Why isn't bash accepting any value for first element of array?

Question

So apparently arrays in bash have specials rules for what might be the first element, any explanation for this?

jgo ~$ myarray=(-n foo) && echo ${myarray[@]}
foo
jgo ~$ myarray=(-v foo) && echo ${myarray[@]}
-v foo
jgo ~$ myarray=(-a foo) && echo ${myarray[@]}
-a foo
jgo ~$ myarray=(-e foo) && echo ${myarray[@]}
foo

e.dan · Accepted Answer · 2016-08-09 08:56:46Z

4

Actually, your output should look more like this:

jgo ~$ myarray=(-n foo) && echo ${myarray[@]}
foojgo ~$ myarray=(-v foo) && echo ${myarray[@]}
-v foo
jgo ~$ myarray=(-a foo) && echo ${myarray[@]}
-a foo
jgo ~$ myarray=(-e foo) && echo ${myarray[@]}
foo

The reason is that -n and -e are valid options to echo whereas the other ones you tried aren't.

answered Aug 9, 2016 at 8:56

e.dan

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4 Comments

gogstad Over a year ago

That's correct. I noticed when I tried it on a proper linux box, the output in the question is from cygwin.

fedorqui Over a year ago

good one. So if you use printf instead of echo, the output is fine. Also if you quote the elements in the array.

Samuel Over a year ago

Or just add space before $myarray --> ` myarray=(-n foo) && echo " ${myarray[@]}"` if you still want to use echo

123 Over a year ago

@Samuel Or as there is no reason to expand the array elements separately you could use echo "${myarray[*]}" which keeps it all as a single arg to echo.

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