A strange error in Fibonacci recursive function in bash: error in the output
I'm trying to write a recursive bash function for Fibonacci sequence. So far I have a code like below:
#!/bin/bash
Fibo() {
case $1 in
0) echo 0;;
1) echo 1;;
*) echo $[$[$Fibo $[$1-1]]+$[$Fibo $[$1-2]]] ;;
esac
}
for (( i=0; i<=$1; i++ )); do
Fibo $i
done
Here's a sample output:
0
1
1
3
5
7
9
11
Of course, it should be 0 1 1 2 3 5 8 13 ...
I tried to write it down in the paper and in my opinion it should work. My code understanding:
Fibo 0 -> = 0
Fibo 1 -> = 1
Fibo 2 -> Fibo 1 + Fibo 0 = 1 + 0 = 1
Fibo 3 -> Fibo 2 + Fibo 1 = 1 + 1 = 2
Fibo 4 -> Fibo 3 + Fibo 2 = 2 + 1 = 3
Fibo 5 -> Fibo 4 + Fibo 3 = 3 + 2 = 5
etc...
What the code "is doing":
Fibo 0 -> = 0
Fibo 1 -> = 1
Fibo 2 -> = 1 + 0 = 1
Fibo 3 -> = 3 (??)
Fibo 4 -> = 5 (this is probably 3+1+1)
Fibo 5 -> = 7 (this is probably once again 5+1+1)
Fibo 6 -> = 9 (once again 7+1+1)
Can you please help me find out where the error lies?
I know there's been many Fibonacci-alike threads already (and I tried to follow the advice given) but I did not find any answer to my problem.
Regards, B
bash function unix recursion fibonacci
asked Nov 24 '18 at 21:35
BloodyMaryBloodyMary
547
add a comment |
I'm trying to write a recursive bash function for Fibonacci sequence. So far I have a code like below:
#!/bin/bash
Fibo() {
case $1 in
0) echo 0;;
1) echo 1;;
*) echo $[$[$Fibo $[$1-1]]+$[$Fibo $[$1-2]]] ;;
esac
}
for (( i=0; i<=$1; i++ )); do
Fibo $i
done
Here's a sample output:
0
1
1
3
5
7
9
11
Of course, it should be 0 1 1 2 3 5 8 13 ...
I tried to write it down in the paper and in my opinion it should work. My code understanding:
Fibo 0 -> = 0
Fibo 1 -> = 1
Fibo 2 -> Fibo 1 + Fibo 0 = 1 + 0 = 1
Fibo 3 -> Fibo 2 + Fibo 1 = 1 + 1 = 2
Fibo 4 -> Fibo 3 + Fibo 2 = 2 + 1 = 3
Fibo 5 -> Fibo 4 + Fibo 3 = 3 + 2 = 5
etc...
What the code "is doing":
Fibo 0 -> = 0
Fibo 1 -> = 1
Fibo 2 -> = 1 + 0 = 1
Fibo 3 -> = 3 (??)
Fibo 4 -> = 5 (this is probably 3+1+1)
Fibo 5 -> = 7 (this is probably once again 5+1+1)
Fibo 6 -> = 9 (once again 7+1+1)
Can you please help me find out where the error lies?
I know there's been many Fibonacci-alike threads already (and I tried to follow the advice given) but I did not find any answer to my problem.
Regards, B
bash function unix recursion fibonacci
asked Nov 24 '18 at 21:35
BloodyMaryBloodyMary
547
add a comment |
I'm trying to write a recursive bash function for Fibonacci sequence. So far I have a code like below:
#!/bin/bash
Fibo() {
case $1 in
0) echo 0;;
1) echo 1;;
*) echo $[$[$Fibo $[$1-1]]+$[$Fibo $[$1-2]]] ;;
esac
}
for (( i=0; i<=$1; i++ )); do
Fibo $i
done
Here's a sample output:
0
1
1
3
5
7
9
11
Of course, it should be 0 1 1 2 3 5 8 13 ...
