Why doesn't mkfifo with a mode of 1755 grant read permissions and sticky bit to the user?
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I'm creating a server and client situation where i want to create a pipe so they can communicate.
I created the pipe in the server code with
mkfifo("fifo",1755);:
- 1 for only user that created and root to be able to delete it or rename it,
- 7 for give read, write and exec to user, and
- 5 for both group and other to only give them read and exec.
The problem is that later in the server code I open the fifo to read from it open("fifo",O_RDONLY); but when i execute it, it shows me an perror that denies me acess to the fifo.
I went to see the permissions of the pipe fifo and it says
p-wx--s--t so:
pstands for pipe,
-means the user has no read. I don't know how when I gave it with the 7,
sgroup executes has user. I don't how if i gave 1 so supposedly it should give to user and others the ability to only read and execute and others have t that was expected.
Do I have a misunderstanding of the permissions?
permissions c mkfifo
edited Apr 11 at 17:21
mosvy
10.1k11237
asked Apr 11 at 10:01
Joao ParenteJoao Parente
333
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Joao Parente is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
I'm creating a server and client situation where i want to create a pipe so they can communicate.
I created the pipe in the server code with
mkfifo("fifo",1755);:
- 1 for only user that created and root to be able to delete it or rename it,
- 7 for give read, write and exec to user, and
- 5 for both group and other to only give them read and exec.
The problem is that later in the server code I open the fifo to read from it open("fifo",O_RDONLY); but when i execute it, it shows me an perror that denies me acess to the fifo.
I went to see the permissions of the pipe fifo and it says
p-wx--s--t so:
pstands for pipe,
-means the user has no read. I don't know how when I gave it with the 7,
sgroup executes has user. I don't how if i gave 1 so supposedly it should give to user and others the ability to only read and execute and others have t that was expected.
Do I have a misunderstanding of the permissions?
permissions c mkfifo
edited Apr 11 at 17:21
mosvy
10.1k11237
asked Apr 11 at 10:01
Joao ParenteJoao Parente
333
New contributor
Joao Parente is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
I'm creating a server and client situation where i want to create a pipe so they can communicate.
I created the pipe in the server code with
mkfifo("fifo",1755);:
- 1 for only user that created and root to be able to delete it or rename it,
- 7 for give read, write and exec to user, and
- 5 for both group and other to only give them read and exec.
The problem is that later in the server code I open the fifo to read from it open("fifo",O_RDONLY); but when i execute it, it shows me an perror that denies me acess to the fifo.
I went to see the permissions of the pipe fifo and it says
p-wx--s--t so:
pstands for pipe,
-means the user has no read. I don't know how when I gave it with the 7,
sgroup executes has user. I don't how if i gave 1 so supposedly it should give to user and others the ability to only read and execute and others have t that was expected.
Do I have a misunderstanding of the permissions?
permissions c mkfifo
edited Apr 11 at 17:21
mosvy
10.1k11237
asked Apr 11 at 10:01
Joao ParenteJoao Parente
333
New contributor
Joao Parente is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I'm creating a server and client situation where i want to create a pipe so they can communicate.
I created the pipe in the server code with
mkfifo("fifo",1755);:
- 1 for only user that created and root to be able to delete it or rename it,
- 7 for give read, write and exec to user, and
- 5 for both group and other to only give them read and exec.
The problem is that later in the server code I open the fifo to read from it open("fifo",O_RDONLY); but when i execute it, it shows me an perror that denies me acess to the fifo.
I went to see the permissions of the pipe fifo and it says
p-wx--s--t so:
pstands for pipe,
-means the user has no read. I don't know how when I gave it with the 7,
sgroup executes has user. I don't how if i gave 1 so supposedly it should give to user and others the ability to only read and execute and others have t that was expected.
