This question already has answers here:
Backticks vs braces in Bash
(3 answers)
Brackets ${}, $(), $[] difference and usage in bash
(1 answer)
Closed 4 years ago.
I have two questions and could use some help understanding them.
What is the difference between ${} and $()? I understand that ()
means running command in separate shell and placing $ means passing
the value to variable. Can someone help me in understanding
this? Please correct me if I am wrong.
If we can use for ((i=0;i<10;i++)); do echo $i; done and it works fine then why can't I use it as while ((i=0;i<10;i++)); do echo $i; done? What is the difference in execution cycle for both?
The syntax is token-level, so the meaning of the dollar sign depends on the token it's in. The expression $(command) is a modern synonym for `command` which stands for command substitution; it means run command and put its output here. So
echo "Today is $(date). A fine day."
will run the date command and include its output in the argument to echo. The parentheses are unrelated to the syntax for running a command in a subshell, although they have something in common (the command substitution also runs in a separate subshell).
By contrast, ${variable} is just a disambiguation mechanism, so you can say ${var}text when you mean the contents of the variable var, followed by text (as opposed to $vartext which means the contents of the variable vartext).
The while loop expects a single argument which should evaluate to true or false (or actually multiple, where the last one's truth value is examined -- thanks Jonathan Leffler for pointing this out); when it's false, the loop is no longer executed. The for loop iterates over a list of items and binds each to a loop variable in turn; the syntax you refer to is one (rather generalized) way to express a loop over a range of arithmetic values.
A for loop like that can be rephrased as a while loop. The expression
for ((init; check; step)); do
body
done
is equivalent to
init
while check; do
body
step
done
It makes sense to keep all the loop control in one place for legibility; but as you can see when it's expressed like this, the for loop does quite a bit more than the while loop.
Of course, this syntax is Bash-specific; classic Bourne shell only has
for variable in token1 token2 ...; do
(Somewhat more elegantly, you could avoid the echo in the first example as long as you are sure that your argument string doesn't contain any % format codes:
date +'Today is %c. A fine day.'
Avoiding a process where you can is an important consideration, even though it doesn't make a lot of difference in this isolated example.)
$() means: "first evaluate this, and then evaluate the rest of the line".
Ex :
echo $(pwd)/myFile.txt
will be interpreted as
echo /my/path/myFile.txt
On the other hand ${} expands a variable.
Ex:
MY_VAR=toto
echo ${MY_VAR}/myFile.txt
will be interpreted as
echo toto/myFile.txt
Why can't I use it as bash$ while ((i=0;i<10;i++)); do echo $i; done
I'm afraid the answer is just that the bash syntax for while just isn't the same as the syntax for for.
your understanding is right. For detailed info on {} see bash ref - parameter expansion
'for' and 'while' have different syntax and offer different styles of programmer control for an iteration. Most non-asm languages offer a similar syntax.
With while, you would probably write i=0; while [ $i -lt 10 ]; do echo $i; i=$(( i + 1 )); done in essence manage everything about the iteration yourself
Related
This question already has answers here:
How to use a variable's value as another variable's name in bash [duplicate]
(6 answers)
Closed 5 years ago.
Let's say I have a variable's name stored in another variable:
myvar=123
varname=myvar
Now, I'd like to get 123 by just using $varname variable.
Is there a direct way for that? I found no such bash builtin for lookup by name, so came up with this:
function var { v="\$$1"; eval "echo "$v; }
so
var $varname # gives 123
Which doesn't look too bad in the end, but I'm wondering if I missed something more obvious.
From the man page of bash:
${!varname}
If the first character of parameter is an exclamation point, a level of
variable indirection is introduced. Bash uses the value of the variable formed from the rest of parameter as the name of the variable;
this variable is then expanded and that value is used in the rest of
the substitution, rather than the value of parameter itself. This is
known as indirect expansion.
There isn't a direct Posix-conforming syntax, only a bashism. I usually do this:
eval t="\$$varname"
This will work on any Posix shell, including those systems where bash is the login shell and /bin/sh is something smaller and faster like ash. I like bash and use it for my login shell but I avoid bashisms in command files.
Note: One problem with writing bash-specific scripts is that even if you can count on bash being installed, it could be anywhere on the path. It might be a good idea in that case to use the fully general /usr/bin/env shebang style, but note that this is still not 100% portable and has security issues.
