unexpected behaviour in shell command substitution

Given the shell script test.sh

for var in "$@"

do

    echo "$var"

done

and the invocation sh test.sh $(echo '"hello " "hi and bye"')

I would expect the output

hello

hi and bye

but instead get:

"hello

"

"hi

and

bye"

If I change to sh test.sh $(echo "hello " "hi and bye")

then I get

hello

hi

and

bye

Neither of these behaviours are desired, how can I get the desired behaviour?

What I want to understand is why sh test.sh $(echo "hello " "hi and bye") and sh test.sh "hello " "hi and bye" are not equivalent?

edited Jan 21 at 11:10

asked Jan 21 at 10:59

user1207217

60849

add a comment |

Given the shell script test.sh

for var in "$@"

do

    echo "$var"

done

and the invocation sh test.sh $(echo '"hello " "hi and bye"')

I would expect the output

hello

hi and bye

but instead get:

"hello

"

"hi

and

bye"

If I change to sh test.sh $(echo "hello " "hi and bye")

then I get

hello

hi

and

bye

Neither of these behaviours are desired, how can I get the desired behaviour?

What I want to understand is why sh test.sh $(echo "hello " "hi and bye") and sh test.sh "hello " "hi and bye" are not equivalent?

edited Jan 21 at 11:10

asked Jan 21 at 10:59

user1207217

60849

add a comment |

Given the shell script test.sh

for var in "$@"

do

    echo "$var"

done

and the invocation sh test.sh $(echo '"hello " "hi and bye"')

I would expect the output

hello

hi and bye

but instead get:

"hello

"

"hi

and

bye"

If I change to sh test.sh $(echo "hello " "hi and bye")

then I get

hello

hi

and

bye

Neither of these behaviours are desired, how can I get the desired behaviour?

What I want to understand is why sh test.sh $(echo "hello " "hi and bye") and sh test.sh "hello " "hi and bye" are not equivalent?

edited Jan 21 at 11:10

asked Jan 21 at 10:59

user1207217

60849

Given the shell script test.sh

for var in "$@"

do

    echo "$var"

done

and the invocation sh test.sh $(echo '"hello " "hi and bye"')

I would expect the output

hello

hi and bye

but instead get:

"hello

"

"hi

and

bye"

If I change to sh test.sh $(echo "hello " "hi and bye")

then I get

hello

hi

and

bye

Neither of these behaviours are desired, how can I get the desired behaviour?

What I want to understand is why sh test.sh $(echo "hello " "hi and bye") and sh test.sh "hello " "hi and bye" are not equivalent?

shell

edited Jan 21 at 11:10

asked Jan 21 at 10:59

user1207217

60849

edited Jan 21 at 11:10

asked Jan 21 at 10:59

user1207217

60849

edited Jan 21 at 11:10

asked Jan 21 at 10:59

user1207217

60849

asked Jan 21 at 10:59

user1207217

60849

asked Jan 21 at 10:59

user1207217

60849

add a comment |

4 Answers
4

active

oldest

votes

Your command substitution will generate a string.

In the case of

$(echo '"hello " "hi and bye"')

this string will be "hello " "hi and bye".

The string is then undergoing word splitting (and filename globbing, but it doesn't affect this example). The word splitting happen on every character that is the same as one of the characters in $IFS (by default, spaces, tabs and newlines).

The words generated by the default value of IFS would be "hello, ", "hi, and, and bye".

These are then given as separate arguments to your script.

In your second command, the command substitution is

$(echo "hello " "hi and bye")

This generates the string hello hi and bye and the word splitting would result in the four words hello, hi, and, and bye.

In your last example, you use the two arguments hello and hi and bye directly with your script. These won't undergo word splitting because they are quoted.

edited Jan 21 at 11:30

answered Jan 21 at 11:20

Kusalananda

134k17255418

Thanks. Is the short answer to this then that you cannot (reliably) use command substitution to generate arguments to another command?

– user1207217
Jan 21 at 11:26

@user1207217 Correct, especially if the arguments contain characters from $IFS or filename globbing characters.

