status=0
$status=1
echo $status
Can anyone tell my what i am doing wrong with this?
It gives me the following error:
0=1: command not found
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4 Answers
This line is OK - it assigns the value 0 to the variable status:
status=0
This is wrong:
$status=1
By putting a $ in front of the variable name you are dereferencing it, i.e. getting its value, which in this case is 0. In other words, bash is expanding what you wrote to:
0=1
Which makes no sense, hence the error.
If your intent is to reassign a new value 1 to the status variable, then just do it the same as the original assignment:
status=1
2 Comments
chepner
0=1 makes perfect sense syntactically, but it will not do what the user intended. If an executable file named 0=1 exists in the user's path, it will be run.
Digital Trauma
@chepner - yes, very true.
Bash assignments can't have a dollar in front. Variable replacements in bash are like macro expansions in C; they occur before any parsing. For example, this horrible thing works:
foof="[ -f"
if $foof .bashrc ] ; then echo "hey"; fi
1 Comment
chepner
Not quite that hairy; parsing is done first, but
[ is not syntax (it's a command name), so it works as expected. foo=if; $foo [ -f .bashrc ]; then echo "hey"; fi would properly fail with a syntax error.Only use the $ when actually using the variable in bash. Omit it when assigning or re-assining.
e.g.
status=0
status2=1
status="$status2"
answered Dec 18, 2013 at 22:56
Reinstate Monica Please
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Comments
also this ugly thing works too :
status='a'
eval $status=1
echo $a
1
1 Comment
Digital Trauma
Or you can achieve extra levels of indirection using a parameter expansion like this:
a=1; status='a'; echo ${!status}