Why C readlink() has ELOOP as a possible error









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Background



I read about the Open Group Specification about readlink(), and there is an error called ELOOP, which indicates "A loop exists in symbolic links encountered during resolution of the path argument.", so I assume this function will continue path resolution until encountering a non-link file.



However, I did an experiment and found that readlink() only resolve the passed in path argument and just stops there but not keep resolving until reaching a non-link file.



My Problem



  1. If it's for realpath(), that makes all the sense to have ELOOP as a possible error. But why does ELOOP even exist for readlink() while it only resolves the path once?

  2. I saw this on the spec "The [ELOOP] optional error condition is added to align with the IEEE P1003.1a draft standard", does that mean the behavior of readlink() (whether it keeps resolving until reaching a non-link file) depends on implementation?

my gcc version is 8.2.1










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    up vote
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    down vote

    favorite












    Background



    I read about the Open Group Specification about readlink(), and there is an error called ELOOP, which indicates "A loop exists in symbolic links encountered during resolution of the path argument.", so I assume this function will continue path resolution until encountering a non-link file.



    However, I did an experiment and found that readlink() only resolve the passed in path argument and just stops there but not keep resolving until reaching a non-link file.



    My Problem



    1. If it's for realpath(), that makes all the sense to have ELOOP as a possible error. But why does ELOOP even exist for readlink() while it only resolves the path once?

    2. I saw this on the spec "The [ELOOP] optional error condition is added to align with the IEEE P1003.1a draft standard", does that mean the behavior of readlink() (whether it keeps resolving until reaching a non-link file) depends on implementation?

    my gcc version is 8.2.1










    share|improve this question

























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Background



      I read about the Open Group Specification about readlink(), and there is an error called ELOOP, which indicates "A loop exists in symbolic links encountered during resolution of the path argument.", so I assume this function will continue path resolution until encountering a non-link file.



      However, I did an experiment and found that readlink() only resolve the passed in path argument and just stops there but not keep resolving until reaching a non-link file.



      My Problem



      1. If it's for realpath(), that makes all the sense to have ELOOP as a possible error. But why does ELOOP even exist for readlink() while it only resolves the path once?

      2. I saw this on the spec "The [ELOOP] optional error condition is added to align with the IEEE P1003.1a draft standard", does that mean the behavior of readlink() (whether it keeps resolving until reaching a non-link file) depends on implementation?

      my gcc version is 8.2.1










      share|improve this question















      Background



      I read about the Open Group Specification about readlink(), and there is an error called ELOOP, which indicates "A loop exists in symbolic links encountered during resolution of the path argument.", so I assume this function will continue path resolution until encountering a non-link file.



      However, I did an experiment and found that readlink() only resolve the passed in path argument and just stops there but not keep resolving until reaching a non-link file.



      My Problem



      1. If it's for realpath(), that makes all the sense to have ELOOP as a possible error. But why does ELOOP even exist for readlink() while it only resolves the path once?

      2. I saw this on the spec "The [ELOOP] optional error condition is added to align with the IEEE P1003.1a draft standard", does that mean the behavior of readlink() (whether it keeps resolving until reaching a non-link file) depends on implementation?

      my gcc version is 8.2.1







      c symlink readlink






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      edited Nov 10 at 14:09









      melpomene

      55.8k54387




      55.8k54387










      asked Nov 10 at 13:58









      David Chen

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      4891614






















          1 Answer
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          readlink gives you the immediate target of a symbolic link. But what if resolving the path to the symbolic link involves another symlink?



          Take readlink("/foo/bar") as an example. It's supposed to return the link target of bar, but if /foo is a symlink pointing to itself, you'll get ELOOP because readlink has to resolve the directory part before getting to the final entry.



          See also man path_resolution.






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            1 Answer
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            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            3
            down vote



            accepted










            readlink gives you the immediate target of a symbolic link. But what if resolving the path to the symbolic link involves another symlink?



            Take readlink("/foo/bar") as an example. It's supposed to return the link target of bar, but if /foo is a symlink pointing to itself, you'll get ELOOP because readlink has to resolve the directory part before getting to the final entry.



            See also man path_resolution.






            share|improve this answer
























              up vote
              3
              down vote



              accepted










              readlink gives you the immediate target of a symbolic link. But what if resolving the path to the symbolic link involves another symlink?



              Take readlink("/foo/bar") as an example. It's supposed to return the link target of bar, but if /foo is a symlink pointing to itself, you'll get ELOOP because readlink has to resolve the directory part before getting to the final entry.



              See also man path_resolution.






              share|improve this answer






















                up vote
                3
                down vote



                accepted







                up vote
                3
                down vote



                accepted






                readlink gives you the immediate target of a symbolic link. But what if resolving the path to the symbolic link involves another symlink?



                Take readlink("/foo/bar") as an example. It's supposed to return the link target of bar, but if /foo is a symlink pointing to itself, you'll get ELOOP because readlink has to resolve the directory part before getting to the final entry.



                See also man path_resolution.






                share|improve this answer












                readlink gives you the immediate target of a symbolic link. But what if resolving the path to the symbolic link involves another symlink?



                Take readlink("/foo/bar") as an example. It's supposed to return the link target of bar, but if /foo is a symlink pointing to itself, you'll get ELOOP because readlink has to resolve the directory part before getting to the final entry.



                See also man path_resolution.







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Nov 10 at 14:03









                melpomene

                55.8k54387




                55.8k54387



























                     

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