Why can I not use variables as prefix to a command to set environment variables?
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3
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Normally, it is possible to set an environment variable for a command by prefixing it like so:
hello=hi bash -c 'echo $hello'
I also know that we can use a variable to substitute any part of a command invocation like the following:
$ cmd=bash
$ $cmd -c "echo hi" # equivalent to bash -c "echo hi"
I was very surprised to find out that you cannot use a variable to prefix a command to set an environment variable. Test case:
$ prefix=hello=hi
$ echo $prefix # prints hello=hi
$ $prefix bash -c 'echo $hello'
hello=hi: command not found
Why can I not set the environment variable using a variable? Is the prefix part a special part? I was able to get it working by using eval in front, but I still do not understand why. I am using bash 4.4.
bash environment-variables variable
add a comment |
up vote
3
down vote
favorite
Normally, it is possible to set an environment variable for a command by prefixing it like so:
hello=hi bash -c 'echo $hello'
I also know that we can use a variable to substitute any part of a command invocation like the following:
$ cmd=bash
$ $cmd -c "echo hi" # equivalent to bash -c "echo hi"
I was very surprised to find out that you cannot use a variable to prefix a command to set an environment variable. Test case:
$ prefix=hello=hi
$ echo $prefix # prints hello=hi
$ $prefix bash -c 'echo $hello'
hello=hi: command not found
Why can I not set the environment variable using a variable? Is the prefix part a special part? I was able to get it working by using eval in front, but I still do not understand why. I am using bash 4.4.
bash environment-variables variable
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Normally, it is possible to set an environment variable for a command by prefixing it like so:
hello=hi bash -c 'echo $hello'
I also know that we can use a variable to substitute any part of a command invocation like the following:
$ cmd=bash
$ $cmd -c "echo hi" # equivalent to bash -c "echo hi"
I was very surprised to find out that you cannot use a variable to prefix a command to set an environment variable. Test case:
$ prefix=hello=hi
$ echo $prefix # prints hello=hi
$ $prefix bash -c 'echo $hello'
hello=hi: command not found
Why can I not set the environment variable using a variable? Is the prefix part a special part? I was able to get it working by using eval in front, but I still do not understand why. I am using bash 4.4.
bash environment-variables variable
Normally, it is possible to set an environment variable for a command by prefixing it like so:
hello=hi bash -c 'echo $hello'
I also know that we can use a variable to substitute any part of a command invocation like the following:
$ cmd=bash
$ $cmd -c "echo hi" # equivalent to bash -c "echo hi"
I was very surprised to find out that you cannot use a variable to prefix a command to set an environment variable. Test case:
$ prefix=hello=hi
$ echo $prefix # prints hello=hi
$ $prefix bash -c 'echo $hello'
hello=hi: command not found
Why can I not set the environment variable using a variable? Is the prefix part a special part? I was able to get it working by using eval in front, but I still do not understand why. I am using bash 4.4.
bash environment-variables variable
bash environment-variables variable
edited 2 hours ago
Jeff Schaller
38.1k1053124
38.1k1053124
asked 2 hours ago
wbkang
1183
1183
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
3
down vote
I suspect this is the part of the sequence that's catching you:
The words that are not variable assignments or redirections are expanded (see Shell Expansions). If any words remain after expansion, the first word is taken to be the name of the command and the remaining words are the arguments
That's from the Bash reference manual in the section on Simple Command Expansion.
In the cmd=bash
example, no environment variables are set, and bash processes the command line up through parameter expansion, leaving bash -c "echo hi"
.
In the prefix=hello=hi
example, there are again no variable assignments in the first pass, so processing continues to parameter expansion, resulting in a first word of hello=hi
.
Once the variable assignments have been processed, they are not re-processed during command execution.
See the processing and its results under set -x
:
$ prefix=hello=hi
+ prefix=hello=hi
$ $prefix bash -c 'echo $hello'
+ hello=hi bash -c 'echo $hello'
-bash: hello=hi: command not found
$ hello=42 bash -c 'echo $hello'
+ hello=42
+ bash -c 'echo $hello'
42
1
Similarly,$foo=bar bash -c ...
won't work, because$foo=bar
isn't a variable assignment (whatever be the value of$foo
), because$foo=bar
doesn't fit thename=value
pattern for variable assignment.
– muru
2 hours ago
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
I suspect this is the part of the sequence that's catching you:
The words that are not variable assignments or redirections are expanded (see Shell Expansions). If any words remain after expansion, the first word is taken to be the name of the command and the remaining words are the arguments
That's from the Bash reference manual in the section on Simple Command Expansion.
In the cmd=bash
example, no environment variables are set, and bash processes the command line up through parameter expansion, leaving bash -c "echo hi"
.
In the prefix=hello=hi
example, there are again no variable assignments in the first pass, so processing continues to parameter expansion, resulting in a first word of hello=hi
.
