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 co

Answer 1

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[@]}.

Answer 2

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.