I tried to write it down in the paper and in my opinion it should work. My code understanding:
Fibo 0 -> = 0
Fibo 1 -> = 1
Fibo 2 -> Fibo 1 + Fibo 0 = 1 + 0 = 1
Fibo 3 -> Fibo 2 + Fibo 1 = 1 + 1 = 2
Fibo 4 -> Fibo 3 + Fibo 2 = 2 + 1 = 3
Fibo 5 -> Fibo 4 + Fibo 3 = 3 + 2 = 5
etc...
What the code "is doing":
Fibo 0 -> = 0
Fibo 1 -> = 1
Fibo 2 -> = 1 + 0 = 1
Fibo 3 -> = 3 (??)
Fibo 4 -> = 5 (this is probably 3+1+1)
Fibo 5 -> = 7 (this is probably once again 5+1+1)
Fibo 6 -> = 9 (once again 7+1+1)
Can you please help me find out where the error lies?
I know there's been many Fibonacci-alike threads already (and I tried to follow the advice given) but I did not find any answer to my problem.
Regards, B
bash function unix recursion fibonacci
asked Nov 24 '18 at 21:35
BloodyMaryBloodyMary
547
I'm trying to write a recursive bash function for Fibonacci sequence. So far I have a code like below:
#!/bin/bash
Fibo() {
case $1 in
0) echo 0;;
1) echo 1;;
*) echo $[$[$Fibo $[$1-1]]+$[$Fibo $[$1-2]]] ;;
esac
}
for (( i=0; i<=$1; i++ )); do
Fibo $i
done
Here's a sample output:
0
1
1
3
5
7
9
11
Of course, it should be 0 1 1 2 3 5 8 13 ...
I tried to write it down in the paper and in my opinion it should work. My code understanding:
Fibo 0 -> = 0
Fibo 1 -> = 1
Fibo 2 -> Fibo 1 + Fibo 0 = 1 + 0 = 1
Fibo 3 -> Fibo 2 + Fibo 1 = 1 + 1 = 2
Fibo 4 -> Fibo 3 + Fibo 2 = 2 + 1 = 3
Fibo 5 -> Fibo 4 + Fibo 3 = 3 + 2 = 5
etc...
What the code "is doing":
Fibo 0 -> = 0
Fibo 1 -> = 1
Fibo 2 -> = 1 + 0 = 1
Fibo 3 -> = 3 (??)
Fibo 4 -> = 5 (this is probably 3+1+1)
Fibo 5 -> = 7 (this is probably once again 5+1+1)
Fibo 6 -> = 9 (once again 7+1+1)
Can you please help me find out where the error lies?
I know there's been many Fibonacci-alike threads already (and I tried to follow the advice given) but I did not find any answer to my problem.
Regards, B
bash function unix recursion fibonacci
bash function unix recursion fibonacci
asked Nov 24 '18 at 21:35
BloodyMaryBloodyMary
547
asked Nov 24 '18 at 21:35
BloodyMaryBloodyMary
547
asked Nov 24 '18 at 21:35
BloodyMaryBloodyMary
547
asked Nov 24 '18 at 21:35
BloodyMaryBloodyMary
547
asked Nov 24 '18 at 21:35
BloodyMaryBloodyMary
547
547
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Here's an easier way to reproduce your problem:
#!/bin/bash
foo() { echo 42; }
echo $[$foo 1]
This prints 1.
This is because $foo is not a valid variable (though there is a function by this name) so it's replaced with nothing. echo $[ 1] naturally prints 1.
To expand to the result of a function, use $(command substitution):
#!/bin/bash
foo() { echo 42; }
echo $[$(foo 1)]
This prints 42. Here it is applied to your code:
#!/bin/bash
Fibo() {
case $1 in
0) echo 0;;
1) echo 1;;
*) echo $[$[$(Fibo $[$1-1])]+$[$(Fibo $[$1-2])]] ;;
esac
}
for (( i=0; i<=$1; i++ )); do
Fibo $i
done
Result:
$ ./foo 10
0
1
1
2
3
5
8
13
21
34
55
answered Nov 24 '18 at 21:43
that other guythat other guy
74.1k885124
Thanks for explaining how to get the result of a bash function. I didn't know that I'm doing it in wrong way. I followed your advice and now my code works fine (but it's slightly different than yours :) - "echo $(( $[$(Fibo $(($1-1)))] + $[$(Fibo $(($1-2)))] )) ;;". Thanks again!
