Is a file creation mask setted by a given shell's umask is typically unique to the operating system entirely or just to that given shell?

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
2
down vote

favorite












Is a file creation mask setted by a given shell's umask is typically unique to the operating system entirely or just to that given shell?



For example, if I change the file creation mask (umask's mask/bitmask) in Bash will it change only for Bash or also for possible other existing shells in my operating system like Dash, ksh, zsh and so forth? ("a case were one shell effects others").










share|improve this question



























    up vote
    2
    down vote

    favorite












    Is a file creation mask setted by a given shell's umask is typically unique to the operating system entirely or just to that given shell?



    For example, if I change the file creation mask (umask's mask/bitmask) in Bash will it change only for Bash or also for possible other existing shells in my operating system like Dash, ksh, zsh and so forth? ("a case were one shell effects others").










    share|improve this question

























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      Is a file creation mask setted by a given shell's umask is typically unique to the operating system entirely or just to that given shell?



      For example, if I change the file creation mask (umask's mask/bitmask) in Bash will it change only for Bash or also for possible other existing shells in my operating system like Dash, ksh, zsh and so forth? ("a case were one shell effects others").










      share|improve this question















      Is a file creation mask setted by a given shell's umask is typically unique to the operating system entirely or just to that given shell?



      For example, if I change the file creation mask (umask's mask/bitmask) in Bash will it change only for Bash or also for possible other existing shells in my operating system like Dash, ksh, zsh and so forth? ("a case were one shell effects others").







      umask






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 4 hours ago

























      asked 6 hours ago









      JohnDoea

      55831




      55831




















          1 Answer
          1






          active

          oldest

          votes

















          up vote
          3
          down vote



          accepted










          To understand umask, you first need to understand the structure of processes in Unix/Linux. Which is to say, they form a tree-like structure. Each process needs to have a parent, which is the process that spawned it. (With the exception of the first process, init). Each process may or may not spawn more processes, called children.



          Each process has a mask property. This is what is queried or set using the umask command.



          Processes inherit the mask of their parent. They can then change their own mask. For example, there is a umask() C function that can change the mask of the program you are writing, without needing to call umask from the shell.



          A child process cannot affect the mask of their parent. Therefore, changing the mask of a process will not affect the entire system. It will only affect any future child processes.



          Since the purpose of a shell is to be able to create and control other processes, a umask command is built in to most shells. This isn't required for a shell, it is possible to write a basic shell that has no umask function. But such a shell would not be considered useful as a general-purpose shell for logging in and administering a system.



          You can test what I'm saying yourself, using the fact that a shell like Bash can spawn other Bash shells (or whatever other shell you like):



          1. Open a terminal

          2. Run the umask command to query the current value

          3. Run bash (or whatever) to spawn a child shell

          4. Run umask to check the value of the child's mask

          5. Set the child's mask to something else, eg run umask 0000

          6. Run umask to check the child's mask again

          7. Exit from the child shell (run exit or press Ctrl-d)

          8. You are now back in the parent shell again. Run umask to check its mask

          Useful references:



          • man 1 umask


          • man 2 umask (This gives the reference for the umask() C function)


          • man bash (and search for umask)

          • https://en.wikipedia.org/wiki/Umask





          share|improve this answer






















          • Hello, I'm sad to say that I don't get an answer to my particular question. You might want to consider editing a bit to touch my question directly.
            – JohnDoea
            3 hours ago










          • This might result in upvotes...
            – JohnDoea
            3 hours ago










          • You want me to put a line at the top saying "Changing umask does not affect the whole operating system"?
            – cryptarch
            3 hours ago






          • 1




            Do I understand correct? --- A Bash process with mask X will inherit it to a Dash process but if we change it in the Dash process to Y, the ksh process will have Y But the Bash will still have X.
            – JohnDoea
            56 mins ago






          • 1




            Yes, in this case I am. Thanks and upvoted.
            – JohnDoea
            54 mins ago











