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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6because you forgot the $1 when calling print_something. Please try to include text in the question and not in the title.Rui F Ribeiro– Rui F Ribeiro2018-08-27 08:45:20 +00:00Commented Aug 27, 2018 at 8:45
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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?pietro letti– pietro letti2018-08-27 09:26:33 +00:00Commented Aug 27, 2018 at 9:26
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13When 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.pipe– pipe2018-08-27 13:59:47 +00:00Commented Aug 27, 2018 at 13:59
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@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.Codingale– Codingale2018-08-27 16:40:47 +00:00Commented Aug 27, 2018 at 16:40
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2 Answers
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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1Passing
"$@"toprint_something, as it's currently written, would still only print the first of the arguments though.2018-08-27 09:21:41 +00:00Commented Aug 27, 2018 at 9:21 -
13But the point was to show how to pass all arguments, I presume. The fact that the function, as it stands, only uses the first of its arguments is a bit irrelevant.weirdan– weirdan2018-08-27 09:36:17 +00:00Commented Aug 27, 2018 at 9:36
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Well, just thinking there's no point in passing all the arguments along if only the first one is being used.2018-08-27 09:50:36 +00:00Commented Aug 27, 2018 at 9:50
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15@allo No.
"$*"would be a single string (joined on the first character of$IFS) while"$@"would be a list of individually quoted elements.2018-08-27 11:41:01 +00:00Commented Aug 27, 2018 at 11:41 -
6@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.Monty Harder– Monty Harder2018-08-27 18:35:18 +00:00Commented Aug 27, 2018 at 18:35
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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You need to become clearer what you are talking about. The caller's
$1is generally different from the function's$1, although they CAN become the same if used like proposed above. If I get you right, theechocan stay the same (echo $1) when the function is called with single parameters (print_something $2takes the caller's $1 and "makes" it$1inside the function)RudiC– RudiC2018-08-27 09:07:33 +00:00Commented Aug 27, 2018 at 9:07 -
6Using
echo $1doesn't make sense unless you want$1to 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$1for values of$1like-nene,-EE...Stéphane Chazelas– Stéphane Chazelas2018-08-27 09:16:10 +00:00Commented Aug 27, 2018 at 9:16