– BloodyMary
Nov 24 '18 at 22:03
@BloodyMary Yours is better since$((..))is the modern, standard way of doing shell arithmetic evaluation. I only used$[..]because your original example did and cringed a little the whole time.
– that other guy
Nov 24 '18 at 22:20
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Here's an easier way to reproduce your problem:
#!/bin/bash
foo() { echo 42; }
echo $[$foo 1]
This prints 1.
This is because $foo is not a valid variable (though there is a function by this name) so it's replaced with nothing. echo $[ 1] naturally prints 1.
To expand to the result of a function, use $(command substitution):
#!/bin/bash
foo() { echo 42; }
echo $[$(foo 1)]
This prints 42. Here it is applied to your code:
#!/bin/bash
Fibo() {
case $1 in
0) echo 0;;
1) echo 1;;
*) echo $[$[$(Fibo $[$1-1])]+$[$(Fibo $[$1-2])]] ;;
esac
}
for (( i=0; i<=$1; i++ )); do
Fibo $i
done
Result:
$ ./foo 10
0
1
1
2
3
5
8
13
21
34
55
answered Nov 24 '18 at 21:43
that other guythat other guy
74.1k885124
Thanks for explaining how to get the result of a bash function. I didn't know that I'm doing it in wrong way. I followed your advice and now my code works fine (but it's slightly different than yours :) - "echo $(( $[$(Fibo $(($1-1)))] + $[$(Fibo $(($1-2)))] )) ;;". Thanks again!
– BloodyMary
Nov 24 '18 at 22:03
@BloodyMary Yours is better since$((..))is the modern, standard way of doing shell arithmetic evaluation. I only used$[..]because your original example did and cringed a little the whole time.
– that other guy
Nov 24 '18 at 22:20
add a comment |
Here's an easier way to reproduce your problem:
#!/bin/bash
foo() { echo 42; }
echo $[$foo 1]
This prints 1.
This is because $foo is not a valid variable (though there is a function by this name) so it's replaced with nothing. echo $[ 1] naturally prints 1.
To expand to the result of a function, use $(command substitution):
#!/bin/bash
foo() { echo 42; }
echo $[$(foo 1)]
This prints 42. Here it is applied to your code:
#!/bin/bash
Fibo() {
case $1 in
0) echo 0;;
1) echo 1;;
*) echo $[$[$(Fibo $[$1-1])]+$[$(Fibo $[$1-2])]] ;;
esac
}
for (( i=0; i<=$1; i++ )); do
Fibo $i
done
Result:
$ ./foo 10
0
1
1
2
3
5
8
13
21
34
55
answered Nov 24 '18 at 21:43
that other guythat other guy
74.1k885124
Thanks for explaining how to get the result of a bash function. I didn't know that I'm doing it in wrong way. I followed your advice and now my code works fine (but it's slightly different than yours :) - "echo $(( $[$(Fibo $(($1-1)))] + $[$(Fibo $(($1-2)))] )) ;;". Thanks again!
– BloodyMary
Nov 24 '18 at 22:03
@BloodyMary Yours is better since$((..))is the modern, standard way of doing shell arithmetic evaluation. I only used$[..]because your original example did and cringed a little the whole time.
– that other guy
Nov 24 '18 at 22:20
add a comment |
Here's an easier way to reproduce your problem:
#!/bin/bash
foo() { echo 42; }
echo $[$foo 1]
This prints 1.
This is because $foo is not a valid variable (though there is a function by this name) so it's replaced with nothing. echo $[ 1] naturally prints 1.
