Is coproc the same as &?

up vote
2
down vote

favorite

I have read that $coproc < command > is different from $< command > & in that coproc will execute command in a sub-shell process.

But when I tested it, it worked just like $< command > &. The test is as follow:

First: test the behavior of $< command > &.

Run $nano & on tty1

On another tty, output from $ps -t tty1 --forest indicates nano process is child process of the -bash process (login bash shell process -> no sub-shell process was created)

Second: test the behavior of $coproc < command >

Run $coproc nano on tty1

On another tty, output from $ps -t tty1 --forest is the same as above (no sub-shell process was created)

So is $coproc < command > simply the same as $< command > &?

The shell used was a bash shell

asked 1 hour ago

Tran Triet

748

Possible duplicate of How do you use the command coproc in various shells?
â€“Â StÃ©phane Chazelas
13 mins ago

add a commentÂ |Â

up vote
2
down vote

favorite

I have read that $coproc < command > is different from $< command > & in that coproc will execute command in a sub-shell process.

But when I tested it, it worked just like $< command > &. The test is as follow:

First: test the behavior of $< command > &.

Run $nano & on tty1

On another tty, output from $ps -t tty1 --forest indicates nano process is child process of the -bash process (login bash shell process -> no sub-shell process was created)

Second: test the behavior of $coproc < command >

Run $coproc nano on tty1

On another tty, output from $ps -t tty1 --forest is the same as above (no sub-shell process was created)

So is $coproc < command > simply the same as $< command > &?

The shell used was a bash shell

asked 1 hour ago

Tran Triet

748

Possible duplicate of How do you use the command coproc in various shells?
â€“Â StÃ©phane Chazelas
13 mins ago

add a commentÂ |Â

up vote
2
down vote

favorite

I have read that $coproc < command > is different from $< command > & in that coproc will execute command in a sub-shell process.

But when I tested it, it worked just like $< command > &. The test is as follow:

First: test the behavior of $< command > &.

Run $nano & on tty1

On another tty, output from $ps -t tty1 --forest indicates nano process is child process of the -bash process (login bash shell process -> no sub-shell process was created)

Second: test the behavior of $coproc < command >

Run $coproc nano on tty1

On another tty, output from $ps -t tty1 --forest is the same as above (no sub-shell process was created)

So is $coproc < command > simply the same as $< command > &?

The shell used was a bash shell

asked 1 hour ago

Tran Triet

748

I have read that $coproc < command > is different from $< command > & in that coproc will execute command in a sub-shell process.

But when I tested it, it worked just like $< command > &. The test is as follow:

First: test the behavior of $< command > &.

Run $nano & on tty1

On another tty, output from $ps -t tty1 --forest indicates nano process is child process of the -bash process (login bash shell process -> no sub-shell process was created)

Second: test the behavior of $coproc < command >

Run $coproc nano on tty1

On another tty, output from $ps -t tty1 --forest is the same as above (no sub-shell process was created)

So is $coproc < command > simply the same as $< command > &?

The shell used was a bash shell

bash coprocesses

asked 1 hour ago

Tran Triet

748

asked 1 hour ago

Tran Triet

748

asked 1 hour ago

Tran Triet

748

asked 1 hour ago

Tran Triet

748

asked 1 hour ago

Tran Triet

748

Possible duplicate of How do you use the command coproc in various shells?
â€“Â StÃ©phane Chazelas
13 mins ago

add a commentÂ |Â

Possible duplicate of How do you use the command coproc in various shells?
â€“Â StÃ©phane Chazelas
13 mins ago

Possible duplicate of How do you use the command coproc in various shells?
â€“Â StÃ©phane Chazelas
13 mins ago

add a commentÂ |Â

2 Answers
2

active

oldest

votes

up vote
2
down vote

coproc utility is not the same as utility & in bash.

With coproc utility you get an array, COPROC, that contains the standard input and output filedescriptors of utility. You may then do things like

#!/bin/bash

coproc bc -l

for (( k = 0; k < 50; ++k )); do
 printf "2.3*%d + 1n" "$k" >&$COPROC[1]
 read -u "$COPROC[0]" a
 printf '%.2fn' "$a"
done

kill "$COPROC_PID"

Here, bc -l is a co-process and acts like a "arithmetic computation service" for the shell loop, taking expressions to compute on its standard input and giving back results on its standard output.

As far as I can tell, bash also only supports one co-process at any one time. The ksh93 shell also supports co-processes, but the syntax is totally different (but somewhat sleeker).

edited 10 mins ago

answered 16 mins ago

Kusalananda

107k14209331

add a commentÂ |Â

up vote
0
down vote

The difference is the creation of two I/O channels, as described in man bash:

A coprocess is executed asynchronously in a subshell, as if the
command had been terminated with
the & control operator, with a two-way pipe established between the executing shell and the coprocess.

answered 1 hour ago

RudiC

1,5979

add a commentÂ |Â

Your Answer

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "106"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: false,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);

);

draft saved

draft discarded

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2funix.stackexchange.com%2fquestions%2f472561%2fis-coproc-command-the-same-as-command%23new-answer', 'question_page');

);

Post as a guest

Name

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
2
down vote

coproc utility is not the same as utility & in bash.

