What does the last “-” (hyphen) mean in options of `bash`?
In this tutorial we need to execute the following command:
# curl -sL https://rpm.nodesource.com/setup_6.x | sudo -E bash -
What does the last -
(hyphen) after bash
mean?
I've seen a lot of commands with this, and couldn't find myself a logical explanation and neither find how to reformulate a google search for it. Is it the output of the piped command?
linux command-line bash pipe
add a comment |
In this tutorial we need to execute the following command:
# curl -sL https://rpm.nodesource.com/setup_6.x | sudo -E bash -
What does the last -
(hyphen) after bash
mean?
I've seen a lot of commands with this, and couldn't find myself a logical explanation and neither find how to reformulate a google search for it. Is it the output of the piped command?
linux command-line bash pipe
add a comment |
In this tutorial we need to execute the following command:
# curl -sL https://rpm.nodesource.com/setup_6.x | sudo -E bash -
What does the last -
(hyphen) after bash
mean?
I've seen a lot of commands with this, and couldn't find myself a logical explanation and neither find how to reformulate a google search for it. Is it the output of the piped command?
linux command-line bash pipe
In this tutorial we need to execute the following command:
# curl -sL https://rpm.nodesource.com/setup_6.x | sudo -E bash -
What does the last -
(hyphen) after bash
mean?
I've seen a lot of commands with this, and couldn't find myself a logical explanation and neither find how to reformulate a google search for it. Is it the output of the piped command?
linux command-line bash pipe
linux command-line bash pipe
edited 22 mins ago
Kamil Maciorowski
24.1k155176
24.1k155176
asked 1 hour ago
Omar BISTAMI
255
255
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
In man bash
, at the end of the single-character options there is:-
-- A -- signals the end of options and disables further option processing.
Any arguments after the -- are treated as filenames and arguments. An
argument of - is equivalent to --.
If you have quoted the complete command, I can see no reason to use -
after bash
in this instance, but it does no harm.
Thank you for your answer, Yes i have quoted the complete command. so anything after - or -- will not be seen as an option but as a file name or arguments, could you please give an example where it is useful ?
– Omar BISTAMI
1 hour ago
It's really to allow for a script whose name begins-
, an unlikely requirement, but-
/--
makes it possible.
– AFH
38 mins ago
add a comment |
Bash behaves in somewhat non-standard way when it comes to -
.
POSIX says:
Guideline 10:
The first--
argument that is not an option-argument should be accepted as a delimiter indicating the end of options. Any following arguments should be treated as operands, even if they begin with the-
character.
[…]
Guideline 13:
For utilities that use operands to represent files to be opened for either reading or writing, the-
operand should be used to mean only standard input (or standard output when it is clear from context that an output file is being specified) or a file named-
.
And
Where a utility described in the Shell and Utilities volume of POSIX.1-2017 as conforming to these guidelines is required to accept, or not to accept, the operand
-
to mean standard input or output, this usage is explained in the OPERANDS section. Otherwise, if such a utility uses operands to represent files, it is implementation-defined whether the operand-
stands for standard input (or standard output), or for a file named-
.
But then man 1 bash
reads:
A
--
signals the end of options and disables further option processing. Any arguments after the--
are treated as filenames and arguments. An argument of-
is equivalent to--
.
So for Bash -
means neither standard input nor a file, hence somewhat non-standard.
Now your particular case:
curl -sL https://rpm.nodesource.com/setup_6.x | sudo -E bash -
I suspect the author of this command may not realize -
is equivalent to --
in this case. I suspect the author wanted to make sure bash
will read from its standard input, they expected -
to work according to the guideline 13.
But even if it worked according to the guideline, -
would be unnecessary here because bash
detects when its standard input is a pipe and acts accordingly (unless -c
is given etc.).
Yet -
doesn't work according to the guideline, it works like --
. Still --
is unnecessary here because there are no arguments after it.
