Bash for loop syntax

Question

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.

Not sure about the explanation for this But when you use {0..7} this gets expanded to 1 2 3 4 5 ... etc But when you you {0..${#letters[*]} this expands to a string "{0..7}" Which you can not iterate on — YMeC
– YMeC, Commented Feb 10, 2017 at 15:17

choroba · Accepted Answer · 2017-02-10 15:19:31Z

5

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.

edited Feb 10, 2017 at 15:19

answered Feb 10, 2017 at 15:13

choroba

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Comments

Tom Fenech · Accepted Answer · 2017-02-10 15:32:58Z

3

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[@]}.

edited Feb 10, 2017 at 15:32

answered Feb 10, 2017 at 15:13

Tom Fenech

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1 Comment

chepner Over a year ago

These also accommodate sparse arrays; you can't always assume that an array defines a value for every index between 0 and ${#letters[@]}.

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