– Kusalananda
Jan 21 at 11:29

@Kusalananda thanks, although thats a huge pain ....

– user1207217
Jan 21 at 11:30

1

@user1207217 You may want to ask a separate question about what it actually is that you are intending or trying to do.

– Kusalananda
Jan 21 at 11:31

Good idea, thanks. unix.stackexchange.com/questions/495769/…

– user1207217
Jan 21 at 11:40

add a comment |

TL;DR: the result of echo is a victim of word splitting on $() result, and sh test.sh "hello " "hi and bye" uses a different rule.

Let's take a peak as what's actually happening, when you add set -x for debugging:

> set -x

> ./main.sh $(echo '"hello " "hi and bye"')

++ echo '"hello " "hi and bye"'

+ ./main.sh '"hello' '"' '"hi' and 'bye"'

"hello

"

"hi

and

bye"

echo receives "hello " "hi and bye" as a single argument (yep, including all spaces and single quote) due to single-quotes in the command itself. But command substitution turns that into "hello " "hi and bye". To quote bash manual:

If the substitution appears within double quotes, word splitting and pathname expansion are not performed on the results.

However, in your case, unquoted command substitution allows exactly for word splitting to occur based on blank character (which is one of the 3 characters set in the $IFS variable, which is what is used for word splitting). Thus, if you break the resulting string at each space, what do you have?

"hello, then space, after which we have next token

" - that one character by itself, isolated by spaces from both sides.

"hi, again with space on left and right

and

bye"

all in all, you get 5 tokens broken at whitespaces. It's important to realize that word splitting, in this case, occurs exactly a result of command substitution. Later, all of these are treated as individual tokens that have been already parsed. As you see in set -x output, the shell script is called with exactly those tokens.

As for why sh test.sh "hello " "hi and bye" behaves differently, that's because the different rule applies. Your command-line is parsed after you enter it, and quoted strings are treated as a single unit, i.e. hello and hi and bye

edited Feb 4 at 15:01

msp9011

4,38044066

answered Jan 21 at 11:19

Sergiy Kolodyazhnyy

10.5k42663

add a comment |

You could change IFS to something other than space (since you want to protect the space in hi and bye), and use that something to separate the two strings:

$ IFS=:

$ sh test.sh $(echo "hello ":"hi and bye")

hello 

hi and bye

The order of operations (those relevant in this example, anyway) is somewhat like this:

substitutions

word splitting

quote removal

Note that quotes created as a result of substitutions aren't special in word splitting or quote removal, so a quote in the output of (1) passes through (2) and (3) like any other character. So:

In echo "hello " "hi and bye", the shell removes these quotes, so echo gets hello and hi and bye, so outputs hello hi and bye, joining the strings with a space.

In echo '"hello " "hi and bye"', the shell removes the outer ', the echo gets "hello " "hi and bye" and outputs that.

In sh test.sh $(echo '"hello " "hi and bye"'), the command substitution is replaced with "hello " "hi and bye", but these quotes are a result of a substitution, and so aren't involved in word splitting or quote removal.

So the shell splits those into "hello, ", "hi, and, bye", hence the output you get.

In sh test.sh $(echo "hello " "hi and bye"), the command substitution is replaced with hello hi and bye, which gets split into hello, hi and bye.

edited Jan 21 at 11:27

answered Jan 21 at 11:07

Olorin

3,8581721

1

Thanks. I guess what I am not understanding is why the usual shell substitution is not occurring, i.e. why is the behaviour not the same as if I do sh test.sh "hello " "hi and bye"?

– user1207217
Jan 21 at 11:09

@user1207217 does the edit help?

– Olorin
Jan 21 at 11:27

In $(echo "hello ":"hi and bye"), you might as well write $(echo "hello :hi and bye"). The colon works the same quoted or not.

– ilkkachu
Jan 21 at 11:31

add a comment |

How about

sh test.sh "hello " "hi and bye"

answered Jan 21 at 11:13

gerhard d.

