How do I use command-line arguments to `sh` in the `-c` command string?
I know that
sh -c 'echo $1' sh 4
will output 4.
and
sh -c 'echo $2' sh 4 5
will output 5.
But I cannot understand how the parameters after the second sh were passed to the command next to sh -c. I read the man page of both bash and dash but could not find the introduction about this kind of syntax.
command-line bash sh dash-shell
edited Jan 7 at 7:32
dessert
25.1k673106
asked Jan 7 at 6:24
gbcatgbcat
185
add a comment |
I know that
sh -c 'echo $1' sh 4
will output 4.
and
sh -c 'echo $2' sh 4 5
will output 5.
But I cannot understand how the parameters after the second sh were passed to the command next to sh -c. I read the man page of both bash and dash but could not find the introduction about this kind of syntax.
command-line bash sh dash-shell
edited Jan 7 at 7:32
dessert
25.1k673106
asked Jan 7 at 6:24
gbcatgbcat
185
add a comment |
I know that
sh -c 'echo $1' sh 4
will output 4.
and
sh -c 'echo $2' sh 4 5
will output 5.
But I cannot understand how the parameters after the second sh were passed to the command next to sh -c. I read the man page of both bash and dash but could not find the introduction about this kind of syntax.
command-line bash sh dash-shell
edited Jan 7 at 7:32
dessert
25.1k673106
asked Jan 7 at 6:24
gbcatgbcat
185
I know that
sh -c 'echo $1' sh 4
will output 4.
and
sh -c 'echo $2' sh 4 5
will output 5.
But I cannot understand how the parameters after the second sh were passed to the command next to sh -c. I read the man page of both bash and dash but could not find the introduction about this kind of syntax.
command-line bash sh dash-shell
command-line bash sh dash-shell
edited Jan 7 at 7:32
dessert
25.1k673106
asked Jan 7 at 6:24
gbcatgbcat
185
edited Jan 7 at 7:32
dessert
25.1k673106
asked Jan 7 at 6:24
gbcatgbcat
185
edited Jan 7 at 7:32
dessert
25.1k673106
edited Jan 7 at 7:32
dessert
25.1k673106
edited Jan 7 at 7:32
dessert
25.1k673106
25.1k673106
asked Jan 7 at 6:24
gbcatgbcat
185
asked Jan 7 at 6:24
gbcatgbcat
185
asked Jan 7 at 6:24
gbcatgbcat
185
185
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
From man sh:
-c string If the -c option is present, then commands are read from
string. If there are arguments after the string, they are
assigned to the positional parameters, starting with $0.
In your command the second sh is just a parameter with position 0 while 4 has position 1 and so on.
You can run this to check:
$ sh -c 'echo $0' sh 4 5
sh
edited Jan 7 at 7:33
dessert
25.1k673106
answered Jan 7 at 6:40
eyadofeyadof
1,24411517
That seems to be the text in Bash's man page. It's actually incorrect in that it implies that$0is one of the positional parameters. It isn't. The POSIX definition states that "A positional parameter is a parameter denoted by the decimal value represented by one or more digits, other than the single digit 0.", and Bash's manual itself contains that same definition.
– ilkkachu
Jan 7 at 11:31
add a comment |
This behavior is actually specified by POSIX standard, which all Bourne-like shells should support to claim themselves portable.
sh -c [-abCefhimnuvx] [-o option]... [+abCefhimnuvx] [+o option]...
command_string [command_name [argument...]]
See the command_string parameter ? Now let's look at -c flag description:
-c
Read commands from the command_string operand. Set the value of special parameter 0 (see Special Parameters) from the value of the command_name operand and the positional parameters ($1, $2, and so on) in sequence from the remaining argument operands. No commands shall be read from the standard input.
In other words, where in normal shell script $0 (which is usually shell name in interactive mode or script name when you run a script) would be set by the shell itself, with -c you have to specify it yourself. Thus,
sh -c 'echo Hi, I am $0 , my first positional parameter is $1' foobar 5
would set the process name to sh foobar.
Just in case you're wondering what $0 is, it's also covered in "Special Parameters" section of Shell Command Language Specifications:
0
(Zero.) Expands to the name of the shell or shell script.
answered Jan 7 at 6:41
Sergiy KolodyazhnyySergiy Kolodyazhnyy
74.6k9155325
add a comment |
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2 Answers
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2 Answers
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From man sh:
-c string If the -c option is present, then commands are read from
string. If there are arguments after the string, they are
assigned to the positional parameters, starting with $0.