To expand to the result of a function, use $(command substitution):
#!/bin/bash
foo() { echo 42; }
echo $[$(foo 1)]
This prints 42. Here it is applied to your code:
#!/bin/bash
Fibo() {
case $1 in
0) echo 0;;
1) echo 1;;
*) echo $[$[$(Fibo $[$1-1])]+$[$(Fibo $[$1-2])]] ;;
esac
}
for (( i=0; i<=$1; i++ )); do
Fibo $i
done
Result:
$ ./foo 10
0
1
1
2
3
5
8
13
21
34
55
answered Nov 24 '18 at 21:43
that other guythat other guy
74.1k885124
Here's an easier way to reproduce your problem:
#!/bin/bash
foo() { echo 42; }
echo $[$foo 1]
This prints 1.
This is because $foo is not a valid variable (though there is a function by this name) so it's replaced with nothing. echo $[ 1] naturally prints 1.
To expand to the result of a function, use $(command substitution):
#!/bin/bash
foo() { echo 42; }
echo $[$(foo 1)]
This prints 42. Here it is applied to your code:
#!/bin/bash
Fibo() {
case $1 in
0) echo 0;;
1) echo 1;;
*) echo $[$[$(Fibo $[$1-1])]+$[$(Fibo $[$1-2])]] ;;
esac
}
for (( i=0; i<=$1; i++ )); do
Fibo $i
done
Result:
$ ./foo 10
0
1
1
2
3
5
8
13
21
34
55
answered Nov 24 '18 at 21:43
that other guythat other guy
74.1k885124
answered Nov 24 '18 at 21:43
that other guythat other guy
74.1k885124
answered Nov 24 '18 at 21:43
that other guythat other guy
74.1k885124
answered Nov 24 '18 at 21:43
that other guythat other guy
74.1k885124
74.1k885124
Thanks for explaining how to get the result of a bash function. I didn't know that I'm doing it in wrong way. I followed your advice and now my code works fine (but it's slightly different than yours :) - "echo $(( $[$(Fibo $(($1-1)))] + $[$(Fibo $(($1-2)))] )) ;;". Thanks again!
– BloodyMary
Nov 24 '18 at 22:03
@BloodyMary Yours is better since$((..))is the modern, standard way of doing shell arithmetic evaluation. I only used$[..]because your original example did and cringed a little the whole time.
– that other guy
Nov 24 '18 at 22:20
add a comment |
Thanks for explaining how to get the result of a bash function. I didn't know that I'm doing it in wrong way. I followed your advice and now my code works fine (but it's slightly different than yours :) - "echo $(( $[$(Fibo $(($1-1)))] + $[$(Fibo $(($1-2)))] )) ;;". Thanks again!
– BloodyMary
Nov 24 '18 at 22:03
@BloodyMary Yours is better since$((..))is the modern, standard way of doing shell arithmetic evaluation. I only used$[..]because your original example did and cringed a little the whole time.
– that other guy
Nov 24 '18 at 22:20
Thanks for explaining how to get the result of a bash function. I didn't know that I'm doing it in wrong way. I followed your advice and now my code works fine (but it's slightly different than yours :) - "echo $(( $[$(Fibo $(($1-1)))] + $[$(Fibo $(($1-2)))] )) ;;". Thanks again!
– BloodyMary
Nov 24 '18 at 22:03
Thanks for explaining how to get the result of a bash function. I didn't know that I'm doing it in wrong way. I followed your advice and now my code works fine (but it's slightly different than yours :) - "echo $(( $[$(Fibo $(($1-1)))] + $[$(Fibo $(($1-2)))] )) ;;". Thanks again!
– BloodyMary
Nov 24 '18 at 22:03
@BloodyMary Yours is better since
$((..)) is the modern, standard way of doing shell arithmetic evaluation. I only used $[..] because your original example did and cringed a little the whole time.– that other guy
Nov 24 '18 at 22:20
@BloodyMary Yours is better since
$((..)) is the modern, standard way of doing shell arithmetic evaluation. I only used $[..] because your original example did and cringed a little the whole time.– that other guy
Nov 24 '18 at 22:20
add a comment |
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