In my opinion the last -
changes nothing. The command would work without it.
To see how --
and -
can be useful in general, study the example below.
cat
in my Kubuntu obeys both guidelines and I will use it to demonstrate usefulness of -
and --
.
Let a file named foo
exist. This will print the file:
cat foo
Let a file named --help
exist. This won't print the file:
cat --help
But this will print the file named --help
:
cat -- --help
This will concatenate the file named --help
with whatever comes from the standard input:
cat -- --help -
It seems you don't really need --
, because you can always pass ./--help
which will be interpreted as a file for sure. But consider
cat "$file"
when you don't know beforehand what the content of the variable is. You cannot just prepend ./
to it, because it may be an absolute path and ./
would break it. On the other hand it may be a file named --help
(because why not?). In this case --
is very useful; this is a lot more robust command:
cat -- "$file"
Thank your for your answer , is there any examples that can help explain it more in bash context ?
– Omar BISTAMI
1 hour ago
@OmarBISTAMI I have expanded my answer.
– Kamil Maciorowski
45 mins ago
add a comment |
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In man bash
, at the end of the single-character options there is:-
-- A -- signals the end of options and disables further option processing.
Any arguments after the -- are treated as filenames and arguments. An
argument of - is equivalent to --.
If you have quoted the complete command, I can see no reason to use -
after bash
in this instance, but it does no harm.
Thank you for your answer, Yes i have quoted the complete command. so anything after - or -- will not be seen as an option but as a file name or arguments, could you please give an example where it is useful ?
– Omar BISTAMI
1 hour ago
It's really to allow for a script whose name begins-
, an unlikely requirement, but-
/--
makes it possible.
– AFH
38 mins ago
add a comment |
In man bash
, at the end of the single-character options there is:-
-- A -- signals the end of options and disables further option processing.
Any arguments after the -- are treated as filenames and arguments. An
argument of - is equivalent to --.
If you have quoted the complete command, I can see no reason to use -
after bash
in this instance, but it does no harm.
Thank you for your answer, Yes i have quoted the complete command. so anything after - or -- will not be seen as an option but as a file name or arguments, could you please give an example where it is useful ?
– Omar BISTAMI
1 hour ago
It's really to allow for a script whose name begins-
, an unlikely requirement, but-
/--
makes it possible.
– AFH
38 mins ago
add a comment |
In man bash
, at the end of the single-character options there is:-
-- A -- signals the end of options and disables further option processing.
Any arguments after the -- are treated as filenames and arguments. An
argument of - is equivalent to --.
If you have quoted the complete command, I can see no reason to use -
after bash
in this instance, but it does no harm.
In man bash
, at the end of the single-character options there is:-
-- A -- signals the end of options and disables further option processing.
Any arguments after the -- are treated as filenames and arguments. An
argument of - is equivalent to --.
If you have quoted the complete command, I can see no reason to use -
after bash
in this instance, but it does no harm.
answered 1 hour ago
AFH
13.9k31938
13.9k31938
Thank you for your answer, Yes i have quoted the complete command. so anything after - or -- will not be seen as an option but as a file name or arguments, could you please give an example where it is useful ?
– Omar BISTAMI
1 hour ago
It's really to allow for a script whose name begins-
, an unlikely requirement, but-
/--
makes it possible.
– AFH
38 mins ago
add a comment |
Thank you for your answer, Yes i have quoted the complete command. so anything after - or -- will not be seen as an option but as a file name or arguments, could you please give an example where it is useful ?
– Omar BISTAMI
1 hour ago
It's really to allow for a script whose name begins-
, an unlikely requirement, but-
/--
makes it possible.
– AFH
38 mins ago
Thank you for your answer, Yes i have quoted the complete command. so anything after - or -- will not be seen as an option but as a file name or arguments, could you please give an example where it is useful ?
