Mdthbs

Deal with toxic manager when you can't quit

How to notate time signature switching consistently every measure

If a sorcerer casts the Banishment spell on a PC while in Avernus, does the PC return to their home plane?

A word that means fill it to the required quantity

How to obtain a position of last non-zero element

What to do when moving next to a bird sanctuary with a loosely-domesticated cat?

Why doesn't UInt have a toDouble()?

Flight paths in orbit around Ceres?

Is it possible for absolutely everyone to attain enlightenment?

Did the UK government pay "millions and millions of dollars" to try to snag Julian Assange?

For what reasons would an animal species NOT cross a *horizontal* land bridge?

Cooking pasta in a water boiler

How do PCB vias affect signal quality?

What could be the right powersource for 15 seconds lifespan disposable giant chainsaw?

If I score a critical hit on an 18 or higher, what are my chances of getting a critical hit if I roll 3d20?

Why can't devices on different VLANs, but on the same subnet, communicate?

Why don't hard Brexiteers insist on a hard border to prevent illegal immigration after Brexit?

How can I add encounters in the Lost Mine of Phandelver campaign without giving PCs too much XP?

"as much details as you can remember"

Button changing its text & action. Good or terrible?

Mathematics of imaging the black hole

Slides for 30 min~1 hr Skype tenure track application interview

What is the most efficient way to store a numeric range?

How to charge AirPods to keep battery healthy?

Why doesn't mkfifo with a mode of 1755 grant read permissions and sticky bit to the user?

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}

6

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:

• 
  p stands for pipe,


• 
  - means the user has no read. I don't know how when I gave it with the 7,


• 
  s group 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

share|improve this question

edited 6 hours ago

mosvy

9,67711036

asked 13 hours ago

Joao ParenteJoao Parente

311

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 | 

6

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:

• 
  p stands for pipe,


• 
  - means the user has no read. I don't know how when I gave it with the 7,


• 
  s group 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

share|improve this question

edited 6 hours ago

mosvy

9,67711036

asked 13 hours ago

Joao ParenteJoao Parente

311

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:

• 
  p stands for pipe,


• 
  - means the user has no read. I don't know how when I gave it with the 7,


• 
  s group 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

share|improve this question

edited 6 hours ago

mosvy

9,67711036

asked 13 hours ago

Joao ParenteJoao Parente

311

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.

share|improve this question

edited 6 hours ago

mosvy

9,67711036

asked 13 hours ago

Joao ParenteJoao Parente

311

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.

share|improve this question

edited 6 hours ago

mosvy

9,67711036

asked 13 hours ago

Joao ParenteJoao Parente

311

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 6 hours ago

mosvy

9,67711036

edited 6 hours ago

mosvy

9,67711036

asked 13 hours ago

Joao ParenteJoao Parente

311

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 13 hours ago

Joao ParenteJoao Parente

311

1 Answer
1

active

oldest

votes

11

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--t so:

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)

share|improve this answer

edited 13 hours ago

answered 13 hours ago

mosvymosvy

9,67711036

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  13 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  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
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thanks very much :P
  
  – Joao Parente
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  6 hours ago
  

  
  
  
  

add a comment | 

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',
autoActivateHeartbeat: false,
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
});


}
});

Joao Parente is a new contributor. Be nice, and check out our Code of Conduct.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2funix.stackexchange.com%2fquestions%2f511874%2fwhy-doesnt-mkfifo-with-a-mode-of-1755-grant-read-permissions-and-sticky-bit-to%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

11

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--t so:

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)

share|improve this answer

edited 13 hours ago

answered 13 hours ago

mosvymosvy

9,67711036

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  13 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  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
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thanks very much :P
  
  – Joao Parente
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  6 hours ago
  

  
  
  
  

add a comment | 

11

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--t so:

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)

share|improve this answer

edited 13 hours ago

answered 13 hours ago

mosvymosvy

9,67711036

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  13 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  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
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thanks very much :P
  
  – Joao Parente
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  6 hours ago
  

  
  
  
  

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--t so:

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)

share|improve this answer

edited 13 hours ago

answered 13 hours ago

mosvymosvy

9,67711036

share|improve this answer

edited 13 hours ago

answered 13 hours ago

mosvymosvy

9,67711036

answered 13 hours ago

mosvymosvy

9,67711036

answered 13 hours ago

mosvymosvy

9,67711036