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0

I have aliased pushd in my bash shell as follows so that it suppresses output:

alias pushd='pushd "$@" > /dev/null'

This works fine most of the time, but I'm running into trouble now using it inside functions that take arguments. For example,

test() {

  pushd .

  ...

}

Running test without arguments is fine. But with arguments:

> test x y z

bash: pushd: too many arguments

I take it that pushd is trying to take . x y z as arguments instead of just .. How can I prevent this? Is there a "local" equivalent of $@ that would only see . and not x y z?

bash alias function pushd

share|improve this question

edited 2 days ago

Jeff Schaller♦

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asked 2 days ago

ThéophileThéophile

10333 bronze badges

New contributor

Théophile is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Why are you using $@ at all?
  
  – Wildcard
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Wildcard Because I had copy/pasted that line from someone else who was apparently also a beginner like me. :S
  
  – Théophile
  yesterday
  

  
  
  
  

add a comment | 

0

I have aliased pushd in my bash shell as follows so that it suppresses output:

alias pushd='pushd "$@" > /dev/null'

This works fine most of the time, but I'm running into trouble now using it inside functions that take arguments. For example,

test() {

  pushd .

  ...

}

Running test without arguments is fine. But with arguments:

> test x y z

bash: pushd: too many arguments

I take it that pushd is trying to take . x y z as arguments instead of just .. How can I prevent this? Is there a "local" equivalent of $@ that would only see . and not x y z?

bash alias function pushd

share|improve this question

edited 2 days ago

Jeff Schaller♦

49k1111 gold badges7272 silver badges162162 bronze badges

asked 2 days ago

ThéophileThéophile

10333 bronze badges

New contributor

Théophile is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Why are you using $@ at all?
  
  – Wildcard
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Wildcard Because I had copy/pasted that line from someone else who was apparently also a beginner like me. :S
  
  – Théophile
  yesterday
  

  
  
  
  

add a comment | 

I have aliased pushd in my bash shell as follows so that it suppresses output:

alias pushd='pushd "$@" > /dev/null'

This works fine most of the time, but I'm running into trouble now using it inside functions that take arguments. For example,

test() {

  pushd .

  ...

}

Running test without arguments is fine. But with arguments:

> test x y z

bash: pushd: too many arguments

I take it that pushd is trying to take . x y z as arguments instead of just .. How can I prevent this? Is there a "local" equivalent of $@ that would only see . and not x y z?

bash alias function pushd

share|improve this question

edited 2 days ago

Jeff Schaller♦

49k1111 gold badges7272 silver badges162162 bronze badges

asked 2 days ago

ThéophileThéophile

10333 bronze badges

New contributor

Théophile is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

alias pushd='pushd "$@" > /dev/null'

test() {

  pushd .

  ...

}

> test x y z

bash: pushd: too many arguments

share|improve this question

edited 2 days ago

Jeff Schaller♦

49k1111 gold badges7272 silver badges162162 bronze badges

asked 2 days ago

ThéophileThéophile

10333 bronze badges

New contributor

Théophile is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited 2 days ago

Jeff Schaller♦

49k1111 gold badges7272 silver badges162162 bronze badges

asked 2 days ago

ThéophileThéophile

10333 bronze badges

New contributor

Théophile is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 2 days ago

Jeff Schaller♦

49k1111 gold badges7272 silver badges162162 bronze badges

edited 2 days ago

Jeff Schaller♦

49k1111 gold badges7272 silver badges162162 bronze badges

asked 2 days ago

ThéophileThéophile

10333 bronze badges

New contributor

Théophile is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 2 days ago

ThéophileThéophile

10333 bronze badges

1 Answer
1

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5

Aliases define a way to replace a shell token with some string before the shell event tries to parse code. It's not a programming structure like a function.

In

alias pushd='pushd "$@" > /dev/null'

and then:

pushd .

What's going on is that the pushd is replaced with pushd "$@" > /dev/null and then the result parsed. So the shell ends up parsing:

pushd "$@" > /dev/null .

Redirections can appear anywhere on the command line, so it's exactly the same as:

pushd "$@" . > /dev/null

or

> /dev/null pushd "$@" .

When you're running that from the prompt, "$@" is the list of arguments your shell received so unless you ran set arg1 arg2, that will likely be empty, so it will be the same as

pushd . > /dev/null

But within a function, that "$@" will be the arguments of the function.

Here, you either want to define pushd as a function like:

pushd() { command pushd "$@" > /dev/null; }

Or an alias like:

alias pushd='> /dev/null pushd'

or

alias pushd='pushd > /dev/null

share|improve this answer

answered 2 days ago

Stéphane ChazelasStéphane Chazelas

330k5858 gold badges642642 silver badges10091009 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Amazing! Thank you.
  