In your command the second sh is just a parameter with position 0 while 4 has position 1 and so on.
You can run this to check:
$ sh -c 'echo $0' sh 4 5
sh
edited Jan 7 at 7:33
dessert
25.1k673106
answered Jan 7 at 6:40
eyadofeyadof
1,24411517
That seems to be the text in Bash's man page. It's actually incorrect in that it implies that$0is one of the positional parameters. It isn't. The POSIX definition states that "A positional parameter is a parameter denoted by the decimal value represented by one or more digits, other than the single digit 0.", and Bash's manual itself contains that same definition.
– ilkkachu
Jan 7 at 11:31
add a comment |
From man sh:
-c string If the -c option is present, then commands are read from
string. If there are arguments after the string, they are
assigned to the positional parameters, starting with $0.
In your command the second sh is just a parameter with position 0 while 4 has position 1 and so on.
You can run this to check:
$ sh -c 'echo $0' sh 4 5
sh
edited Jan 7 at 7:33
dessert
25.1k673106
answered Jan 7 at 6:40
eyadofeyadof
1,24411517
That seems to be the text in Bash's man page. It's actually incorrect in that it implies that$0is one of the positional parameters. It isn't. The POSIX definition states that "A positional parameter is a parameter denoted by the decimal value represented by one or more digits, other than the single digit 0.", and Bash's manual itself contains that same definition.
– ilkkachu
Jan 7 at 11:31
add a comment |
From man sh:
-c string If the -c option is present, then commands are read from
string. If there are arguments after the string, they are
assigned to the positional parameters, starting with $0.
In your command the second sh is just a parameter with position 0 while 4 has position 1 and so on.
You can run this to check:
$ sh -c 'echo $0' sh 4 5
sh
edited Jan 7 at 7:33
dessert
25.1k673106
answered Jan 7 at 6:40
eyadofeyadof
1,24411517
From man sh:
-c string If the -c option is present, then commands are read from
string. If there are arguments after the string, they are
assigned to the positional parameters, starting with $0.
In your command the second sh is just a parameter with position 0 while 4 has position 1 and so on.
You can run this to check:
$ sh -c 'echo $0' sh 4 5
sh
edited Jan 7 at 7:33
dessert
25.1k673106
answered Jan 7 at 6:40
eyadofeyadof
1,24411517
edited Jan 7 at 7:33
dessert
25.1k673106
edited Jan 7 at 7:33
dessert
25.1k673106
edited Jan 7 at 7:33
dessert
25.1k673106
25.1k673106
answered Jan 7 at 6:40
eyadofeyadof
1,24411517
answered Jan 7 at 6:40
eyadofeyadof
1,24411517
answered Jan 7 at 6:40
eyadofeyadof
1,24411517
1,24411517
That seems to be the text in Bash's man page. It's actually incorrect in that it implies that$0is one of the positional parameters. It isn't. The POSIX definition states that "A positional parameter is a parameter denoted by the decimal value represented by one or more digits, other than the single digit 0.", and Bash's manual itself contains that same definition.
– ilkkachu
Jan 7 at 11:31
add a comment |
That seems to be the text in Bash's man page. It's actually incorrect in that it implies that$0is one of the positional parameters. It isn't. The POSIX definition states that "A positional parameter is a parameter denoted by the decimal value represented by one or more digits, other than the single digit 0.", and Bash's manual itself contains that same definition.
– ilkkachu
Jan 7 at 11:31
That seems to be the text in Bash's man page. It's actually incorrect in that it implies that
$0 is one of the positional parameters. It isn't. The POSIX definition states that "A positional parameter is a parameter denoted by the decimal value represented by one or more digits, other than the single digit 0.", and Bash's manual itself contains that same definition.– ilkkachu
Jan 7 at 11:31
That seems to be the text in Bash's man page. It's actually incorrect in that it implies that
$0 is one of the positional parameters. It isn't. The POSIX definition states that "A positional parameter is a parameter denoted by the decimal value represented by one or more digits, other than the single digit 0.", and Bash's manual itself contains that same definition.– ilkkachu
Jan 7 at 11:31
add a comment |
This behavior is actually specified by POSIX standard, which all Bourne-like shells should support to claim themselves portable.
sh -c [-abCefhimnuvx] [-o option]... [+abCefhimnuvx] [+o option]...
command_string [command_name [argument...]]
See the command_string parameter ? Now let's look at -c flag description:
-c
Read commands from the command_string operand. Set the value of special parameter 0 (see Special Parameters) from the value of the command_name operand and the positional parameters ($1, $2, and so on) in sequence from the remaining argument operands. No commands shall be read from the standard input.
