Arguments passed into for loop in bash script [duplicate]

Question

I am trying to pass the argument as max limit for the for loop like this:

#!/bin/bash  for i in {1..$1} do     echo $i done 

This however returns {1..2} when called with argument 2, instead of executing the script and giving me

1 2 

Answer 1

Variable substitutions are not done inside of curly braces. You can use fixed numbers but not variables.

Brace Expansion

A sequence expression takes the form {x..y}, where x and y are either integers or single characters. ...

Brace expansion is performed before any other expansions, and any characters special to other expansions are preserved in the result. It is strictly textual. Bash does not apply any syntactic interpretation to the context of the expansion or the text between the braces.

A correctly-formed brace expansion must contain unquoted opening and closing braces, and at least one unquoted comma or a valid sequence expression. Any incorrectly formed brace expansion is left unchanged.

Try one of these alternatives:

for ((i = 1; i <= $1; i++)); do     echo $i done  # Not recommended with large sequences. for i in $(seq 1 $1); do     echo $i done 

Answer 2

This will cycle through all true arguments (a.k.a. "testo mesto" is one argument)

#cycle through all args for (( i=1; i<=$1; i++ )); do     eval arg=\$$i     echo "$arg" done 

OR

#cycle through all args for (( i=1; i<=$1; i++ )); do     echo "${!i}" done