          Your Answer








          StackExchange.ready(function()
          var channelOptions =
          tags: "".split(" "),
          id: "106"
          ;
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function()
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled)
          StackExchange.using("snippets", function()
          createEditor();
          );

          else
          createEditor();

          );

          function createEditor()
          StackExchange.prepareEditor(
          heartbeatType: 'answer',
          convertImagesToLinks: false,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: null,
          bindNavPrevention: true,
          postfix: "",
          imageUploader:
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          ,
          onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          );



          );













           

          draft saved


          draft discarded


















          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2funix.stackexchange.com%2fquestions%2f481146%2fis-a-file-creation-mask-setted-by-a-given-shells-umask-is-typically-unique-to-t%23new-answer', 'question_page');

          );

          Post as a guest






























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          3
          down vote



          accepted










          To understand umask, you first need to understand the structure of processes in Unix/Linux. Which is to say, they form a tree-like structure. Each process needs to have a parent, which is the process that spawned it. (With the exception of the first process, init). Each process may or may not spawn more processes, called children.



          Each process has a mask property. This is what is queried or set using the umask command.



          Processes inherit the mask of their parent. They can then change their own mask. For example, there is a umask() C function that can change the mask of the program you are writing, without needing to call umask from the shell.



          A child process cannot affect the mask of their parent. Therefore, changing the mask of a process will not affect the entire system. It will only affect any future child processes.



          Since the purpose of a shell is to be able to create and control other processes, a umask command is built in to most shells. This isn't required for a shell, it is possible to write a basic shell that has no umask function. But such a shell would not be considered useful as a general-purpose shell for logging in and administering a system.



          You can test what I'm saying yourself, using the fact that a shell like Bash can spawn other Bash shells (or whatever other shell you like):



          1. Open a terminal

          2. Run the umask command to query the current value

          3. Run bash (or whatever) to spawn a child shell

          4. Run umask to check the value of the child's mask

          5. Set the child's mask to something else, eg run umask 0000

          6. Run umask to check the child's mask again

          7. Exit from the child shell (run exit or press Ctrl-d)

          8. You are now back in the parent shell again. Run umask to check its mask

          Useful references:



          • man 1 umask


          • man 2 umask (This gives the reference for the umask() C function)


          • man bash (and search for umask)

          • https://en.wikipedia.org/wiki/Umask





          share|improve this answer






















          • Hello, I'm sad to say that I don't get an answer to my particular question. You might want to consider editing a bit to touch my question directly.
            – JohnDoea
            3 hours ago










          • This might result in upvotes...
            – JohnDoea
            3 hours ago










          • You want me to put a line at the top saying "Changing umask does not affect the whole operating system"?
            – cryptarch
            3 hours ago






          • 1




            Do I understand correct? --- A Bash process with mask X will inherit it to a Dash process but if we change it in the Dash process to Y, the ksh process will have Y But the Bash will still have X.
            – JohnDoea
            56 mins ago






          • 1




            Yes, in this case I am. Thanks and upvoted.
            – JohnDoea
            54 mins ago















          up vote
          3
          down vote



          accepted










          To understand umask, you first need to understand the structure of processes in Unix/Linux. Which is to say, they form a tree-like structure. Each process needs to have a parent, which is the process that spawned it. (With the exception of the first process, init). Each process may or may not spawn more processes, called children.



          Each process has a mask property. This is what is queried or set using the umask command.



          Processes inherit the mask of their parent. They can then change their own mask. For example, there is a umask() C function that can change the mask of the program you are writing, without needing to call umask from the shell.



          A child process cannot affect the mask of their parent. Therefore, changing the mask of a process will not affect the entire system. It will only affect any future child processes.



          Since the purpose of a shell is to be able to create and control other processes, a umask command is built in to most shells. This isn't required for a shell, it is possible to write a basic shell that has no umask function. But such a shell would not be considered useful as a general-purpose shell for logging in and administering a system.