${!varname} should do the trick
$ var="content"
$ myvar=var
$ echo ${!myvar}
content
I usually look at Advance Bash-Scripting Guide when I need to freshen up my Bash skills.
Regarding your question look at Indirect References
Notation is:
Version < 2
\$$var
Version >= 2
${!varname}
# bmuSetIndirectVar()
# TO DOUBLE CHECK THIS COMMENT AND DEMO
# This function is an helper to read indirect variables.
# i.e. get the content of a variable whose name is saved
# within an other variable. Like:
# MYDIR="/tmp"
# WHICHDIR="MYDIR"
# bmuSetIndirectVar "WHICHDIR" "$MYDIR"
#
bmuSetIndirectVar(){
tmpVarName=$1
locVarName=$1
extVarName=$2
#echo "debug Ind Input >$1< >$2<"
eval tmpVarName=\$$extVarName
#echo "debug Ind Output >$tmpVarName< >$extVarName<"
export $locVarName="${tmpVarName}"
}
I am currently using this little function. I am not fully happy with it, and I have seen different solutions on the web (if I could recall I would write them here), but it seems to work. Within these few lines there is already some redundancy and extra data but it was helpful for debugging.
If you want to see it in place, i.e. where I am using it, check:
https://github.com/mariotti/bmu/blob/master/bin/backmeup.shellfunctions.sh
Of course it is not the best solution, but made me going on with the work, in
the hope I can replace it with something a bit more general soon.
I'm working on getting accustomed to shell scripting and ran across a behavior I found interesting and unexplained. In the following code the first for loop will execute correctly but the second will not.
declare letters=(a b c d e f g)
for i in {0..7}; do
echo ${letters[i]}
done
for i in {0..${#letters[*]}}; do
echo ${letters[i]}
done
The second for loop results in the following error:
syntax error: operand expected (error token is "{0..7}")
What confuses me is that ${#letters[*]} is clearly getting evaluated, correctly, to the number 7. But despite this the code fails even though we just saw that the same loop with {0..7} works perfectly fine.
What is the reason for this?
I am running OS X 10.12.2, GNU bash version 3.2.57.
The bracket expansion happens before parameter expansion (see EXPANSIONS in man bash), therefore it works for literals only. In other words, you can't use brace expansion with variables.
You can use a C-style loop:
for ((i=0; i<${#letters[#]}; i++)) ; do
echo ${letters[i]}
done
or an external command like seq:
for i in $(seq 1 ${#letters[#]}) ; do
echo ${letters[i-1]}
done
But you usually don't need the indices, instead one loops over the elements themselves, see #TomFenech's answer below. He also shows another way of getting the list of indices.
Note that it should be {0..6}, not 7.
Brace expansion occurs before parameter expansion, so you can't use a variable as part of a range.
Expand the array into a list of values:
for letter in "${letters[#]}"; do
echo "$letter"
done
Or, expand the indices of the array into a list:
for i in ${!letters[#]}; do
echo "${letters[i]}"
done
As mentioned in the comments (thanks), these two approaches also accommodate sparse arrays; you can't always assume that an array defines a value for every index between 0 and ${#letters[#]}.
I would like to achieve this in Bash: echo $(a=1)and print the value of variable a
I test eval, $$a,{}, $() but none of them work as most of the times either I got literally a=1 or in one case (I don't remember which) it tried to execute the value.
I known that I can do: a=1;echo $a but because I'm little fun one command per line (even if sometimes is getting little complicated) I was wondering if is possible to do this either with echo or with printf
If you know that $a is previously unset, you can do this using the following syntax:
echo ${a:=1}
This, and other types of parameter expansion, are defined in the POSIX shell command language specification.
If you want to assign a numeric value, another option, which doesn't depend on the value previously being unset, would be to use an arithmetic expansion:
echo $(( a = 1 ))
This assigns the value and echoes the number that has been assigned.
It's worth mentioning that what you're trying to do cannot be done in a subshell by design because a child process cannot modify the environment of its parent.
I'm writing a script that looks through a series of directories for the presence of a file, and when found, pushes it onto the directory stack via pushd. The dirs command is insanely obnoxious, and I stumbled on the bash variable form of it's contents, $DIRSTACK
$DIRSTACK is an array of directories in the stack. It's always guaranteed to have 1 entry, the current working directory, and then pushed directories follow.
