Length of a sequence of numbers using seq in shell

Question

I am new to shell scripting and I am trying a simple task of getting the length of a sequence of numbers generated using seq.

With the help of a related post here: How to find the array length in unix shell? I was able to do this -

a=(1 2 3 4 5)
echo ${#a[@]}  #length of a
5              #length of a = 5 (This is fine !!)

However when I try to do a similar thing using seq ..

b=$(seq 1 1 10)
echo $b
1 2 3 4 5 6 7 8 9 10
echo ${#b[@]}
1              #the length of b is 1, while I expect it to be 10

Why does this happen ? Are the variable types a and b different? is b not an array ?

I am sure I am missing something very trivial here, help is greatly appreciated.

Thanks

Ashwin

Answer 1

You need to store the output in an array to find the length of the array:

$ b=($(seq 1 1 10))
$ echo ${#b[@]}
10

Saying b=$(seq 1 1 10) doesn't produce an array.

Answer 2

Try

echo ${b[0]}

It will be 1 2 3 4 5 6 7 8 9 10 because all your values are stored in first element of array a as a string.

b=($(seq 1 1 10)) 

will do what you want.