– Omar BISTAMI
1 hour ago
Thank you for your answer, Yes i have quoted the complete command. so anything after - or -- will not be seen as an option but as a file name or arguments, could you please give an example where it is useful ?
– Omar BISTAMI
1 hour ago
It's really to allow for a script whose name begins
-
, an unlikely requirement, but -
/--
makes it possible.– AFH
38 mins ago
It's really to allow for a script whose name begins
-
, an unlikely requirement, but -
/--
makes it possible.– AFH
38 mins ago
add a comment |
Bash behaves in somewhat non-standard way when it comes to -
.
POSIX says:
Guideline 10:
The first--
argument that is not an option-argument should be accepted as a delimiter indicating the end of options. Any following arguments should be treated as operands, even if they begin with the-
character.
[…]
Guideline 13:
For utilities that use operands to represent files to be opened for either reading or writing, the-
operand should be used to mean only standard input (or standard output when it is clear from context that an output file is being specified) or a file named-
.
And
Where a utility described in the Shell and Utilities volume of POSIX.1-2017 as conforming to these guidelines is required to accept, or not to accept, the operand
-
to mean standard input or output, this usage is explained in the OPERANDS section. Otherwise, if such a utility uses operands to represent files, it is implementation-defined whether the operand-
stands for standard input (or standard output), or for a file named-
.
But then man 1 bash
reads:
A
--
signals the end of options and disables further option processing. Any arguments after the--
are treated as filenames and arguments. An argument of-
is equivalent to--
.
So for Bash -
means neither standard input nor a file, hence somewhat non-standard.
Now your particular case:
curl -sL https://rpm.nodesource.com/setup_6.x | sudo -E bash -
I suspect the author of this command may not realize -
is equivalent to --
in this case. I suspect the author wanted to make sure bash
will read from its standard input, they expected -
to work according to the guideline 13.
But even if it worked according to the guideline, -
would be unnecessary here because bash
detects when its standard input is a pipe and acts accordingly (unless -c
is given etc.).
Yet -
doesn't work according to the guideline, it works like --
. Still --
is unnecessary here because there are no arguments after it.
In my opinion the last -
changes nothing. The command would work without it.
To see how --
and -
can be useful in general, study the example below.
cat
in my Kubuntu obeys both guidelines and I will use it to demonstrate usefulness of -
and --
.
Let a file named foo
exist. This will print the file:
cat foo
Let a file named --help
exist. This won't print the file:
cat --help
But this will print the file named --help
:
cat -- --help
This will concatenate the file named --help
with whatever comes from the standard input:
cat -- --help -
It seems you don't really need --
, because you can always pass ./--help
which will be interpreted as a file for sure. But consider
cat "$file"
when you don't know beforehand what the content of the variable is. You cannot just prepend ./
to it, because it may be an absolute path and ./
would break it. On the other hand it may be a file named --help
(because why not?). In this case --
is very useful; this is a lot more robust command:
cat -- "$file"
Thank your for your answer , is there any examples that can help explain it more in bash context ?
– Omar BISTAMI
1 hour ago
@OmarBISTAMI I have expanded my answer.
– Kamil Maciorowski
45 mins ago
add a comment |
Bash behaves in somewhat non-standard way when it comes to -
.
POSIX says:
Guideline 10:
The first--
argument that is not an option-argument should be accepted as a delimiter indicating the end of options. Any following arguments should be treated as operands, even if they begin with the-
character.
[…]
Guideline 13:
For utilities that use operands to represent files to be opened for either reading or writing, the-
operand should be used to mean only standard input (or standard output when it is clear from context that an output file is being specified) or a file named-
.
And
Where a utility described in the Shell and Utilities volume of POSIX.1-2017 as conforming to these guidelines is required to accept, or not to accept, the operand
-
to mean standard input or output, this usage is explained in the OPERANDS section. Otherwise, if such a utility uses operands to represent files, it is implementation-defined whether the operand-
stands for standard input (or standard output), or for a file named-
.
