How to loop through the provided parameter?

Question

I have a following code:

function test() {
    for ft in "$1"; do
        echo $ft
    done
}

test 5

I am trying to loop through number 12, therefore I want ft to be 1,2,3,4,5 respectively in each loop. But it is just printing 5. I also tried {1..$1} but result is still same. Please help!

Duplicate of https://unix.stackexchange.com/questions/63931/how-to-iterate-through-a-comma-separated-list-and-execute-a-command-for-each-ent — steve, Nov 30 '19 at 15:23
Looks more like How to convert an input parameter to integer value in a for loop in bash? to me — steeldriver, Nov 30 '19 at 15:26
You are only providing one argument to your function. Looping over that would just print 5 (not 12). It's unclear what it is you are actually wanting to do. How is the sequence 1,2,3,4,5 related to 12? — Kusalananda, Nov 30 '19 at 16:14

Andy Dalton · Accepted Answer · 2019-12-01T01:32:23.250

I'm not totally sure what you're asking. Do you want to supply a parameter, and loop over the values from 1 to that value? If so, you can do this with bash (and probably other shells as well:

function test() {
    for ((i = 1; i <= $1; ++i)); do
        echo $i
    done
}

test 12
1
2
3
4
5
6
7
8
9
10
11
12

If you don't have a shell that supports the for(...) syntax, you can do the same thing with:

function test() {
    i=1

    while [ $i -le $1 ]; do
        echo $i
        i=$(expr $i + 1)
    done
}

Kusalananda · Answer 2 · 2019-12-02T11:03:34.133

You can solve this in two ways:

Give the function a single integer and let it output the integers between some start value (1) and that integer:

count_up () { seq "$1"; }

or

count_up () {
    for (( i = 1; i <= $1; ++i )); do
        echo "$i"
    done
}

or

count_up () {
    value=1

    while [ "$value" -le "$1" ]; do
        echo "$value"
    done
}

or any other number of variations.

Give the actual integers as arguments to the function using e.g. count_up {1..5} and define the function as simply
```
count_up () { printf '%s\n' "$@"; }
```
or
```
count_up () {
    for value do
        echo "$value"
    done
}
```
Although, there would not be much point in doing it this way as you could just call seq 5 or use printf '%s\n' {1..5} directly (also, the function should probably be called something like list_args instead).

How to loop through the provided parameter?

2 Answers2