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?









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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").










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    up vote
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    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
      1
      down vote

      favorite









      up vote
      1
      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






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      edited Nov 11 at 22:36

























      asked Nov 11 at 21:08









      JohnDoea

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          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


















          • 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
            Nov 12 at 2:33










          • If you use Dash to launch ksh, yeah, that's what will happen
            – cryptarch
            Nov 12 at 2:35






          • 1




            Yes, in this case I am. Thanks and upvoted.
            – JohnDoea
            Nov 12 at 2:35











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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          5
          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


















          • 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
            Nov 12 at 2:33










          • If you use Dash to launch ksh, yeah, that's what will happen
            – cryptarch
            Nov 12 at 2:35






          • 1




            Yes, in this case I am. Thanks and upvoted.
            – JohnDoea
            Nov 12 at 2:35















          up vote
          5
          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


















          • 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
            Nov 12 at 2:33










          • If you use Dash to launch ksh, yeah, that's what will happen
            – cryptarch
            Nov 12 at 2:35






          • 1




            Yes, in this case I am. Thanks and upvoted.
            – JohnDoea
            Nov 12 at 2:35













          up vote
          5
          down vote



          accepted







          up vote
          5
          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 Nov 12 at 0:14

























          answered Nov 11 at 21:35









          cryptarch

          3856




          3856







          • 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
            Nov 12 at 2:33










          • If you use Dash to launch ksh, yeah, that's what will happen
            – cryptarch
            Nov 12 at 2:35






          • 1




            Yes, in this case I am. Thanks and upvoted.
            – JohnDoea
            Nov 12 at 2:35













          • 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
            Nov 12 at 2:33










          • If you use Dash to launch ksh, yeah, that's what will happen
            – cryptarch
            Nov 12 at 2:35






          • 1




            Yes, in this case I am. Thanks and upvoted.
            – JohnDoea
            Nov 12 at 2:35








          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
          Nov 12 at 2:33




          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
          Nov 12 at 2:33












          If you use Dash to launch ksh, yeah, that's what will happen
          – cryptarch
          Nov 12 at 2:35




          If you use Dash to launch ksh, yeah, that's what will happen
          – cryptarch
          Nov 12 at 2:35




          1




          1




          Yes, in this case I am. Thanks and upvoted.
          – JohnDoea
          Nov 12 at 2:35





          Yes, in this case I am. Thanks and upvoted.
          – JohnDoea
          Nov 12 at 2:35


















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