Revisions to Can a shell script prints its argument, quoted as you would write them on the shell prompt?

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edited Aug 31, 2016 at 23:03

Gilles 'SO- stop being evil'

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occurred Aug 31, 2016 at 17:36

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edited Aug 31, 2016 at 17:09

Rahul

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In a shell script, my understanding is that $@"$@" expands to the script arguments, quoting them as needed. For instance this forwards the script arguments to gcc:

gcc -fPIC "$@"

When using the bash pass-to-stdin syntax <<< though, @$"@$" doesn't work as I would expect it to.

#!/bin/bash cat <<< "$@"

Calling the script as ./test.sh foo "bar baz" gives

foo bar baz

I would expect

foo "bar baz"

Is there a way to write a shell script that prints it's arguments as you would write them on the shell prompt? For instance: a hint as to what command to use next, including the script arguments in the hint.

In a shell script, my understanding is that $@ expands to the script arguments, quoting them as needed. For instance this forwards the script arguments to gcc:

gcc -fPIC "$@"

When using the bash pass-to-stdin syntax <<< though, @$ doesn't work as I would expect it to.

#!/bin/bash cat <<< "$@"

Calling the script as ./test.sh foo "bar baz" gives

foo bar baz

I would expect

foo "bar baz"

Is there a way to write a shell script that prints it's arguments as you would write them on the shell prompt? For instance: a hint as to what command to use next, including the script arguments in the hint.

In a shell script, my understanding is that "$@" expands to the script arguments, quoting them as needed. For instance this forwards the script arguments to gcc:

gcc -fPIC "$@"

When using the bash pass-to-stdin syntax <<< though, "@$" doesn't work as I would expect it to.

#!/bin/bash cat <<< "$@"

Calling the script as ./test.sh foo "bar baz" gives

foo bar baz

I would expect

foo "bar baz"

Is there a way to write a shell script that prints it's arguments as you would write them on the shell prompt? For instance: a hint as to what command to use next, including the script arguments in the hint.

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edited Aug 31, 2016 at 17:01

Rahul

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4
46
56

In a shell script, my understanding is that "$@"$@ expands to the script arguments, quoting them as needed. For instance this forwards the script arguments to gcc:

gcc -fPIC "$@"

When using the bash pass-to-stdin syntax <<< though, "@$"@$ doesn't work as I would expect it to.

#!/bin/bash cat <<< "$@"

Calling the script as ./test.sh foo "bar baz" gives

foo bar baz

I would expect

foo "bar baz"

Is there a way to write a shell script that prints it's arguments as you would write them on the shell prompt? For instance: a hint as to what command to use next, including the script arguments in the hint.

In a shell script, my understanding is that "$@" expands to the script arguments, quoting them as needed. For instance this forwards the script arguments to gcc:

gcc -fPIC "$@"

When using the bash pass-to-stdin syntax <<< though, "@$" doesn't work as I would expect it to.

#!/bin/bash cat <<< "$@"

Calling the script as ./test.sh foo "bar baz" gives

foo bar baz

I would expect

foo "bar baz"

Is there a way to write a shell script that prints it's arguments as you would write them on the shell prompt? For instance: a hint as to what command to use next, including the script arguments in the hint.

In a shell script, my understanding is that $@ expands to the script arguments, quoting them as needed. For instance this forwards the script arguments to gcc:

gcc -fPIC "$@"

When using the bash pass-to-stdin syntax <<< though, @$ doesn't work as I would expect it to.

#!/bin/bash cat <<< "$@"

Calling the script as ./test.sh foo "bar baz" gives

foo bar baz

I would expect

foo "bar baz"

Is there a way to write a shell script that prints it's arguments as you would write them on the shell prompt? For instance: a hint as to what command to use next, including the script arguments in the hint.