Do I have a misunderstanding of the permissions?
permissions c mkfifo
permissions c mkfifo
edited Apr 11 at 17:21
mosvy
10.1k11237
asked Apr 11 at 10:01
Joao ParenteJoao Parente
333
New contributor
Joao Parente is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Apr 11 at 17:21
mosvy
10.1k11237
asked Apr 11 at 10:01
Joao ParenteJoao Parente
333
New contributor
Joao Parente is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Apr 11 at 17:21
mosvy
10.1k11237
edited Apr 11 at 17:21
mosvy
10.1k11237
edited Apr 11 at 17:21
mosvy
10.1k11237
10.1k11237
asked Apr 11 at 10:01
Joao ParenteJoao Parente
333
New contributor
Joao Parente is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 11 at 10:01
Joao ParenteJoao Parente
333
asked Apr 11 at 10:01
Joao ParenteJoao Parente
333
333
New contributor
Joao Parente is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Joao Parente is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Joao Parente is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
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You cannot simply exec a binary from a pipe: Is there a way to execute a native binary from a pipe?. Also I don't think the sticky bit on executables is worth anything on modern systems.
I created the pipe in the server code with
mkfifo("fifo",1755);
I went to see the permissions of the pipe fifo and it says
p-wx--s--tso:
Your error is to have written the 1755 permission without the leading 0, which means that 1755 has been treated as a decimal instead of octal (1755 & ~022 = 03311 = p-wx--s--t; where 022 is your umask)
answered Apr 11 at 10:18
mosvymosvy
10.1k11237
So a better solution would be create a dir and chmod 1755 the dir then create the pipe in that dir with ``` mkfifo ("fifo",755); ``` ?
– Joao Parente
Apr 11 at 10:19
3
The sticky bit has different semantics for files and directories: for directories it prevents users from removing or renaming files they don't own; for regular files it doesn't mean much nowadays. Read the chmod(2) manpage for all the details. Also,mkfifo("fifo", 0755)with the leading0;-).
– mosvy
Apr 11 at 10:24
thanks very much :P
– Joao Parente
Apr 11 at 10:26
i tried mkfifo("fifo",0777); but i only gotprwxr-xr-xbut if i do chmod 0777 fifo after i created fifo i getprwxrwxrwxany idea why it doesnt work when i create it
– Joao Parente
Apr 11 at 17:09
Because of the umask -- the umask which is 022 will mask out the write permissions from group (020) and other (002) anytime a file is created with open(2), mkfifo(2), etc. Read the umask(2) manpage. The umask doesn't affect the chmod syscall or command.
– mosvy
Apr 11 at 17:15
add a comment |
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1 Answer
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1 Answer
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You cannot simply exec a binary from a pipe: Is there a way to execute a native binary from a pipe?. Also I don't think the sticky bit on executables is worth anything on modern systems.
I created the pipe in the server code with
mkfifo("fifo",1755);
I went to see the permissions of the pipe fifo and it says
p-wx--s--tso:
Your error is to have written the 1755 permission without the leading 0, which means that 1755 has been treated as a decimal instead of octal (1755 & ~022 = 03311 = p-wx--s--t; where 022 is your umask)
answered Apr 11 at 10:18
mosvymosvy
10.1k11237
So a better solution would be create a dir and chmod 1755 the dir then create the pipe in that dir with ``` mkfifo ("fifo",755); ``` ?
– Joao Parente
Apr 11 at 10:19
3
The sticky bit has different semantics for files and directories: for directories it prevents users from removing or renaming files they don't own; for regular files it doesn't mean much nowadays. Read the chmod(2) manpage for all the details. Also,mkfifo("fifo", 0755)with the leading0;-).
– mosvy
Apr 11 at 10:24
thanks very much :P
– Joao Parente
Apr 11 at 10:26
i tried mkfifo("fifo",0777); but i only gotprwxr-xr-xbut if i do chmod 0777 fifo after i created fifo i getprwxrwxrwxany idea why it doesnt work when i create it
– Joao Parente
Apr 11 at 17:09
Because of the umask -- the umask which is 022 will mask out the write permissions from group (020) and other (002) anytime a file is created with open(2), mkfifo(2), etc. Read the umask(2) manpage. The umask doesn't affect the chmod syscall or command.