But then man 1 bash
reads:
A
--
signals the end of options and disables further option processing. Any arguments after the--
are treated as filenames and arguments. An argument of-
is equivalent to--
.
So for Bash -
means neither standard input nor a file, hence somewhat non-standard.
Now your particular case:
curl -sL https://rpm.nodesource.com/setup_6.x | sudo -E bash -
I suspect the author of this command may not realize -
is equivalent to --
in this case. I suspect the author wanted to make sure bash
will read from its standard input, they expected -
to work according to the guideline 13.
But even if it worked according to the guideline, -
would be unnecessary here because bash
detects when its standard input is a pipe and acts accordingly (unless -c
is given etc.).
Yet -
doesn't work according to the guideline, it works like --
. Still --
is unnecessary here because there are no arguments after it.
In my opinion the last -
changes nothing. The command would work without it.
To see how --
and -
can be useful in general, study the example below.
cat
in my Kubuntu obeys both guidelines and I will use it to demonstrate usefulness of -
and --
.
Let a file named foo
exist. This will print the file:
cat foo
Let a file named --help
exist. This won't print the file:
cat --help
But this will print the file named --help
:
cat -- --help
This will concatenate the file named --help
with whatever comes from the standard input:
cat -- --help -
It seems you don't really need --
, because you can always pass ./--help
which will be interpreted as a file for sure. But consider
cat "$file"
when you don't know beforehand what the content of the variable is. You cannot just prepend ./
to it, because it may be an absolute path and ./
would break it. On the other hand it may be a file named --help
(because why not?). In this case --
is very useful; this is a lot more robust command:
cat -- "$file"
Thank your for your answer , is there any examples that can help explain it more in bash context ?
– Omar BISTAMI
1 hour ago
@OmarBISTAMI I have expanded my answer.
– Kamil Maciorowski
45 mins ago
add a comment |
Bash behaves in somewhat non-standard way when it comes to -
.
POSIX says:
Guideline 10:
The first--
argument that is not an option-argument should be accepted as a delimiter indicating the end of options. Any following arguments should be treated as operands, even if they begin with the-
character.
[…]
Guideline 13:
For utilities that use operands to represent files to be opened for either reading or writing, the-
operand should be used to mean only standard input (or standard output when it is clear from context that an output file is being specified) or a file named-
.
And
Where a utility described in the Shell and Utilities volume of POSIX.1-2017 as conforming to these guidelines is required to accept, or not to accept, the operand
-
to mean standard input or output, this usage is explained in the OPERANDS section. Otherwise, if such a utility uses operands to represent files, it is implementation-defined whether the operand-
stands for standard input (or standard output), or for a file named-
.
But then man 1 bash
reads:
A
--
signals the end of options and disables further option processing. Any arguments after the--
are treated as filenames and arguments. An argument of-
is equivalent to--
.
So for Bash -
means neither standard input nor a file, hence somewhat non-standard.
Now your particular case:
curl -sL https://rpm.nodesource.com/setup_6.x | sudo -E bash -
I suspect the author of this command may not realize -
is equivalent to --
in this case. I suspect the author wanted to make sure bash
will read from its standard input, they expected -
to work according to the guideline 13.
But even if it worked according to the guideline, -
would be unnecessary here because bash
detects when its standard input is a pipe and acts accordingly (unless -c
is given etc.).
Yet -
doesn't work according to the guideline, it works like --
. Still --
is unnecessary here because there are no arguments after it.
In my opinion the last -
changes nothing. The command would work without it.
To see how --
and -
can be useful in general, study the example below.
cat
in my Kubuntu obeys both guidelines and I will use it to demonstrate usefulness of -
and --
.