Once the variable assignments have been processed, they are not re-processed during command execution.
See the processing and its results under set -x
:
$ prefix=hello=hi
+ prefix=hello=hi
$ $prefix bash -c 'echo $hello'
+ hello=hi bash -c 'echo $hello'
-bash: hello=hi: command not found
$ hello=42 bash -c 'echo $hello'
+ hello=42
+ bash -c 'echo $hello'
42
1
Similarly,$foo=bar bash -c ...
won't work, because$foo=bar
isn't a variable assignment (whatever be the value of$foo
), because$foo=bar
doesn't fit thename=value
pattern for variable assignment.
– muru
2 hours ago
add a comment |
up vote
3
down vote
I suspect this is the part of the sequence that's catching you:
The words that are not variable assignments or redirections are expanded (see Shell Expansions). If any words remain after expansion, the first word is taken to be the name of the command and the remaining words are the arguments
That's from the Bash reference manual in the section on Simple Command Expansion.
In the cmd=bash
example, no environment variables are set, and bash processes the command line up through parameter expansion, leaving bash -c "echo hi"
.
In the prefix=hello=hi
example, there are again no variable assignments in the first pass, so processing continues to parameter expansion, resulting in a first word of hello=hi
.
Once the variable assignments have been processed, they are not re-processed during command execution.
See the processing and its results under set -x
:
$ prefix=hello=hi
+ prefix=hello=hi
$ $prefix bash -c 'echo $hello'
+ hello=hi bash -c 'echo $hello'
-bash: hello=hi: command not found
$ hello=42 bash -c 'echo $hello'
+ hello=42
+ bash -c 'echo $hello'
42
1
Similarly,$foo=bar bash -c ...
won't work, because$foo=bar
isn't a variable assignment (whatever be the value of$foo
), because$foo=bar
doesn't fit thename=value
pattern for variable assignment.
– muru
2 hours ago
add a comment |
up vote
3
down vote
up vote
3
down vote
I suspect this is the part of the sequence that's catching you:
The words that are not variable assignments or redirections are expanded (see Shell Expansions). If any words remain after expansion, the first word is taken to be the name of the command and the remaining words are the arguments
That's from the Bash reference manual in the section on Simple Command Expansion.
In the cmd=bash
example, no environment variables are set, and bash processes the command line up through parameter expansion, leaving bash -c "echo hi"
.
In the prefix=hello=hi
example, there are again no variable assignments in the first pass, so processing continues to parameter expansion, resulting in a first word of hello=hi
.
Once the variable assignments have been processed, they are not re-processed during command execution.
See the processing and its results under set -x
:
$ prefix=hello=hi
+ prefix=hello=hi
$ $prefix bash -c 'echo $hello'
+ hello=hi bash -c 'echo $hello'
-bash: hello=hi: command not found
$ hello=42 bash -c 'echo $hello'
+ hello=42
+ bash -c 'echo $hello'
42
I suspect this is the part of the sequence that's catching you:
The words that are not variable assignments or redirections are expanded (see Shell Expansions). If any words remain after expansion, the first word is taken to be the name of the command and the remaining words are the arguments
That's from the Bash reference manual in the section on Simple Command Expansion.
In the cmd=bash
example, no environment variables are set, and bash processes the command line up through parameter expansion, leaving bash -c "echo hi"
.
In the prefix=hello=hi
example, there are again no variable assignments in the first pass, so processing continues to parameter expansion, resulting in a first word of hello=hi
.
Once the variable assignments have been processed, they are not re-processed during command execution.
See the processing and its results under set -x
:
$ prefix=hello=hi
+ prefix=hello=hi
$ $prefix bash -c 'echo $hello'
+ hello=hi bash -c 'echo $hello'
-bash: hello=hi: command not found
$ hello=42 bash -c 'echo $hello'
+ hello=42
+ bash -c 'echo $hello'
42
answered 2 hours ago
Jeff Schaller
38.1k1053124
38.1k1053124
1
Similarly,$foo=bar bash -c ...
won't work, because$foo=bar
isn't a variable assignment (whatever be the value of$foo
), because$foo=bar
doesn't fit thename=value
pattern for variable assignment.
– muru
2 hours ago
add a comment |
1
Similarly,$foo=bar bash -c ...
won't work, because$foo=bar
isn't a variable assignment (whatever be the value of$foo
), because$foo=bar
doesn't fit thename=value
pattern for variable assignment.
– muru
2 hours ago
1
1
Similarly,
$foo=bar bash -c ...
won't work, because $foo=bar
isn't a variable assignment (whatever be the value of $foo
), because $foo=bar
doesn't fit the name=value
pattern for variable assignment.– muru
2 hours ago
Similarly,
$foo=bar bash -c ...
won't work, because $foo=bar
isn't a variable assignment (whatever be the value of $foo
), because $foo=bar
doesn't fit the name=value
pattern for variable assignment.– muru
2 hours ago
add a comment |
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