– mosvy
Apr 11 at 17:15
add a comment |
You cannot simply exec a binary from a pipe: Is there a way to execute a native binary from a pipe?. Also I don't think the sticky bit on executables is worth anything on modern systems.
I created the pipe in the server code with
mkfifo("fifo",1755);
I went to see the permissions of the pipe fifo and it says
p-wx--s--tso:
Your error is to have written the 1755 permission without the leading 0, which means that 1755 has been treated as a decimal instead of octal (1755 & ~022 = 03311 = p-wx--s--t; where 022 is your umask)
answered Apr 11 at 10:18
mosvymosvy
10.1k11237
So a better solution would be create a dir and chmod 1755 the dir then create the pipe in that dir with ``` mkfifo ("fifo",755); ``` ?
– Joao Parente
Apr 11 at 10:19
3
The sticky bit has different semantics for files and directories: for directories it prevents users from removing or renaming files they don't own; for regular files it doesn't mean much nowadays. Read the chmod(2) manpage for all the details. Also,mkfifo("fifo", 0755)with the leading0;-).
– mosvy
Apr 11 at 10:24
thanks very much :P
– Joao Parente
Apr 11 at 10:26
i tried mkfifo("fifo",0777); but i only gotprwxr-xr-xbut if i do chmod 0777 fifo after i created fifo i getprwxrwxrwxany idea why it doesnt work when i create it
– Joao Parente
Apr 11 at 17:09
Because of the umask -- the umask which is 022 will mask out the write permissions from group (020) and other (002) anytime a file is created with open(2), mkfifo(2), etc. Read the umask(2) manpage. The umask doesn't affect the chmod syscall or command.
– mosvy
Apr 11 at 17:15
add a comment |
You cannot simply exec a binary from a pipe: Is there a way to execute a native binary from a pipe?. Also I don't think the sticky bit on executables is worth anything on modern systems.
I created the pipe in the server code with
mkfifo("fifo",1755);
I went to see the permissions of the pipe fifo and it says
p-wx--s--tso:
Your error is to have written the 1755 permission without the leading 0, which means that 1755 has been treated as a decimal instead of octal (1755 & ~022 = 03311 = p-wx--s--t; where 022 is your umask)
answered Apr 11 at 10:18
mosvymosvy
10.1k11237
You cannot simply exec a binary from a pipe: Is there a way to execute a native binary from a pipe?. Also I don't think the sticky bit on executables is worth anything on modern systems.
I created the pipe in the server code with
mkfifo("fifo",1755);
I went to see the permissions of the pipe fifo and it says
p-wx--s--tso:
Your error is to have written the 1755 permission without the leading 0, which means that 1755 has been treated as a decimal instead of octal (1755 & ~022 = 03311 = p-wx--s--t; where 022 is your umask)
answered Apr 11 at 10:18
mosvymosvy
10.1k11237
edited Apr 11 at 10:38
answered Apr 11 at 10:18
mosvymosvy
10.1k11237
answered Apr 11 at 10:18
mosvymosvy
10.1k11237
answered Apr 11 at 10:18
mosvymosvy
10.1k11237
10.1k11237
So a better solution would be create a dir and chmod 1755 the dir then create the pipe in that dir with ``` mkfifo ("fifo",755); ``` ?
– Joao Parente
Apr 11 at 10:19
3
The sticky bit has different semantics for files and directories: for directories it prevents users from removing or renaming files they don't own; for regular files it doesn't mean much nowadays. Read the chmod(2) manpage for all the details. Also,mkfifo("fifo", 0755)with the leading0;-).