1,256311

I'm aware I can do this, the shell script is a MWE of an unexpected behaviour -- I'm trying to use a command substitution to generate a long list of args (some of which contain spaces) but it doesn't work correctly because even with quoting in the command output the above is occurring, i.e. the shell is not doing its usual parsing and treating everything inside a single or double quote pair as a single arg

– user1207217
Jan 21 at 11:15

add a comment |

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

Your command substitution will generate a string.

In the case of

$(echo '"hello " "hi and bye"')

this string will be "hello " "hi and bye".

The words generated by the default value of IFS would be "hello, ", "hi, and, and bye".

These are then given as separate arguments to your script.

In your second command, the command substitution is

$(echo "hello " "hi and bye")

This generates the string hello hi and bye and the word splitting would result in the four words hello, hi, and, and bye.

In your last example, you use the two arguments hello and hi and bye directly with your script. These won't undergo word splitting because they are quoted.

edited Jan 21 at 11:30

answered Jan 21 at 11:20

Kusalananda

134k17255418

Thanks. Is the short answer to this then that you cannot (reliably) use command substitution to generate arguments to another command?

– user1207217
Jan 21 at 11:26

@user1207217 Correct, especially if the arguments contain characters from $IFS or filename globbing characters.

– Kusalananda
Jan 21 at 11:29

@Kusalananda thanks, although thats a huge pain ....

– user1207217
Jan 21 at 11:30

1

@user1207217 You may want to ask a separate question about what it actually is that you are intending or trying to do.

– Kusalananda
Jan 21 at 11:31

Good idea, thanks. unix.stackexchange.com/questions/495769/…

– user1207217
Jan 21 at 11:40

add a comment |

Your command substitution will generate a string.

In the case of

$(echo '"hello " "hi and bye"')

this string will be "hello " "hi and bye".

The words generated by the default value of IFS would be "hello, ", "hi, and, and bye".

These are then given as separate arguments to your script.

In your second command, the command substitution is

$(echo "hello " "hi and bye")

This generates the string hello hi and bye and the word splitting would result in the four words hello, hi, and, and bye.

In your last example, you use the two arguments hello and hi and bye directly with your script. These won't undergo word splitting because they are quoted.

edited Jan 21 at 11:30

answered Jan 21 at 11:20

Kusalananda

134k17255418

Thanks. Is the short answer to this then that you cannot (reliably) use command substitution to generate arguments to another command?

– user1207217
Jan 21 at 11:26

@user1207217 Correct, especially if the arguments contain characters from $IFS or filename globbing characters.

– Kusalananda
Jan 21 at 11:29

@Kusalananda thanks, although thats a huge pain ....

– user1207217
Jan 21 at 11:30

1

@user1207217 You may want to ask a separate question about what it actually is that you are intending or trying to do.

– Kusalananda
Jan 21 at 11:31

Good idea, thanks. unix.stackexchange.com/questions/495769/…

– user1207217
Jan 21 at 11:40

add a comment |

Your command substitution will generate a string.

In the case of

$(echo '"hello " "hi and bye"')

this string will be "hello " "hi and bye".

The words generated by the default value of IFS would be "hello, ", "hi, and, and bye".

These are then given as separate arguments to your script.

In your second command, the command substitution is

$(echo "hello " "hi and bye")

This generates the string hello hi and bye and the word splitting would result in the four words hello, hi, and, and bye.

In your last example, you use the two arguments hello and hi and bye directly with your script. These won't undergo word splitting because they are quoted.

edited Jan 21 at 11:30

answered Jan 21 at 11:20

Kusalananda

134k17255418

Your command substitution will generate a string.

In the case of

$(echo '"hello " "hi and bye"')

this string will be "hello " "hi and bye".

The words generated by the default value of IFS would be "hello, ", "hi, and, and bye".

These are then given as separate arguments to your script.

In your second command, the command substitution is

$(echo "hello " "hi and bye")

This generates the string hello hi and bye and the word splitting would result in the four words hello, hi, and, and bye.