  – Théophile
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I had actually thought of writing it as a function, but I didn't know about command yet, so of course it created an infinite loop.
  
  – Théophile
  2 days ago
  
  
  

  
  
  
  

add a comment | 

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5

Aliases define a way to replace a shell token with some string before the shell event tries to parse code. It's not a programming structure like a function.

In

alias pushd='pushd "$@" > /dev/null'

and then:

pushd .

What's going on is that the pushd is replaced with pushd "$@" > /dev/null and then the result parsed. So the shell ends up parsing:

pushd "$@" > /dev/null .

Redirections can appear anywhere on the command line, so it's exactly the same as:

pushd "$@" . > /dev/null

or

> /dev/null pushd "$@" .

When you're running that from the prompt, "$@" is the list of arguments your shell received so unless you ran set arg1 arg2, that will likely be empty, so it will be the same as

pushd . > /dev/null

But within a function, that "$@" will be the arguments of the function.

Here, you either want to define pushd as a function like:

pushd() { command pushd "$@" > /dev/null; }

Or an alias like:

alias pushd='> /dev/null pushd'

or

alias pushd='pushd > /dev/null

share|improve this answer

answered 2 days ago

Stéphane ChazelasStéphane Chazelas

330k5858 gold badges642642 silver badges10091009 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Amazing! Thank you.
  
  – Théophile
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I had actually thought of writing it as a function, but I didn't know about command yet, so of course it created an infinite loop.
  
  – Théophile
  2 days ago
  
  
  

  
  
  
  

add a comment | 

5

Aliases define a way to replace a shell token with some string before the shell event tries to parse code. It's not a programming structure like a function.

In

alias pushd='pushd "$@" > /dev/null'

and then:

pushd .

What's going on is that the pushd is replaced with pushd "$@" > /dev/null and then the result parsed. So the shell ends up parsing:

pushd "$@" > /dev/null .

Redirections can appear anywhere on the command line, so it's exactly the same as:

pushd "$@" . > /dev/null

or

> /dev/null pushd "$@" .

When you're running that from the prompt, "$@" is the list of arguments your shell received so unless you ran set arg1 arg2, that will likely be empty, so it will be the same as

pushd . > /dev/null

But within a function, that "$@" will be the arguments of the function.

Here, you either want to define pushd as a function like:

pushd() { command pushd "$@" > /dev/null; }

Or an alias like:

alias pushd='> /dev/null pushd'

or

alias pushd='pushd > /dev/null

share|improve this answer

answered 2 days ago

Stéphane ChazelasStéphane Chazelas

330k5858 gold badges642642 silver badges10091009 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Amazing! Thank you.
  
  – Théophile
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I had actually thought of writing it as a function, but I didn't know about command yet, so of course it created an infinite loop.
  
  – Théophile
  2 days ago
  
  
  

  
  
  
  

add a comment | 

Aliases define a way to replace a shell token with some string before the shell event tries to parse code. It's not a programming structure like a function.

In

alias pushd='pushd "$@" > /dev/null'

and then:

pushd .

What's going on is that the pushd is replaced with pushd "$@" > /dev/null and then the result parsed. So the shell ends up parsing:

pushd "$@" > /dev/null .

Redirections can appear anywhere on the command line, so it's exactly the same as:

pushd "$@" . > /dev/null

or

> /dev/null pushd "$@" .

When you're running that from the prompt, "$@" is the list of arguments your shell received so unless you ran set arg1 arg2, that will likely be empty, so it will be the same as

pushd . > /dev/null

But within a function, that "$@" will be the arguments of the function.

Here, you either want to define pushd as a function like:

pushd() { command pushd "$@" > /dev/null; }

Or an alias like:

alias pushd='> /dev/null pushd'

or

alias pushd='pushd > /dev/null

share|improve this answer

answered 2 days ago

Stéphane ChazelasStéphane Chazelas

330k5858 gold badges642642 silver badges10091009 bronze badges

alias pushd='pushd "$@" > /dev/null'

pushd .

pushd "$@" > /dev/null .

pushd "$@" . > /dev/null

> /dev/null pushd "$@" .

pushd . > /dev/null

pushd() { command pushd "$@" > /dev/null; }

alias pushd='> /dev/null pushd'

alias pushd='pushd > /dev/null

share|improve this answer

answered 2 days ago

Stéphane ChazelasStéphane Chazelas

330k5858 gold badges642642 silver badges10091009 bronze badges

answered 2 days ago

Stéphane ChazelasStéphane Chazelas

330k5858 gold badges642642 silver badges10091009 bronze badges

answered 2 days ago

Stéphane ChazelasStéphane Chazelas

330k5858 gold badges642642 silver badges10091009 bronze badges