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using 'echo' & 'printf' in bash function calls

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2

I have a simple shell function call and I'm using echo and printf commands to print the parameter I'm passing. I have noticed the following:

1. 
   echo is printing the output


2. 
   printf is not printing the output

Am I missing something here?

check_host(){

    # prints output

    echo $1



    # does not print the output

    printf $1



    }

check_host $(hostname)

bash shell-script

share|improve this question

edited 3 hours ago

terdon♦

131k32258436

asked 3 hours ago

sqlcheckpointsqlcheckpoint

1134

New contributor

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  that works :) any explanation on this ?
  
  – sqlcheckpoint
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  See also unix.stackexchange.com/questions/131766/…
  
  – Jeff Schaller
  3 hours ago
  

  
  
  
  

add a comment | 

2

I have a simple shell function call and I'm using echo and printf commands to print the parameter I'm passing. I have noticed the following:

1. 
   echo is printing the output


2. 
   printf is not printing the output

Am I missing something here?

check_host(){

    # prints output

    echo $1



    # does not print the output

    printf $1



    }

check_host $(hostname)

bash shell-script

share|improve this question

edited 3 hours ago

terdon♦

131k32258436

asked 3 hours ago

sqlcheckpointsqlcheckpoint

1134

New contributor

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  that works :) any explanation on this ?
  
  – sqlcheckpoint
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  See also unix.stackexchange.com/questions/131766/…
  
  – Jeff Schaller
  3 hours ago
  

  
  
  
  

add a comment | 

I have a simple shell function call and I'm using echo and printf commands to print the parameter I'm passing. I have noticed the following:

1. 
   echo is printing the output


2. 
   printf is not printing the output

Am I missing something here?

check_host(){

    # prints output

    echo $1



    # does not print the output

    printf $1



    }

check_host $(hostname)

bash shell-script

share|improve this question

edited 3 hours ago

terdon♦

131k32258436

asked 3 hours ago

sqlcheckpointsqlcheckpoint

1134

New contributor

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

check_host(){

    # prints output

    echo $1



    # does not print the output

    printf $1



    }

check_host $(hostname)

share|improve this question

edited 3 hours ago

terdon♦

131k32258436

asked 3 hours ago

sqlcheckpointsqlcheckpoint

1134

New contributor

sqlcheckpoint 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 3 hours ago

terdon♦

131k32258436

asked 3 hours ago

sqlcheckpointsqlcheckpoint

1134

New contributor

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

edited 3 hours ago

terdon♦

131k32258436

edited 3 hours ago

terdon♦

131k32258436

asked 3 hours ago

sqlcheckpointsqlcheckpoint

1134

New contributor

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

asked 3 hours ago

sqlcheckpointsqlcheckpoint

1134

1 Answer
1

active

oldest

votes

6

The function you show would print the first argument twice, once with a newline appended, and with no newline at the end of the of second output.

E.g. in an interactive Bash shell, you'd get something like this

user@foo /tmp$ check_host foo

foo

foouser@foo /tmp$ 

The output from printf is there, just not on a line of its own.

The difference between echo and printf is that echo prints a newline at the end even if you don't ask for it (you can prevent that in Bash by using echo -n), and printf works more like the printf() function in C, in that it only prints what you ask. You'll have to explicitly use n in the printf format string to get the newline.

Note that in general, you'd want to quote those variables and the command substitution to prevent issues with word splitting. This probably isn't a problem with the hostname, but if you have values with whitespace, you'll need it.

So:

check_host() {

    echo "$1"

    printf "%sn" "$1"

}

check_host "$(hostname)"

Printing arbitrary data with printf should also be done through the %s format specifier as above. Otherwise any % signs in the data would be interpreted by printf.

Also see: Why is printf better than echo?

share|improve this answer

answered 3 hours ago

ilkkachuilkkachu

60.1k997169

add a comment | 

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6

The function you show would print the first argument twice, once with a newline appended, and with no newline at the end of the of second output.

E.g. in an interactive Bash shell, you'd get something like this

user@foo /tmp$ check_host foo

foo

foouser@foo /tmp$ 

The output from printf is there, just not on a line of its own.

The difference between echo and printf is that echo prints a newline at the end even if you don't ask for it (you can prevent that in Bash by using echo -n), and printf works more like the printf() function in C, in that it only prints what you ask. You'll have to explicitly use n in the printf format string to get the newline.

Note that in general, you'd want to quote those variables and the command substitution to prevent issues with word splitting. This probably isn't a problem with the hostname, but if you have values with whitespace, you'll need it.

So:

check_host() {

    echo "$1"

    printf "%sn" "$1"

}

check_host "$(hostname)"

Printing arbitrary data with printf should also be done through the %s format specifier as above. Otherwise any % signs in the data would be interpreted by printf.

Also see: Why is printf better than echo?

share|improve this answer

answered 3 hours ago

ilkkachuilkkachu

60.1k997169

add a comment | 

6

The function you show would print the first argument twice, once with a newline appended, and with no newline at the end of the of second output.

E.g. in an interactive Bash shell, you'd get something like this

user@foo /tmp$ check_host foo

foo

foouser@foo /tmp$ 

The output from printf is there, just not on a line of its own.

The difference between echo and printf is that echo prints a newline at the end even if you don't ask for it (you can prevent that in Bash by using echo -n), and printf works more like the printf() function in C, in that it only prints what you ask. You'll have to explicitly use n in the printf format string to get the newline.

Note that in general, you'd want to quote those variables and the command substitution to prevent issues with word splitting. This probably isn't a problem with the hostname, but if you have values with whitespace, you'll need it.

So:

check_host() {

    echo "$1"

    printf "%sn" "$1"

}

check_host "$(hostname)"

Printing arbitrary data with printf should also be done through the %s format specifier as above. Otherwise any % signs in the data would be interpreted by printf.

Also see: Why is printf better than echo?

share|improve this answer

answered 3 hours ago

ilkkachuilkkachu

60.1k997169

add a comment | 

The function you show would print the first argument twice, once with a newline appended, and with no newline at the end of the of second output.

E.g. in an interactive Bash shell, you'd get something like this

user@foo /tmp$ check_host foo

foo

foouser@foo /tmp$ 

The output from printf is there, just not on a line of its own.

The difference between echo and printf is that echo prints a newline at the end even if you don't ask for it (you can prevent that in Bash by using echo -n), and printf works more like the printf() function in C, in that it only prints what you ask. You'll have to explicitly use n in the printf format string to get the newline.

Note that in general, you'd want to quote those variables and the command substitution to prevent issues with word splitting. This probably isn't a problem with the hostname, but if you have values with whitespace, you'll need it.

So:

check_host() {

    echo "$1"

    printf "%sn" "$1"

}

check_host "$(hostname)"

Printing arbitrary data with printf should also be done through the %s format specifier as above. Otherwise any % signs in the data would be interpreted by printf.

Also see: Why is printf better than echo?

share|improve this answer

answered 3 hours ago

ilkkachuilkkachu

60.1k997169

user@foo /tmp$ check_host foo

foo

foouser@foo /tmp$ 

check_host() {

    echo "$1"

    printf "%sn" "$1"

}

check_host "$(hostname)"

share|improve this answer

answered 3 hours ago

ilkkachuilkkachu

60.1k997169

answered 3 hours ago

ilkkachuilkkachu

60.1k997169

answered 3 hours ago

ilkkachuilkkachu

60.1k997169