In your last example, you use the two arguments hello and hi and bye directly with your script. These won't undergo word splitting because they are quoted.

edited Jan 21 at 11:30

answered Jan 21 at 11:20

Kusalananda

134k17255418

edited Jan 21 at 11:30

answered Jan 21 at 11:20

Kusalananda

134k17255418

answered Jan 21 at 11:20

Kusalananda

134k17255418

answered Jan 21 at 11:20

Kusalananda

134k17255418

Thanks. Is the short answer to this then that you cannot (reliably) use command substitution to generate arguments to another command?

– user1207217
Jan 21 at 11:26

@user1207217 Correct, especially if the arguments contain characters from $IFS or filename globbing characters.

– Kusalananda
Jan 21 at 11:29

@Kusalananda thanks, although thats a huge pain ....

– user1207217
Jan 21 at 11:30

1

@user1207217 You may want to ask a separate question about what it actually is that you are intending or trying to do.

– Kusalananda
Jan 21 at 11:31

Good idea, thanks. unix.stackexchange.com/questions/495769/…

– user1207217
Jan 21 at 11:40

add a comment |

Thanks. Is the short answer to this then that you cannot (reliably) use command substitution to generate arguments to another command?

– user1207217
Jan 21 at 11:26

@user1207217 Correct, especially if the arguments contain characters from $IFS or filename globbing characters.

– Kusalananda
Jan 21 at 11:29

@Kusalananda thanks, although thats a huge pain ....

– user1207217
Jan 21 at 11:30

1

@user1207217 You may want to ask a separate question about what it actually is that you are intending or trying to do.

– Kusalananda
Jan 21 at 11:31

Good idea, thanks. unix.stackexchange.com/questions/495769/…

– user1207217
Jan 21 at 11:40

Thanks. Is the short answer to this then that you cannot (reliably) use command substitution to generate arguments to another command?

– user1207217
Jan 21 at 11:26

@user1207217 Correct, especially if the arguments contain characters from $IFS or filename globbing characters.

– Kusalananda
Jan 21 at 11:29

@Kusalananda thanks, although thats a huge pain ....

– user1207217
Jan 21 at 11:30

@user1207217 You may want to ask a separate question about what it actually is that you are intending or trying to do.

– Kusalananda
Jan 21 at 11:31

Good idea, thanks. unix.stackexchange.com/questions/495769/…

– user1207217
Jan 21 at 11:40

add a comment |

TL;DR: the result of echo is a victim of word splitting on $() result, and sh test.sh "hello " "hi and bye" uses a different rule.

Let's take a peak as what's actually happening, when you add set -x for debugging:

> set -x

> ./main.sh $(echo '"hello " "hi and bye"')

++ echo '"hello " "hi and bye"'

+ ./main.sh '"hello' '"' '"hi' and 'bye"'

"hello

"

"hi

and

bye"

If the substitution appears within double quotes, word splitting and pathname expansion are not performed on the results.

"hello, then space, after which we have next token

" - that one character by itself, isolated by spaces from both sides.

"hi, again with space on left and right

and

bye"

edited Feb 4 at 15:01

msp9011

4,38044066

answered Jan 21 at 11:19

Sergiy Kolodyazhnyy

10.5k42663

add a comment |

TL;DR: the result of echo is a victim of word splitting on $() result, and sh test.sh "hello " "hi and bye" uses a different rule.

Let's take a peak as what's actually happening, when you add set -x for debugging:

> set -x

> ./main.sh $(echo '"hello " "hi and bye"')

++ echo '"hello " "hi and bye"'

+ ./main.sh '"hello' '"' '"hi' and 'bye"'

"hello

"

"hi

and

bye"

If the substitution appears within double quotes, word splitting and pathname expansion are not performed on the results.

"hello, then space, after which we have next token

" - that one character by itself, isolated by spaces from both sides.

"hi, again with space on left and right

and

bye"

edited Feb 4 at 15:01

msp9011

4,38044066

answered Jan 21 at 11:19

Sergiy Kolodyazhnyy

10.5k42663

add a comment |

TL;DR: the result of echo is a victim of word splitting on $() result, and sh test.sh "hello " "hi and bye" uses a different rule.

