[SOLVED] using printf with a variable

ip_address · 03-17-2013, 09:23 PM

I want to control the number of times a string (in this case "hi") could be printed using a variable (x). How it could be achieved?

For example, I would like to print "hi" 5 times for now but i want to control the number of times "hi" could be printed using a variable.

Code:

x=5

printf "hi\n%.0s" {1.."$x"}

I want result to be like this:

hi
hi
hi
hi
hi

danielbmartin · 03-17-2013, 10:27 PM

This code ...

Code:

num=5
yes "hi" |head -n$num

... produces this result ...

Code:

hi
hi
hi
hi
hi

There are many ways to achieve this.

Daniel B. Martin

danielbmartin · 03-17-2013, 10:40 PM

This code ...

Code:

n=5
awk -v n=$n 'BEGIN{for (i=1;i<=n;i++) print "hi"}'

... produces this result ...

Code:

hi
hi
hi
hi
hi

There are many ways to achieve this.

Daniel B. Martin

cfajohnson · 03-17-2013, 11:03 PM

Code:

repeat()
{
 local num=$1
 shift
 while [ $(( num -= 1 )) -ge 0 ]
 do
   eval "$@"
 done
}

Example:

Code:

$ n=5
$ repeat "$n" 'echo hi'
hi
hi
hi
hi
hi

millgates · 03-18-2013, 02:25 AM

Quote:

Originally Posted by ip_address

I want to control the number of times a string (in this case "hi") could be printed using a variable (x). How it could be achieved?

For example, I would like to print "hi" 5 times for now but i want to control the number of times "hi" could be printed using a variable.

Code:

x=5

printf "hi\n%.0s" {1.."$x"}

I want result to be like this:

hi
hi
hi
hi
hi

The problem with your code is that in bash, brace expansion happens before variable substitution and bash
therefore the braces containing a variable cannot be properly expanded. You can use seq instead, though:

Code:

printf "hi\n%.0s" $(seq 1.."$x")

danielbmartin · 03-18-2013, 07:22 AM

Quote:

Originally Posted by millgates

Code:

printf "hi\n%.0s" $(seq 1.."$x")

I tried this in a bash shell with this unexpected result...

Code:

seq: invalid floating point argument: 1..
Try `seq --help' for more information.
hi

Daniel B. Martin

millgates · 03-18-2013, 08:07 AM

Quote:

Originally Posted by danielbmartin

I tried this in a bash shell with this unexpected result...

Code:

seq: invalid floating point argument: 1..
Try `seq --help' for more information.
hi

Daniel B. Martin

Oops... those dots are not supposed to be there, of course. Sorry.

Code:

printf "hi\n%.0s" $(seq 1 "$x")

Ramurd · 03-18-2013, 08:16 AM

or

Code:

x=5

for((i=0;i<$x;i++))
do
   printf "Hi\n"
done

danielbmartin · 03-18-2013, 09:52 PM

Quote:

Originally Posted by millgates

Code:

printf "hi\n%.0s" $(seq 1 "$x")

I had better luck with this ...

Code:

x=5
printf "hi\n%.0s" $(seq "$x")

Daniel B. Martin

David the H. · 03-20-2013, 05:26 AM

As millgates said, brace expansion with a variable doesn't work because the braces are expanded before the variable.

Bash Pitfall #33
brace expansion

You're generally much better off using a loop of some kind, as posted above. Although here's another technique you can use with only bash built-ins:

Code:

x=5
printf -v repvar '%0*d' "$x"
printf '%b' "${repvar//0/hi\n}"

The first printf generates a string of zeros equal in length to the input number. The '*' in the format string takes the first (and only) argument as the padding number to use, instead of printing it directly. printf always prints at least once, so no other arguments are needed. '-v' captures the result in a variable instead of printing it directly.

Then a global replacement parameter substitution is used to replace each zero with the desired string. The '%b' format token in the second printf expands backslash escapes like '\n' in the input (making it equivalent to 'echo -en'), thus giving you a newline after each 'hi'.