In other words, where in normal shell script $0 (which is usually shell name in interactive mode or script name when you run a script) would be set by the shell itself, with -c you have to specify it yourself. Thus,
sh -c 'echo Hi, I am $0 , my first positional parameter is $1' foobar 5
would set the process name to sh foobar.
Just in case you're wondering what $0 is, it's also covered in "Special Parameters" section of Shell Command Language Specifications:
0
(Zero.) Expands to the name of the shell or shell script.
answered Jan 7 at 6:41
Sergiy KolodyazhnyySergiy Kolodyazhnyy
74.6k9155325
add a comment |
This behavior is actually specified by POSIX standard, which all Bourne-like shells should support to claim themselves portable.
sh -c [-abCefhimnuvx] [-o option]... [+abCefhimnuvx] [+o option]...
command_string [command_name [argument...]]
See the command_string parameter ? Now let's look at -c flag description:
-c
Read commands from the command_string operand. Set the value of special parameter 0 (see Special Parameters) from the value of the command_name operand and the positional parameters ($1, $2, and so on) in sequence from the remaining argument operands. No commands shall be read from the standard input.
In other words, where in normal shell script $0 (which is usually shell name in interactive mode or script name when you run a script) would be set by the shell itself, with -c you have to specify it yourself. Thus,
sh -c 'echo Hi, I am $0 , my first positional parameter is $1' foobar 5
would set the process name to sh foobar.
Just in case you're wondering what $0 is, it's also covered in "Special Parameters" section of Shell Command Language Specifications:
0
(Zero.) Expands to the name of the shell or shell script.
answered Jan 7 at 6:41
Sergiy KolodyazhnyySergiy Kolodyazhnyy
74.6k9155325
add a comment |
This behavior is actually specified by POSIX standard, which all Bourne-like shells should support to claim themselves portable.
sh -c [-abCefhimnuvx] [-o option]... [+abCefhimnuvx] [+o option]...
command_string [command_name [argument...]]
See the command_string parameter ? Now let's look at -c flag description:
-c
Read commands from the command_string operand. Set the value of special parameter 0 (see Special Parameters) from the value of the command_name operand and the positional parameters ($1, $2, and so on) in sequence from the remaining argument operands. No commands shall be read from the standard input.
In other words, where in normal shell script $0 (which is usually shell name in interactive mode or script name when you run a script) would be set by the shell itself, with -c you have to specify it yourself. Thus,
sh -c 'echo Hi, I am $0 , my first positional parameter is $1' foobar 5
would set the process name to sh foobar.
Just in case you're wondering what $0 is, it's also covered in "Special Parameters" section of Shell Command Language Specifications:
0
(Zero.) Expands to the name of the shell or shell script.
answered Jan 7 at 6:41
Sergiy KolodyazhnyySergiy Kolodyazhnyy
74.6k9155325
This behavior is actually specified by POSIX standard, which all Bourne-like shells should support to claim themselves portable.
sh -c [-abCefhimnuvx] [-o option]... [+abCefhimnuvx] [+o option]...
command_string [command_name [argument...]]
See the command_string parameter ? Now let's look at -c flag description:
-c
Read commands from the command_string operand. Set the value of special parameter 0 (see Special Parameters) from the value of the command_name operand and the positional parameters ($1, $2, and so on) in sequence from the remaining argument operands. No commands shall be read from the standard input.
In other words, where in normal shell script $0 (which is usually shell name in interactive mode or script name when you run a script) would be set by the shell itself, with -c you have to specify it yourself. Thus,
sh -c 'echo Hi, I am $0 , my first positional parameter is $1' foobar 5
would set the process name to sh foobar.
Just in case you're wondering what $0 is, it's also covered in "Special Parameters" section of Shell Command Language Specifications:
0
(Zero.) Expands to the name of the shell or shell script.
answered Jan 7 at 6:41
Sergiy KolodyazhnyySergiy Kolodyazhnyy
74.6k9155325
edited Jan 7 at 7:40
answered Jan 7 at 6:41
Sergiy KolodyazhnyySergiy Kolodyazhnyy
74.6k9155325
answered Jan 7 at 6:41
Sergiy KolodyazhnyySergiy Kolodyazhnyy
74.6k9155325
answered Jan 7 at 6:41
Sergiy KolodyazhnyySergiy Kolodyazhnyy
74.6k9155325
74.6k9155325
add a comment |
add a comment |
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