Let's take a peak as what's actually happening, when you add set -x for debugging:

> set -x

> ./main.sh $(echo '"hello " "hi and bye"')

++ echo '"hello " "hi and bye"'

+ ./main.sh '"hello' '"' '"hi' and 'bye"'

"hello

"

"hi

and

bye"

If the substitution appears within double quotes, word splitting and pathname expansion are not performed on the results.

"hello, then space, after which we have next token

" - that one character by itself, isolated by spaces from both sides.

"hi, again with space on left and right

and

bye"

edited Feb 4 at 15:01

msp9011

4,38044066

answered Jan 21 at 11:19

Sergiy Kolodyazhnyy

10.5k42663

TL;DR: the result of echo is a victim of word splitting on $() result, and sh test.sh "hello " "hi and bye" uses a different rule.

Let's take a peak as what's actually happening, when you add set -x for debugging:

> set -x

> ./main.sh $(echo '"hello " "hi and bye"')

++ echo '"hello " "hi and bye"'

+ ./main.sh '"hello' '"' '"hi' and 'bye"'

"hello

"

"hi

and

bye"

If the substitution appears within double quotes, word splitting and pathname expansion are not performed on the results.

"hello, then space, after which we have next token

" - that one character by itself, isolated by spaces from both sides.

"hi, again with space on left and right

and

bye"

edited Feb 4 at 15:01

msp9011

4,38044066

answered Jan 21 at 11:19

Sergiy Kolodyazhnyy

10.5k42663

edited Feb 4 at 15:01

msp9011

4,38044066

edited Feb 4 at 15:01

msp9011

4,38044066

edited Feb 4 at 15:01

msp9011

4,38044066

answered Jan 21 at 11:19

Sergiy Kolodyazhnyy

10.5k42663

answered Jan 21 at 11:19

Sergiy Kolodyazhnyy

10.5k42663

answered Jan 21 at 11:19

Sergiy Kolodyazhnyy

10.5k42663

add a comment |

You could change IFS to something other than space (since you want to protect the space in hi and bye), and use that something to separate the two strings:

$ IFS=:

$ sh test.sh $(echo "hello ":"hi and bye")

hello 

hi and bye

The order of operations (those relevant in this example, anyway) is somewhat like this:

substitutions

word splitting

quote removal

Note that quotes created as a result of substitutions aren't special in word splitting or quote removal, so a quote in the output of (1) passes through (2) and (3) like any other character. So:

In echo "hello " "hi and bye", the shell removes these quotes, so echo gets hello and hi and bye, so outputs hello hi and bye, joining the strings with a space.

In echo '"hello " "hi and bye"', the shell removes the outer ', the echo gets "hello " "hi and bye" and outputs that.

In sh test.sh $(echo '"hello " "hi and bye"'), the command substitution is replaced with "hello " "hi and bye", but these quotes are a result of a substitution, and so aren't involved in word splitting or quote removal.

So the shell splits those into "hello, ", "hi, and, bye", hence the output you get.

In sh test.sh $(echo "hello " "hi and bye"), the command substitution is replaced with hello hi and bye, which gets split into hello, hi and bye.

edited Jan 21 at 11:27

answered Jan 21 at 11:07

Olorin

3,8581721

1

Thanks. I guess what I am not understanding is why the usual shell substitution is not occurring, i.e. why is the behaviour not the same as if I do sh test.sh "hello " "hi and bye"?

– user1207217
Jan 21 at 11:09

@user1207217 does the edit help?

– Olorin
Jan 21 at 11:27

In $(echo "hello ":"hi and bye"), you might as well write $(echo "hello :hi and bye"). The colon works the same quoted or not.

– ilkkachu
Jan 21 at 11:31

add a comment |

You could change IFS to something other than space (since you want to protect the space in hi and bye), and use that something to separate the two strings:

$ IFS=:

$ sh test.sh $(echo "hello ":"hi and bye")

hello 

hi and bye

The order of operations (those relevant in this example, anyway) is somewhat like this:

substitutions

word splitting

quote removal

Note that quotes created as a result of substitutions aren't special in word splitting or quote removal, so a quote in the output of (1) passes through (2) and (3) like any other character. So:

In echo "hello " "hi and bye", the shell removes these quotes, so echo gets hello and hi and bye, so outputs hello hi and bye, joining the strings with a space.

