what is diff bentween xargs with braces and without in linux

12,756

Solution 1

xargs rm will invoke rm with all arguments as parameter departed with spaces.

xargs -i{} rm {} will invoke rm {} for each of the argument and {} will be replaced by the current argument.

If you have 2 arguments a.txt and b.txt, xargs rm will call this

rm a.txt b.txt

But xargs -i{} rm {} will call

rm a.txt
rm b.txt

This is because -i option implies -L 1 option which mean the command rm will take only 1 line each time. And here each line contains only 1 argument.

Check this Ideone link to get more idea about it.

Solution 2

With braces it will spawn one rm process per file. Without the braces, xargs will pass as many filenames as possible to each rm command.

Compare

ls | xargs echo

and

ls | xargs -i echo '{}'

Solution 3

-i option (equivalent to --replace) creates a sort of placeholder where xargs stores the input it just received. In your second command, the placeholder is "{}", it works like find -exec option. Once defined, xargs will replace this placeholder with the entire line of input. If you don´t like the "{}" name, you can define your own:

ls | xargs -iPLACEHOLDER echo PLACEHOLDER

In your case, both commands are producing the same result. In the second form, you are just making explicit the default behaviour with the -i option.

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Updated on September 18, 2022

Comments

  • Admin
    Admin over 1 year

    I want to know what is the difference between this

    ls | xargs rm

    ls | xargs -i{} rm {}

    Both are working for me

  • Shiplu Mokaddim
    Shiplu Mokaddim over 11 years
    @user19140477031 depends one what operation you are performing. for rm it does not matter