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There is a difference in the output of commands that I don't understand, as stated below:

INPUT="$@"
echo "$INPUT ${#INPUT} ${#INPUT[@]} ${#@}"
# outputs: a b c 5 1 3

arr=(a b c)
echo "$arr ${#arr} ${#arr[@]}"
# outputs: a 1 3

I run a script with ./my_script.sh a b c.

I understand that echo "$arr" is dereferencing $arr to the first element, and then printing a. At the same time for the command $INPUT it prints a b c.

For the command ${#INPUT} and ${#INPUT[@]} why is the output 5 and 1 respectively. Shouldn't it be similar to output of the commands ${#arr} and ${#arr[@]}?

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1 Answer 1

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5

You assigned the array $@ to a scalar INPUT, which resulted in the string a b c, to which the length computation says 5 and the number of elements in the "array" is 1.

Perhaps you meant to assign the array as an array?

input=( "$@" )
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  • 1
    ok, So the shell aggregates the positional parameters and two spaces to output 5. Commented Dec 7, 2018 at 16:05
  • Does "$#" expand to a number or a string? Commented Dec 7, 2018 at 16:47
  • 1
    What's the practical difference? It's a "string" of bytes that corresponds to a number. You can use it in numerical comparisons. Commented Dec 7, 2018 at 16:48
  • I am trying to execute a programme with command ./my_script 1992. In the logic I do a if conditional if [[ "$#" -eq 0 || "$#" -gt 1 ]] then ` echo "Usage: error` else .... I always get only the error message, whereas echo "$#" outputs 1. Commented Dec 7, 2018 at 16:54
  • the || operators aren't valid inside [[ -- move them outside, or simplify the test: if [[ "$#" -ne 1 ]]... Commented Dec 7, 2018 at 16:55

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