In echo '"hello " "hi and bye"', the shell removes the outer ', the echo gets "hello " "hi and bye" and outputs that.

In sh test.sh $(echo '"hello " "hi and bye"'), the command substitution is replaced with "hello " "hi and bye", but these quotes are a result of a substitution, and so aren't involved in word splitting or quote removal.

So the shell splits those into "hello, ", "hi, and, bye", hence the output you get.

In sh test.sh $(echo "hello " "hi and bye"), the command substitution is replaced with hello hi and bye, which gets split into hello, hi and bye.

edited Jan 21 at 11:27

answered Jan 21 at 11:07

Olorin

3,8581721

1

Thanks. I guess what I am not understanding is why the usual shell substitution is not occurring, i.e. why is the behaviour not the same as if I do sh test.sh "hello " "hi and bye"?

– user1207217
Jan 21 at 11:09

@user1207217 does the edit help?

– Olorin
Jan 21 at 11:27

In $(echo "hello ":"hi and bye"), you might as well write $(echo "hello :hi and bye"). The colon works the same quoted or not.

– ilkkachu
Jan 21 at 11:31

add a comment |

You could change IFS to something other than space (since you want to protect the space in hi and bye), and use that something to separate the two strings:

$ IFS=:

$ sh test.sh $(echo "hello ":"hi and bye")

hello 

hi and bye

The order of operations (those relevant in this example, anyway) is somewhat like this:

substitutions

word splitting

quote removal

Note that quotes created as a result of substitutions aren't special in word splitting or quote removal, so a quote in the output of (1) passes through (2) and (3) like any other character. So:

In echo "hello " "hi and bye", the shell removes these quotes, so echo gets hello and hi and bye, so outputs hello hi and bye, joining the strings with a space.

In echo '"hello " "hi and bye"', the shell removes the outer ', the echo gets "hello " "hi and bye" and outputs that.

In sh test.sh $(echo '"hello " "hi and bye"'), the command substitution is replaced with "hello " "hi and bye", but these quotes are a result of a substitution, and so aren't involved in word splitting or quote removal.

So the shell splits those into "hello, ", "hi, and, bye", hence the output you get.

In sh test.sh $(echo "hello " "hi and bye"), the command substitution is replaced with hello hi and bye, which gets split into hello, hi and bye.

edited Jan 21 at 11:27

answered Jan 21 at 11:07

Olorin

3,8581721

You could change IFS to something other than space (since you want to protect the space in hi and bye), and use that something to separate the two strings:

$ IFS=:

$ sh test.sh $(echo "hello ":"hi and bye")

hello 

hi and bye

The order of operations (those relevant in this example, anyway) is somewhat like this:

substitutions

word splitting

quote removal

Note that quotes created as a result of substitutions aren't special in word splitting or quote removal, so a quote in the output of (1) passes through (2) and (3) like any other character. So:

In echo "hello " "hi and bye", the shell removes these quotes, so echo gets hello and hi and bye, so outputs hello hi and bye, joining the strings with a space.

In echo '"hello " "hi and bye"', the shell removes the outer ', the echo gets "hello " "hi and bye" and outputs that.

In sh test.sh $(echo '"hello " "hi and bye"'), the command substitution is replaced with "hello " "hi and bye", but these quotes are a result of a substitution, and so aren't involved in word splitting or quote removal.

So the shell splits those into "hello, ", "hi, and, bye", hence the output you get.

In sh test.sh $(echo "hello " "hi and bye"), the command substitution is replaced with hello hi and bye, which gets split into hello, hi and bye.

edited Jan 21 at 11:27

answered Jan 21 at 11:07

Olorin

3,8581721

edited Jan 21 at 11:27

answered Jan 21 at 11:07

Olorin

3,8581721

answered Jan 21 at 11:07

Olorin

3,8581721

answered Jan 21 at 11:07

Olorin

3,8581721

1

Thanks. I guess what I am not understanding is why the usual shell substitution is not occurring, i.e. why is the behaviour not the same as if I do sh test.sh "hello " "hi and bye"?

