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Why doesn't echo $1 print $1 in this simple bash script? 

#!/bin/bash
# function.sh
print_something () {
echo $1
}
print_something

$ ./function.sh 123  -> why doesn't it print '123' as a result?
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  • 6
    because you forgot the $1 when calling print_something. Please try to include text in the question and not in the title. Commented Aug 27, 2018 at 8:45
  • thanks. What should be the echo argument if I need to insert in the script the commands: print_something "$1"; print_something "$2"; and maybe more? Commented Aug 27, 2018 at 9:26
  • 13
    When I saw this in HNQ I thought you were printing dollar-bills and hoped to read some juicy story about how your printer detects that you're printing fake money. Commented Aug 27, 2018 at 13:59
  • @pipe I've had that happen to me before, for some reason it refuses to print anything not just the ones with the anti-printing preventions. Commented Aug 27, 2018 at 16:40

2 Answers 2

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43

Positional parameters refer to the script's arguments in the main level of the script, but to function arguments in function body. So

print_something Something

would actually print Something.

If you want to pass the script's arguments to a function, you must do that explicitly. Use

print_something "$1"

to pass the first argument, or

print_something "$@"

to pass all of them, though the function in example only uses the first one.

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  • 1
    Passing "$@" to print_something, as it's currently written, would still only print the first of the arguments though. Commented Aug 27, 2018 at 9:21
  • 13
    But the point was to show how to pass all arguments, I presume. The fact that the function, as it stands, only uses the first of its arguments is a bit irrelevant. Commented Aug 27, 2018 at 9:36
  • Well, just thinking there's no point in passing all the arguments along if only the first one is being used. Commented Aug 27, 2018 at 9:50
  • 15
    @allo No. "$*" would be a single string (joined on the first character of $IFS) while "$@" would be a list of individually quoted elements. Commented Aug 27, 2018 at 11:41
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    @Kusalananda the point of telling someone who wants to pass the command-line parameters to a function to use "$@", even if in this case there is only one such parameter, is to cover all such cases. If the OP decides to add a second parameter, there's nothing to change in the function invocation. And everyone else who reads this will learn the right way to do it to avoid having to re-do it later, too. Commented Aug 27, 2018 at 18:35
11

This is because a called function gets its own set of positional parameters, independent of the parent's / caller's set. Try 

print_something "$1"

(and echo "$1", or even better printf '%s\n' "$1", remember to quote parameter expansions and that echo can't be used for arbitrary data).

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2
  • You need to become clearer what you are talking about. The caller's $1 is generally different from the function's $1, although they CAN become the same if used like proposed above. If I get you right, the echo can stay the same (echo $1) when the function is called with single parameters (print_something $2 takes the caller's $1 and "makes" it $1 inside the function) Commented Aug 27, 2018 at 9:07
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    Using echo $1 doesn't make sense unless you want $1 to be treated as a $IFS-delimited list of file patterns to expand. echo "$1" would make more sense, though would not output the content of $1 for values of $1 like -nene, -EE... Commented Aug 27, 2018 at 9:16

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