echo, pipe and command substitution
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I want to substitute an echo command followed with | (pipe). For example,
$(echo "echo 'hello' | cat")
returns
'hello' | cat
I expect this to behave like
echo 'hello' | cat
which returns
hello
but that's not the case. Why?
PS. I'm aware of eval, and
eval $(echo "echo 'hello' | cat")
works as expected
bash pipe echo command-substitution
asked Mar 17 at 18:19
xafizoffxafizoff
84
add a comment |
I want to substitute an echo command followed with | (pipe). For example,
$(echo "echo 'hello' | cat")
returns
'hello' | cat
I expect this to behave like
echo 'hello' | cat
which returns
hello
but that's not the case. Why?
PS. I'm aware of eval, and
eval $(echo "echo 'hello' | cat")
works as expected
bash pipe echo command-substitution
asked Mar 17 at 18:19
xafizoffxafizoff
84
add a comment |
I want to substitute an echo command followed with | (pipe). For example,
$(echo "echo 'hello' | cat")
returns
'hello' | cat
I expect this to behave like
echo 'hello' | cat
which returns
hello
but that's not the case. Why?
PS. I'm aware of eval, and
eval $(echo "echo 'hello' | cat")
works as expected
bash pipe echo command-substitution
asked Mar 17 at 18:19
xafizoffxafizoff
84
I want to substitute an echo command followed with | (pipe). For example,
$(echo "echo 'hello' | cat")
returns
'hello' | cat
I expect this to behave like
echo 'hello' | cat
which returns
hello
but that's not the case. Why?
PS. I'm aware of eval, and
eval $(echo "echo 'hello' | cat")
works as expected
bash pipe echo command-substitution
bash pipe echo command-substitution
asked Mar 17 at 18:19
xafizoffxafizoff
84
asked Mar 17 at 18:19
xafizoffxafizoff
84
edited Mar 17 at 18:27
xafizoff
asked Mar 17 at 18:19
xafizoffxafizoff
84
asked Mar 17 at 18:19
xafizoffxafizoff
84
asked Mar 17 at 18:19
xafizoffxafizoff
84
84
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Because syntax elements like | (or &&, or ; etc.) are recognized as the first thing of command line parsing, and are not processed again after parameters/variables are expanded.
Pretty much the only things that happen after parameter expansions, command substitutions and arithmetic expansions are word splitting and filename globbing. The outputs of expansions also don't expand again: this doesn't print 6:
$ var='$((1+2+3))' # $((..)) not expanded here (single-quotes)
$ echo $var # $((..)) not expanded here either
$((1+2+3))
Having another pass of parsing and expansions is exactly what eval is there for.
Related:
- Behaviour of bash command substitution with command from string in variable
- Why can I not use variables as prefix to a command to set environment variables?
- What is the "eval" command in bash?
answered Mar 17 at 18:40
ilkkachuilkkachu
63.4k10104181
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Because syntax elements like | (or &&, or ; etc.) are recognized as the first thing of command line parsing, and are not processed again after parameters/variables are expanded.
Pretty much the only things that happen after parameter expansions, command substitutions and arithmetic expansions are word splitting and filename globbing. The outputs of expansions also don't expand again: this doesn't print 6:
$ var='$((1+2+3))' # $((..)) not expanded here (single-quotes)
$ echo $var # $((..)) not expanded here either
$((1+2+3))
Having another pass of parsing and expansions is exactly what eval is there for.
Related:
- Behaviour of bash command substitution with command from string in variable
- Why can I not use variables as prefix to a command to set environment variables?
- What is the "eval" command in bash?
answered Mar 17 at 18:40
ilkkachuilkkachu
63.4k10104181
add a comment |
Because syntax elements like | (or &&, or ; etc.) are recognized as the first thing of command line parsing, and are not processed again after parameters/variables are expanded.
Pretty much the only things that happen after parameter expansions, command substitutions and arithmetic expansions are word splitting and filename globbing. The outputs of expansions also don't expand again: this doesn't print 6:
$ var='$((1+2+3))' # $((..)) not expanded here (single-quotes)
$ echo $var # $((..)) not expanded here either
$((1+2+3))
Having another pass of parsing and expansions is exactly what eval is there for.
Related:
- Behaviour of bash command substitution with command from string in variable
- Why can I not use variables as prefix to a command to set environment variables?
- What is the "eval" command in bash?
answered Mar 17 at 18:40
ilkkachuilkkachu
63.4k10104181
add a comment |
Because syntax elements like | (or &&, or ; etc.) are recognized as the first thing of command line parsing, and are not processed again after parameters/variables are expanded.
Pretty much the only things that happen after parameter expansions, command substitutions and arithmetic expansions are word splitting and filename globbing. The outputs of expansions also don't expand again: this doesn't print 6:
$ var='$((1+2+3))' # $((..)) not expanded here (single-quotes)
$ echo $var # $((..)) not expanded here either
$((1+2+3))
Having another pass of parsing and expansions is exactly what eval is there for.
Related:
- Behaviour of bash command substitution with command from string in variable
- Why can I not use variables as prefix to a command to set environment variables?
- What is the "eval" command in bash?
answered Mar 17 at 18:40
ilkkachuilkkachu
63.4k10104181
Because syntax elements like | (or &&, or ; etc.) are recognized as the first thing of command line parsing, and are not processed again after parameters/variables are expanded.
Pretty much the only things that happen after parameter expansions, command substitutions and arithmetic expansions are word splitting and filename globbing. The outputs of expansions also don't expand again: this doesn't print 6:
$ var='$((1+2+3))' # $((..)) not expanded here (single-quotes)
$ echo $var # $((..)) not expanded here either
$((1+2+3))
Having another pass of parsing and expansions is exactly what eval is there for.
Related:
- Behaviour of bash command substitution with command from string in variable
- Why can I not use variables as prefix to a command to set environment variables?
- What is the "eval" command in bash?
answered Mar 17 at 18:40
ilkkachuilkkachu
63.4k10104181
edited Mar 17 at 18:45
answered Mar 17 at 18:40
ilkkachuilkkachu
63.4k10104181
answered Mar 17 at 18:40
ilkkachuilkkachu
63.4k10104181
answered Mar 17 at 18:40
ilkkachuilkkachu
63.4k10104181
63.4k10104181
add a comment |
add a comment |
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