In '127.0.1.1:+xxxxx' What is the plus character for?
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On my linux filesystem, a symlinks points to 127.0.1.1:+xxxxx.
Why the plus sign? Could there also be a minus? Why not just 127.0.1.1:xxxxx?
linux networking symbolic-link ip-address
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up vote
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On my linux filesystem, a symlinks points to 127.0.1.1:+xxxxx.
Why the plus sign? Could there also be a minus? Why not just 127.0.1.1:xxxxx?
linux networking symbolic-link ip-address
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add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
On my linux filesystem, a symlinks points to 127.0.1.1:+xxxxx.
Why the plus sign? Could there also be a minus? Why not just 127.0.1.1:xxxxx?
linux networking symbolic-link ip-address
New contributor
On my linux filesystem, a symlinks points to 127.0.1.1:+xxxxx.
Why the plus sign? Could there also be a minus? Why not just 127.0.1.1:xxxxx?
linux networking symbolic-link ip-address
linux networking symbolic-link ip-address
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New contributor
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asked 36 mins ago
myMethod
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2 Answers
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Symbolic links which don't point to a file have no generic meaning at all. In this case it might be the process ID, or a port with some special protocol spoken over it, or another identifier. It all depends on what program made it.
Software which creates these links simply takes advantage of the facts that 1) a symlink's target may be non-existent or even total nonsense; 2) creating a symlink is a single-syscall completely atomic operation (as is reading its target), unlike creating a regular file which takes at least 3 separate system calls.
Thus symlink creation can be abused as a way of locking (ensuring single instance of a program) even when other mechanisms may be unreliable. The program doesn't need the symlink to actually resolve to a real file: it only cares about whether creating the link succeeds, or whether it fails due to it already existing.
Thank you for the detailed answer.
â myMethod
13 mins ago
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As far as I know the "+" means that the number after the IP (the "xxxxx") refers to a "process ID" (not a port which usually uses the notation [IP-address]:[portnumber]).
I just checked it. In my case it really does.
â myMethod
10 mins ago
@myMethod glad to help, feel free to "accept" my answer, if it answered you're question.
â Albin
9 mins ago
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Symbolic links which don't point to a file have no generic meaning at all. In this case it might be the process ID, or a port with some special protocol spoken over it, or another identifier. It all depends on what program made it.
Software which creates these links simply takes advantage of the facts that 1) a symlink's target may be non-existent or even total nonsense; 2) creating a symlink is a single-syscall completely atomic operation (as is reading its target), unlike creating a regular file which takes at least 3 separate system calls.
Thus symlink creation can be abused as a way of locking (ensuring single instance of a program) even when other mechanisms may be unreliable. The program doesn't need the symlink to actually resolve to a real file: it only cares about whether creating the link succeeds, or whether it fails due to it already existing.
Thank you for the detailed answer.
â myMethod
13 mins ago
add a comment |Â
up vote
0
down vote
accepted
Symbolic links which don't point to a file have no generic meaning at all. In this case it might be the process ID, or a port with some special protocol spoken over it, or another identifier. It all depends on what program made it.
Software which creates these links simply takes advantage of the facts that 1) a symlink's target may be non-existent or even total nonsense; 2) creating a symlink is a single-syscall completely atomic operation (as is reading its target), unlike creating a regular file which takes at least 3 separate system calls.
Thus symlink creation can be abused as a way of locking (ensuring single instance of a program) even when other mechanisms may be unreliable. The program doesn't need the symlink to actually resolve to a real file: it only cares about whether creating the link succeeds, or whether it fails due to it already existing.
Thank you for the detailed answer.
â myMethod
13 mins ago
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Symbolic links which don't point to a file have no generic meaning at all. In this case it might be the process ID, or a port with some special protocol spoken over it, or another identifier. It all depends on what program made it.
Software which creates these links simply takes advantage of the facts that 1) a symlink's target may be non-existent or even total nonsense; 2) creating a symlink is a single-syscall completely atomic operation (as is reading its target), unlike creating a regular file which takes at least 3 separate system calls.
Thus symlink creation can be abused as a way of locking (ensuring single instance of a program) even when other mechanisms may be unreliable. The program doesn't need the symlink to actually resolve to a real file: it only cares about whether creating the link succeeds, or whether it fails due to it already existing.
Symbolic links which don't point to a file have no generic meaning at all. In this case it might be the process ID, or a port with some special protocol spoken over it, or another identifier. It all depends on what program made it.
Software which creates these links simply takes advantage of the facts that 1) a symlink's target may be non-existent or even total nonsense; 2) creating a symlink is a single-syscall completely atomic operation (as is reading its target), unlike creating a regular file which takes at least 3 separate system calls.
Thus symlink creation can be abused as a way of locking (ensuring single instance of a program) even when other mechanisms may be unreliable. The program doesn't need the symlink to actually resolve to a real file: it only cares about whether creating the link succeeds, or whether it fails due to it already existing.
answered 18 mins ago
grawity
220k32449513
220k32449513
Thank you for the detailed answer.
â myMethod
13 mins ago
add a comment |Â
Thank you for the detailed answer.
â myMethod
13 mins ago
Thank you for the detailed answer.
â myMethod
13 mins ago
Thank you for the detailed answer.
â myMethod
13 mins ago
add a comment |Â
up vote
4
down vote
As far as I know the "+" means that the number after the IP (the "xxxxx") refers to a "process ID" (not a port which usually uses the notation [IP-address]:[portnumber]).
I just checked it. In my case it really does.
â myMethod
10 mins ago
@myMethod glad to help, feel free to "accept" my answer, if it answered you're question.
â Albin
9 mins ago
add a comment |Â
up vote
4
down vote
As far as I know the "+" means that the number after the IP (the "xxxxx") refers to a "process ID" (not a port which usually uses the notation [IP-address]:[portnumber]).
I just checked it. In my case it really does.
â myMethod
10 mins ago
@myMethod glad to help, feel free to "accept" my answer, if it answered you're question.
â Albin
9 mins ago
add a comment |Â
up vote
4
down vote
up vote
4
down vote
As far as I know the "+" means that the number after the IP (the "xxxxx") refers to a "process ID" (not a port which usually uses the notation [IP-address]:[portnumber]).
As far as I know the "+" means that the number after the IP (the "xxxxx") refers to a "process ID" (not a port which usually uses the notation [IP-address]:[portnumber]).
answered 25 mins ago
Albin
1,204621
1,204621
I just checked it. In my case it really does.
â myMethod
10 mins ago
@myMethod glad to help, feel free to "accept" my answer, if it answered you're question.
â Albin
9 mins ago
add a comment |Â
I just checked it. In my case it really does.
â myMethod
10 mins ago
@myMethod glad to help, feel free to "accept" my answer, if it answered you're question.
â Albin
9 mins ago
I just checked it. In my case it really does.
â myMethod
10 mins ago
I just checked it. In my case it really does.
â myMethod
10 mins ago
@myMethod glad to help, feel free to "accept" my answer, if it answered you're question.
â Albin
9 mins ago
@myMethod glad to help, feel free to "accept" my answer, if it answered you're question.
â Albin
9 mins ago
add a comment |Â
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myMethod is a new contributor. Be nice, and check out our Code of Conduct.
myMethod is a new contributor. Be nice, and check out our Code of Conduct.
myMethod is a new contributor. Be nice, and check out our Code of Conduct.
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