– user1207217
Jan 21 at 11:09

@user1207217 does the edit help?

– Olorin
Jan 21 at 11:27

In $(echo "hello ":"hi and bye"), you might as well write $(echo "hello :hi and bye"). The colon works the same quoted or not.

– ilkkachu
Jan 21 at 11:31

add a comment |

1

Thanks. I guess what I am not understanding is why the usual shell substitution is not occurring, i.e. why is the behaviour not the same as if I do sh test.sh "hello " "hi and bye"?

– user1207217
Jan 21 at 11:09

@user1207217 does the edit help?

– Olorin
Jan 21 at 11:27

In $(echo "hello ":"hi and bye"), you might as well write $(echo "hello :hi and bye"). The colon works the same quoted or not.

– ilkkachu
Jan 21 at 11:31

Thanks. I guess what I am not understanding is why the usual shell substitution is not occurring, i.e. why is the behaviour not the same as if I do sh test.sh "hello " "hi and bye"?

– user1207217
Jan 21 at 11:09

@user1207217 does the edit help?

– Olorin
Jan 21 at 11:27

In $(echo "hello ":"hi and bye"), you might as well write $(echo "hello :hi and bye"). The colon works the same quoted or not.

– ilkkachu
Jan 21 at 11:31

add a comment |

How about

sh test.sh "hello " "hi and bye"

answered Jan 21 at 11:13

gerhard d.

1,256311

I'm aware I can do this, the shell script is a MWE of an unexpected behaviour -- I'm trying to use a command substitution to generate a long list of args (some of which contain spaces) but it doesn't work correctly because even with quoting in the command output the above is occurring, i.e. the shell is not doing its usual parsing and treating everything inside a single or double quote pair as a single arg

– user1207217
Jan 21 at 11:15

add a comment |

How about

sh test.sh "hello " "hi and bye"

answered Jan 21 at 11:13

gerhard d.

1,256311

I'm aware I can do this, the shell script is a MWE of an unexpected behaviour -- I'm trying to use a command substitution to generate a long list of args (some of which contain spaces) but it doesn't work correctly because even with quoting in the command output the above is occurring, i.e. the shell is not doing its usual parsing and treating everything inside a single or double quote pair as a single arg

– user1207217
Jan 21 at 11:15

add a comment |

How about

sh test.sh "hello " "hi and bye"

answered Jan 21 at 11:13

gerhard d.

1,256311

How about

sh test.sh "hello " "hi and bye"

answered Jan 21 at 11:13

gerhard d.

1,256311

answered Jan 21 at 11:13

gerhard d.

1,256311

answered Jan 21 at 11:13

gerhard d.

1,256311

answered Jan 21 at 11:13

gerhard d.

1,256311

I'm aware I can do this, the shell script is a MWE of an unexpected behaviour -- I'm trying to use a command substitution to generate a long list of args (some of which contain spaces) but it doesn't work correctly because even with quoting in the command output the above is occurring, i.e. the shell is not doing its usual parsing and treating everything inside a single or double quote pair as a single arg

– user1207217
Jan 21 at 11:15

add a comment |

I'm aware I can do this, the shell script is a MWE of an unexpected behaviour -- I'm trying to use a command substitution to generate a long list of args (some of which contain spaces) but it doesn't work correctly because even with quoting in the command output the above is occurring, i.e. the shell is not doing its usual parsing and treating everything inside a single or double quote pair as a single arg

– user1207217
Jan 21 at 11:15

I'm aware I can do this, the shell script is a MWE of an unexpected behaviour -- I'm trying to use a command substitution to generate a long list of args (some of which contain spaces) but it doesn't work correctly because even with quoting in the command output the above is occurring, i.e. the shell is not doing its usual parsing and treating everything inside a single or double quote pair as a single arg

– user1207217
Jan 21 at 11:15

add a comment |

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