          You can test what I'm saying yourself, using the fact that a shell like Bash can spawn other Bash shells (or whatever other shell you like):



          1. Open a terminal

          2. Run the umask command to query the current value

          3. Run bash (or whatever) to spawn a child shell

          4. Run umask to check the value of the child's mask

          5. Set the child's mask to something else, eg run umask 0000

          6. Run umask to check the child's mask again

          7. Exit from the child shell (run exit or press Ctrl-d)

          8. You are now back in the parent shell again. Run umask to check its mask

          Useful references:



          • man 1 umask


          • man 2 umask (This gives the reference for the umask() C function)


          • man bash (and search for umask)

          • https://en.wikipedia.org/wiki/Umask





          share|improve this answer






















          • Hello, I'm sad to say that I don't get an answer to my particular question. You might want to consider editing a bit to touch my question directly.
            – JohnDoea
            3 hours ago










          • This might result in upvotes...
            – JohnDoea
            3 hours ago










          • You want me to put a line at the top saying "Changing umask does not affect the whole operating system"?
            – cryptarch
            3 hours ago






          • 1




            Do I understand correct? --- A Bash process with mask X will inherit it to a Dash process but if we change it in the Dash process to Y, the ksh process will have Y But the Bash will still have X.
            – JohnDoea
            56 mins ago






          • 1




            Yes, in this case I am. Thanks and upvoted.
            – JohnDoea
            54 mins ago













          up vote
          3
          down vote



          accepted







          up vote
          3
          down vote



          accepted






          To understand umask, you first need to understand the structure of processes in Unix/Linux. Which is to say, they form a tree-like structure. Each process needs to have a parent, which is the process that spawned it. (With the exception of the first process, init). Each process may or may not spawn more processes, called children.



          Each process has a mask property. This is what is queried or set using the umask command.



          Processes inherit the mask of their parent. They can then change their own mask. For example, there is a umask() C function that can change the mask of the program you are writing, without needing to call umask from the shell.



          A child process cannot affect the mask of their parent. Therefore, changing the mask of a process will not affect the entire system. It will only affect any future child processes.



          Since the purpose of a shell is to be able to create and control other processes, a umask command is built in to most shells. This isn't required for a shell, it is possible to write a basic shell that has no umask function. But such a shell would not be considered useful as a general-purpose shell for logging in and administering a system.



          You can test what I'm saying yourself, using the fact that a shell like Bash can spawn other Bash shells (or whatever other shell you like):



          1. Open a terminal

          2. Run the umask command to query the current value

          3. Run bash (or whatever) to spawn a child shell

          4. Run umask to check the value of the child's mask

          5. Set the child's mask to something else, eg run umask 0000

          6. Run umask to check the child's mask again

          7. Exit from the child shell (run exit or press Ctrl-d)

          8. You are now back in the parent shell again. Run umask to check its mask

          Useful references:



          • man 1 umask


          • man 2 umask (This gives the reference for the umask() C function)


          • man bash (and search for umask)

          • https://en.wikipedia.org/wiki/Umask





          share|improve this answer














          To understand umask, you first need to understand the structure of processes in Unix/Linux. Which is to say, they form a tree-like structure. Each process needs to have a parent, which is the process that spawned it. (With the exception of the first process, init). Each process may or may not spawn more processes, called children.



          Each process has a mask property. This is what is queried or set using the umask command.



          Processes inherit the mask of their parent. They can then change their own mask. For example, there is a umask() C function that can change the mask of the program you are writing, without needing to call umask from the shell.



          A child process cannot affect the mask of their parent. Therefore, changing the mask of a process will not affect the entire system. It will only affect any future child processes.



          Since the purpose of a shell is to be able to create and control other processes, a umask command is built in to most shells. This isn't required for a shell, it is possible to write a basic shell that has no umask function. But such a shell would not be considered useful as a general-purpose shell for logging in and administering a system.