I'm attempting to iterate over the list of directories, but cannot seem to get the for-loop to accept the sequence length I'm attempting to automatically generate:
for i in {1..${#DIRSTACK[*]}}; do
echo ${DIRSTACK[$i]}
done
When executed, bash fails with the following error:
line 72: {1..2}: syntax error: operand expected (error token is "{1..2}")
I'm honestly stumped, because I've manually written for i in {1..5} in scripts a number of times without issue, and given the error message, it seems like the number of array items expansion is working exactly as I want it to.
Why is this error occurring?
Brace expansion will not work correctly if you have a parameter within it. This is because the parameter, DIRSTACK, in this case, won't be expanded until AFTER the brace has been expanded.
From the bash man page:
Brace expansion is performed before any other expansions, and any
characters special to other expansions are preserved in the result.
It is strictly textual. Bash does not apply any syntactic
interpretation to the context of the expansion or the text between the
braces.
If you simply want to loop over the array, why not use the following?
for i in "${DIRSTACK[#]}"
do
echo $i
done
Or, if you want to explicitly use the length of the array:
for (( i = 0 ; i < ${#DIRSTACK[#]} ; i++ ))
do
echo ${DIRSTACK[$i]}
done
A similar construct to a sequence expression is to use the seq(1) command.
For your specific case, you can use:
for i in $(seq ${#DIRSTACK[*]}); do
echo ${DIRSTACK[$i]}
done
However, given your comment to #dogbane's answer, this will still not do what you want, since you are still iterating the number of elements in the array, but indexing past the end.
What you want is easily achieved by using bash's substring expansion, which also works on arrays.
for dir in "${DIRSTACK[#]:1}" ; do
echo $dir
done
This question already has answers here:
How to use a variable's value as another variable's name in bash [duplicate]
(6 answers)
Closed 5 years ago.
Let's say I have a variable's name stored in another variable:
myvar=123
varname=myvar
Now, I'd like to get 123 by just using $varname variable.
Is there a direct way for that? I found no such bash builtin for lookup by name, so came up with this:
function var { v="\$$1"; eval "echo "$v; }
so
var $varname # gives 123
Which doesn't look too bad in the end, but I'm wondering if I missed something more obvious.
From the man page of bash:
${!varname}
If the first character of parameter is an exclamation point, a level of
variable indirection is introduced. Bash uses the value of the variable formed from the rest of parameter as the name of the variable;
this variable is then expanded and that value is used in the rest of
the substitution, rather than the value of parameter itself. This is
known as indirect expansion.
There isn't a direct Posix-conforming syntax, only a bashism. I usually do this:
eval t="\$$varname"
This will work on any Posix shell, including those systems where bash is the login shell and /bin/sh is something smaller and faster like ash. I like bash and use it for my login shell but I avoid bashisms in command files.
Note: One problem with writing bash-specific scripts is that even if you can count on bash being installed, it could be anywhere on the path. It might be a good idea in that case to use the fully general /usr/bin/env shebang style, but note that this is still not 100% portable and has security issues.
${!varname} should do the trick
$ var="content"
$ myvar=var
$ echo ${!myvar}
content
I usually look at Advance Bash-Scripting Guide when I need to freshen up my Bash skills.
Regarding your question look at Indirect References
Notation is:
Version < 2
\$$var
Version >= 2
${!varname}
# bmuSetIndirectVar()
# TO DOUBLE CHECK THIS COMMENT AND DEMO
# This function is an helper to read indirect variables.
# i.e. get the content of a variable whose name is saved
# within an other variable. Like:
# MYDIR="/tmp"
# WHICHDIR="MYDIR"
# bmuSetIndirectVar "WHICHDIR" "$MYDIR"
#
bmuSetIndirectVar(){
tmpVarName=$1
locVarName=$1
extVarName=$2
#echo "debug Ind Input >$1< >$2<"
eval tmpVarName=\$$extVarName
#echo "debug Ind Output >$tmpVarName< >$extVarName<"
export $locVarName="${tmpVarName}"
}
I am currently using this little function. I am not fully happy with it, and I have seen different solutions on the web (if I could recall I would write them here), but it seems to work. Within these few lines there is already some redundancy and extra data but it was helpful for debugging.
If you want to see it in place, i.e. where I am using it, check:
https://github.com/mariotti/bmu/blob/master/bin/backmeup.shellfunctions.sh
Of course it is not the best solution, but made me going on with the work, in
the hope I can replace it with something a bit more general soon.