Let a file named foo
exist. This will print the file:
cat foo
Let a file named --help
exist. This won't print the file:
cat --help
But this will print the file named --help
:
cat -- --help
This will concatenate the file named --help
with whatever comes from the standard input:
cat -- --help -
It seems you don't really need --
, because you can always pass ./--help
which will be interpreted as a file for sure. But consider
cat "$file"
when you don't know beforehand what the content of the variable is. You cannot just prepend ./
to it, because it may be an absolute path and ./
would break it. On the other hand it may be a file named --help
(because why not?). In this case --
is very useful; this is a lot more robust command:
cat -- "$file"
Bash behaves in somewhat non-standard way when it comes to -
.
POSIX says:
Guideline 10:
The first--
argument that is not an option-argument should be accepted as a delimiter indicating the end of options. Any following arguments should be treated as operands, even if they begin with the-
character.
[…]
Guideline 13:
For utilities that use operands to represent files to be opened for either reading or writing, the-
operand should be used to mean only standard input (or standard output when it is clear from context that an output file is being specified) or a file named-
.
And
Where a utility described in the Shell and Utilities volume of POSIX.1-2017 as conforming to these guidelines is required to accept, or not to accept, the operand
-
to mean standard input or output, this usage is explained in the OPERANDS section. Otherwise, if such a utility uses operands to represent files, it is implementation-defined whether the operand-
stands for standard input (or standard output), or for a file named-
.
But then man 1 bash
reads:
A
--
signals the end of options and disables further option processing. Any arguments after the--
are treated as filenames and arguments. An argument of-
is equivalent to--
.
So for Bash -
means neither standard input nor a file, hence somewhat non-standard.
Now your particular case:
curl -sL https://rpm.nodesource.com/setup_6.x | sudo -E bash -
I suspect the author of this command may not realize -
is equivalent to --
in this case. I suspect the author wanted to make sure bash
will read from its standard input, they expected -
to work according to the guideline 13.
But even if it worked according to the guideline, -
would be unnecessary here because bash
detects when its standard input is a pipe and acts accordingly (unless -c
is given etc.).
Yet -
doesn't work according to the guideline, it works like --
. Still --
is unnecessary here because there are no arguments after it.
In my opinion the last -
changes nothing. The command would work without it.
To see how --
and -
can be useful in general, study the example below.
cat
in my Kubuntu obeys both guidelines and I will use it to demonstrate usefulness of -
and --
.
Let a file named foo
exist. This will print the file:
cat foo
Let a file named --help
exist. This won't print the file:
cat --help
But this will print the file named --help
:
cat -- --help
This will concatenate the file named --help
with whatever comes from the standard input:
cat -- --help -
It seems you don't really need --
, because you can always pass ./--help
which will be interpreted as a file for sure. But consider
cat "$file"
when you don't know beforehand what the content of the variable is. You cannot just prepend ./
to it, because it may be an absolute path and ./
would break it. On the other hand it may be a file named --help
(because why not?). In this case --
is very useful; this is a lot more robust command:
cat -- "$file"
edited 46 mins ago
answered 1 hour ago
Kamil Maciorowski
24.1k155176
24.1k155176
Thank your for your answer , is there any examples that can help explain it more in bash context ?
– Omar BISTAMI
1 hour ago
@OmarBISTAMI I have expanded my answer.
– Kamil Maciorowski
45 mins ago
add a comment |
Thank your for your answer , is there any examples that can help explain it more in bash context ?
– Omar BISTAMI
1 hour ago
@OmarBISTAMI I have expanded my answer.
– Kamil Maciorowski
45 mins ago
Thank your for your answer , is there any examples that can help explain it more in bash context ?
– Omar BISTAMI
1 hour ago
Thank your for your answer , is there any examples that can help explain it more in bash context ?
– Omar BISTAMI
1 hour ago
@OmarBISTAMI I have expanded my answer.
– Kamil Maciorowski
45 mins ago
@OmarBISTAMI I have expanded my answer.
– Kamil Maciorowski
45 mins ago
add a comment |
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