          You can test what I'm saying yourself, using the fact that a shell like Bash can spawn other Bash shells (or whatever other shell you like):



          1. Open a terminal

          2. Run the umask command to query the current value

          3. Run bash (or whatever) to spawn a child shell

          4. Run umask to check the value of the child's mask

          5. Set the child's mask to something else, eg run umask 0000

          6. Run umask to check the child's mask again

          7. Exit from the child shell (run exit or press Ctrl-d)

          8. You are now back in the parent shell again. Run umask to check its mask

          Useful references:



          • man 1 umask


          • man 2 umask (This gives the reference for the umask() C function)


          • man bash (and search for umask)

          • https://en.wikipedia.org/wiki/Umask






          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 3 hours ago

























          answered 5 hours ago









          cryptarch

          2513




          2513











          • Hello, I'm sad to say that I don't get an answer to my particular question. You might want to consider editing a bit to touch my question directly.
            – JohnDoea
            3 hours ago










          • This might result in upvotes...
            – JohnDoea
            3 hours ago










          • You want me to put a line at the top saying "Changing umask does not affect the whole operating system"?
            – cryptarch
            3 hours ago






          • 1




            Do I understand correct? --- A Bash process with mask X will inherit it to a Dash process but if we change it in the Dash process to Y, the ksh process will have Y But the Bash will still have X.
            – JohnDoea
            56 mins ago






          • 1




            Yes, in this case I am. Thanks and upvoted.
            – JohnDoea
            54 mins ago

















          • Hello, I'm sad to say that I don't get an answer to my particular question. You might want to consider editing a bit to touch my question directly.
            – JohnDoea
            3 hours ago










          • This might result in upvotes...
            – JohnDoea
            3 hours ago










          • You want me to put a line at the top saying "Changing umask does not affect the whole operating system"?
            – cryptarch
            3 hours ago






          • 1




            Do I understand correct? --- A Bash process with mask X will inherit it to a Dash process but if we change it in the Dash process to Y, the ksh process will have Y But the Bash will still have X.
            – JohnDoea
            56 mins ago






          • 1




            Yes, in this case I am. Thanks and upvoted.
            – JohnDoea
            54 mins ago
















          Hello, I'm sad to say that I don't get an answer to my particular question. You might want to consider editing a bit to touch my question directly.
          – JohnDoea
          3 hours ago




          Hello, I'm sad to say that I don't get an answer to my particular question. You might want to consider editing a bit to touch my question directly.
          – JohnDoea
          3 hours ago












          This might result in upvotes...
          – JohnDoea
          3 hours ago




          This might result in upvotes...
          – JohnDoea
          3 hours ago












          You want me to put a line at the top saying "Changing umask does not affect the whole operating system"?
          – cryptarch
          3 hours ago




          You want me to put a line at the top saying "Changing umask does not affect the whole operating system"?
          – cryptarch
          3 hours ago




          1




          1




          Do I understand correct? --- A Bash process with mask X will inherit it to a Dash process but if we change it in the Dash process to Y, the ksh process will have Y But the Bash will still have X.
          – JohnDoea
          56 mins ago




          Do I understand correct? --- A Bash process with mask X will inherit it to a Dash process but if we change it in the Dash process to Y, the ksh process will have Y But the Bash will still have X.
          – JohnDoea
          56 mins ago




          1




          1




          Yes, in this case I am. Thanks and upvoted.
          – JohnDoea
          54 mins ago





          Yes, in this case I am. Thanks and upvoted.
          – JohnDoea
          54 mins ago


















           

          draft saved


          draft discarded















































           


          draft saved


          draft discarded














          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2funix.stackexchange.com%2fquestions%2f481146%2fis-a-file-creation-mask-setted-by-a-given-shells-umask-is-typically-unique-to-t%23new-answer', 'question_page');

          );

          Post as a guest













































































          Popular posts from this blog

          How to check contact read email or not when send email to Individual?

          Bahrain

          Postfix configuration issue with fips on centos 7; mailgun relay