5:34
1
1
@KhushrajRathod Why? This is the correct site for it.
– Kusalananda
Dec 19 '18 at 6:49
@KhushrajRathod Why? This is the correct site for it.
– Kusalananda
Dec 19 '18 at 6:49
|
show 2 more comments
1 Answer
1
active
oldest
votes
#!/bin/sh
M="Wrong"
P="Wrong"
while [ "$#" -gt 0 ]; do
case $1 in
-m|--multiplying)
M=$(( $2 * $3 ))
;;
-p|--plus)
P=$(( $2 + $3 ))
;;
*) echo 'Error in command line options' >&2
exit 1
esac
shift 3
done
printf 'Multiplying=%s, Adding=%sn' "$M" "$P"
Your code is a bit difficult to follow due to the shifts being spread out, but the main issue is that you are lacking one shift operation per loop. Your script calculates the result of the first arithmetic operation correctly, but when it comes to looking at the second arithmetic operation, it has only performed two shift operations (each -p A B requires three shift to be processed and for the following -m C D to be in $1, $2 and $3).
My modified version of your code only does a single shift 3 after each operation, while the operation itself uses $1, $2 and $3. I'm also using the more modern arithmetic expansion $(( ... )) rather than the antiquated expr utility. I'm testing on $# whether there are more command line arguments to process and I've update the output slightly so that printf is used for variable data and so that diagnostic messages are outputted onto the standard error stream.
Also note that having options that take more than one option argument is a bit unusual.
answered Dec 19 '18 at 6:55
Kusalananda
121k16229372
add a comment |
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1 Answer
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1 Answer
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oldest
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oldest
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oldest
votes
#!/bin/sh
M="Wrong"
P="Wrong"
while [ "$#" -gt 0 ]; do
case $1 in
-m|--multiplying)
M=$(( $2 * $3 ))
;;
-p|--plus)
P=$(( $2 + $3 ))
;;
*) echo 'Error in command line options' >&2
exit 1
esac
shift 3
done
printf 'Multiplying=%s, Adding=%sn' "$M" "$P"
Your code is a bit difficult to follow due to the shifts being spread out, but the main issue is that you are lacking one shift operation per loop. Your script calculates the result of the first arithmetic operation correctly, but when it comes to looking at the second arithmetic operation, it has only performed two shift operations (each -p A B requires three shift to be processed and for the following -m C D to be in $1, $2 and $3).
My modified version of your code only does a single shift 3 after each operation, while the operation itself uses $1, $2 and $3. I'm also using the more modern arithmetic expansion $(( ... )) rather than the antiquated expr utility. I'm testing on $# whether there are more command line arguments to process and I've update the output slightly so that printf is used for variable data and so that diagnostic messages are outputted onto the standard error stream.
Also note that having options that take more than one option argument is a bit unusual.
answered Dec 19 '18 at 6:55
Kusalananda
121k16229372
add a comment |
#!/bin/sh
M="Wrong"
P="Wrong"
while [ "$#" -gt 0 ]; do
case $1 in
-m|--multiplying)
M=$(( $2 * $3 ))
;;
-p|--plus)
P=$(( $2 + $3 ))
;;
*) echo 'Error in command line options' >&2
exit 1
esac
shift 3
done
printf 'Multiplying=%s, Adding=%sn' "$M" "$P"
Your code is a bit difficult to follow due to the shifts being spread out, but the main issue is that you are lacking one shift operation per loop. Your script calculates the result of the first arithmetic operation correctly, but when it comes to looking at the second arithmetic operation, it has only performed two shift operations (each -p A B requires three shift to be processed and for the following -m C D to be in $1, $2 and $3).
My modified version of your code only does a single shift 3 after each operation, while the operation itself uses $1, $2 and $3. I'm also using the more modern arithmetic expansion $(( ... )) rather than the antiquated expr utility. I'm testing on $# whether there are more command line arguments to process and I've update the output slightly so that printf is used for variable data and so that diagnostic messages are outputted onto the standard error stream.
Also note that having options that take more than one option argument is a bit unusual.
answered Dec 19 '18 at 6:55
Kusalananda
121k16229372
add a comment |
#!/bin/sh
M="Wrong"
P="Wrong"
while [ "$#" -gt 0 ]; do
case $1 in
-m|--multiplying)
M=$(( $2 * $3 ))
;;
-p|--plus)
P=$(( $2 + $3 ))
;;
*) echo 'Error in command line options' >&2
exit 1
esac
shift 3
done
printf 'Multiplying=%s, Adding=%sn' "$M" "$P"
Your code is a bit difficult to follow due to the shifts being spread out, but the main issue is that you are lacking one shift operation per loop. Your script calculates the result of the first arithmetic operation correctly, but when it comes to looking at the second arithmetic operation, it has only performed two shift operations (each -p A B requires three shift to be processed and for the following -m C D to be in $1, $2 and $3).
My modified version of your code only does a single shift 3 after each operation, while the operation itself uses $1, $2 and $3. I'm also using the more modern arithmetic expansion $(( ... )) rather than the antiquated expr utility. I'm testing on $# whether there are more command line arguments to process and I've update the output slightly so that printf is used for variable data and so that diagnostic messages are outputted onto the standard error stream.
Also note that having options that take more than one option argument is a bit unusual.
answered Dec 19 '18 at 6:55
Kusalananda
121k16229372
#!/bin/sh
M="Wrong"
P="Wrong"
while [ "$#" -gt 0 ]; do
case $1 in
-m|--multiplying)
M=$(( $2 * $3 ))
;;
-p|--plus)
P=$(( $2 + $3 ))
;;
*) echo 'Error in command line options' >&2
exit 1
esac
shift 3
done
printf 'Multiplying=%s, Adding=%sn' "$M" "$P"
Your code is a bit difficult to follow due to the shifts being spread out, but the main issue is that you are lacking one shift operation per loop. Your script calculates the result of the first arithmetic operation correctly, but when it comes to looking at the second arithmetic operation, it has only performed two shift operations (each -p A B requires three shift to be processed and for the following -m C D to be in $1, $2 and $3).
My modified version of your code only does a single shift 3 after each operation, while the operation itself uses $1, $2 and $3. I'm also using the more modern arithmetic expansion $(( ... )) rather than the antiquated expr utility. I'm testing on $# whether there are more command line arguments to process and I've update the output slightly so that printf is used for variable data and so that diagnostic messages are outputted onto the standard error stream.
Also note that having options that take more than one option argument is a bit unusual.
answered Dec 19 '18 at 6:55
Kusalananda
121k16229372
edited Dec 19 '18 at 7:00
answered Dec 19 '18 at 6:55
Kusalananda
121k16229372
answered Dec 19 '18 at 6:55
Kusalananda
121k16229372
answered Dec 19 '18 at 6:55
Kusalananda
121k16229372
121k16229372
add a comment |
add a comment |
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(1) Thank you for describing the behavior you want. But, when you ask a "What am I doing wrong?" question, you should also describe the behavior that you are getting. (2) Add
set -xto your script to see what is happening.– G-Man
Dec 19 '18 at 5:30
my code simply skip the and -p part and goes straigh to *) echo "Something wrong"
– Knap
Dec 19 '18 at 5:32
and doesnt display any result
– Knap
Dec 19 '18 at 5:33
(1) Thank you. I forgot to say, "Please do not respond in comments; edit your question to make it clearer and more complete." (2) The
set -xshould illuminate what's happening.– G-Man
Dec 19 '18 at 5:34
1
@KhushrajRathod Why? This is the correct site for it.
– Kusalananda
Dec 19 '18 at 6:49