– mosvy
Apr 11 at 10:24
thanks very much :P
– Joao Parente
Apr 11 at 10:26
i tried mkfifo("fifo",0777); but i only gotprwxr-xr-xbut if i do chmod 0777 fifo after i created fifo i getprwxrwxrwxany idea why it doesnt work when i create it
– Joao Parente
Apr 11 at 17:09
Because of the umask -- the umask which is 022 will mask out the write permissions from group (020) and other (002) anytime a file is created with open(2), mkfifo(2), etc. Read the umask(2) manpage. The umask doesn't affect the chmod syscall or command.
– mosvy
Apr 11 at 17:15
add a comment |
So a better solution would be create a dir and chmod 1755 the dir then create the pipe in that dir with ``` mkfifo ("fifo",755); ``` ?
– Joao Parente
Apr 11 at 10:19
3
The sticky bit has different semantics for files and directories: for directories it prevents users from removing or renaming files they don't own; for regular files it doesn't mean much nowadays. Read the chmod(2) manpage for all the details. Also,mkfifo("fifo", 0755)with the leading0;-).
– mosvy
Apr 11 at 10:24
thanks very much :P
– Joao Parente
Apr 11 at 10:26
i tried mkfifo("fifo",0777); but i only gotprwxr-xr-xbut if i do chmod 0777 fifo after i created fifo i getprwxrwxrwxany idea why it doesnt work when i create it
– Joao Parente
Apr 11 at 17:09
Because of the umask -- the umask which is 022 will mask out the write permissions from group (020) and other (002) anytime a file is created with open(2), mkfifo(2), etc. Read the umask(2) manpage. The umask doesn't affect the chmod syscall or command.
– mosvy
Apr 11 at 17:15
So a better solution would be create a dir and chmod 1755 the dir then create the pipe in that dir with ``` mkfifo ("fifo",755); ``` ?
– Joao Parente
Apr 11 at 10:19
So a better solution would be create a dir and chmod 1755 the dir then create the pipe in that dir with ``` mkfifo ("fifo",755); ``` ?
– Joao Parente
Apr 11 at 10:19
3
3
The sticky bit has different semantics for files and directories: for directories it prevents users from removing or renaming files they don't own; for regular files it doesn't mean much nowadays. Read the chmod(2) manpage for all the details. Also,
mkfifo("fifo", 0755) with the leading 0 ;-).– mosvy
Apr 11 at 10:24
The sticky bit has different semantics for files and directories: for directories it prevents users from removing or renaming files they don't own; for regular files it doesn't mean much nowadays. Read the chmod(2) manpage for all the details. Also,
mkfifo("fifo", 0755) with the leading 0 ;-).– mosvy
Apr 11 at 10:24
thanks very much :P
– Joao Parente
Apr 11 at 10:26
thanks very much :P
– Joao Parente
Apr 11 at 10:26
i tried mkfifo("fifo",0777); but i only got
prwxr-xr-xbut if i do chmod 0777 fifo after i created fifo i get prwxrwxrwxany idea why it doesnt work when i create it– Joao Parente
Apr 11 at 17:09
i tried mkfifo("fifo",0777); but i only got
prwxr-xr-xbut if i do chmod 0777 fifo after i created fifo i get prwxrwxrwxany idea why it doesnt work when i create it– Joao Parente
Apr 11 at 17:09
Because of the umask -- the umask which is 022 will mask out the write permissions from group (020) and other (002) anytime a file is created with open(2), mkfifo(2), etc. Read the umask(2) manpage. The umask doesn't affect the chmod syscall or command.
– mosvy
Apr 11 at 17:15
Because of the umask -- the umask which is 022 will mask out the write permissions from group (020) and other (002) anytime a file is created with open(2), mkfifo(2), etc. Read the umask(2) manpage. The umask doesn't affect the chmod syscall or command.
– mosvy
Apr 11 at 17:15
add a comment |
Joao Parente is a new contributor. Be nice, and check out our Code of Conduct.
Joao Parente is a new contributor. Be nice, and check out our Code of Conduct.
Joao Parente is a new contributor. Be nice, and check out our Code of Conduct.
Joao Parente is a new contributor. Be nice, and check out our Code of Conduct.
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