With coproc utility you get an array, COPROC, that contains the standard input and output filedescriptors of utility. You may then do things like

#!/bin/bash

coproc bc -l

for (( k = 0; k < 50; ++k )); do
 printf "2.3*%d + 1n" "$k" >&$COPROC[1]
 read -u "$COPROC[0]" a
 printf '%.2fn' "$a"
done

kill "$COPROC_PID"

Here, bc -l is a co-process and acts like a "arithmetic computation service" for the shell loop, taking expressions to compute on its standard input and giving back results on its standard output.

As far as I can tell, bash also only supports one co-process at any one time. The ksh93 shell also supports co-processes, but the syntax is totally different (but somewhat sleeker).

edited 10 mins ago

answered 16 mins ago

Kusalananda

107k14209331

add a commentÂ |Â

up vote
2
down vote

coproc utility is not the same as utility & in bash.

With coproc utility you get an array, COPROC, that contains the standard input and output filedescriptors of utility. You may then do things like

#!/bin/bash

coproc bc -l

for (( k = 0; k < 50; ++k )); do
 printf "2.3*%d + 1n" "$k" >&$COPROC[1]
 read -u "$COPROC[0]" a
 printf '%.2fn' "$a"
done

kill "$COPROC_PID"

Here, bc -l is a co-process and acts like a "arithmetic computation service" for the shell loop, taking expressions to compute on its standard input and giving back results on its standard output.

As far as I can tell, bash also only supports one co-process at any one time. The ksh93 shell also supports co-processes, but the syntax is totally different (but somewhat sleeker).

edited 10 mins ago

answered 16 mins ago

Kusalananda

107k14209331

add a commentÂ |Â

up vote
2
down vote

coproc utility is not the same as utility & in bash.

With coproc utility you get an array, COPROC, that contains the standard input and output filedescriptors of utility. You may then do things like

#!/bin/bash

coproc bc -l

for (( k = 0; k < 50; ++k )); do
 printf "2.3*%d + 1n" "$k" >&$COPROC[1]
 read -u "$COPROC[0]" a
 printf '%.2fn' "$a"
done

kill "$COPROC_PID"

Here, bc -l is a co-process and acts like a "arithmetic computation service" for the shell loop, taking expressions to compute on its standard input and giving back results on its standard output.

As far as I can tell, bash also only supports one co-process at any one time. The ksh93 shell also supports co-processes, but the syntax is totally different (but somewhat sleeker).

edited 10 mins ago

answered 16 mins ago

Kusalananda

107k14209331

coproc utility is not the same as utility & in bash.

With coproc utility you get an array, COPROC, that contains the standard input and output filedescriptors of utility. You may then do things like

#!/bin/bash

coproc bc -l

for (( k = 0; k < 50; ++k )); do
 printf "2.3*%d + 1n" "$k" >&$COPROC[1]
 read -u "$COPROC[0]" a
 printf '%.2fn' "$a"
done

kill "$COPROC_PID"

Here, bc -l is a co-process and acts like a "arithmetic computation service" for the shell loop, taking expressions to compute on its standard input and giving back results on its standard output.

As far as I can tell, bash also only supports one co-process at any one time. The ksh93 shell also supports co-processes, but the syntax is totally different (but somewhat sleeker).

edited 10 mins ago

answered 16 mins ago

Kusalananda

107k14209331

edited 10 mins ago

answered 16 mins ago

Kusalananda

107k14209331

answered 16 mins ago

Kusalananda

107k14209331

answered 16 mins ago

Kusalananda

107k14209331

add a commentÂ |Â

up vote
0
down vote

The difference is the creation of two I/O channels, as described in man bash:

A coprocess is executed asynchronously in a subshell, as if the
command had been terminated with
the & control operator, with a two-way pipe established between the executing shell and the coprocess.

answered 1 hour ago

RudiC

1,5979

add a commentÂ |Â

up vote
0
down vote

The difference is the creation of two I/O channels, as described in man bash:

A coprocess is executed asynchronously in a subshell, as if the
command had been terminated with
the & control operator, with a two-way pipe established between the executing shell and the coprocess.

answered 1 hour ago

RudiC

1,5979

add a commentÂ |Â

up vote
0
down vote

The difference is the creation of two I/O channels, as described in man bash:

A coprocess is executed asynchronously in a subshell, as if the
command had been terminated with
the & control operator, with a two-way pipe established between the executing shell and the coprocess.

answered 1 hour ago

RudiC

1,5979

The difference is the creation of two I/O channels, as described in man bash:

A coprocess is executed asynchronously in a subshell, as if the
command had been terminated with
the & control operator, with a two-way pipe established between the executing shell and the coprocess.

answered 1 hour ago

RudiC

1,5979

answered 1 hour ago

RudiC

1,5979

answered 1 hour ago

RudiC

1,5979

answered 1 hour ago

RudiC

1,5979

add a commentÂ |Â

draft saved

draft discarded

draft saved

draft discarded

Post as a guest

Name

Search This Blog

Iyfjky