I just did a chmod -x chmod

Question

So I did a chmod -x chmod. How I can fix this problem? How do I give execute rights back to chmod?

See also the [question on Unix & Linux SE](http://unix.stackexchange.com/q/83862/21962). — unor, Jul 27 '13 at 11:51

score 45 · Accepted Answer · answered Oct 11 '10 at 07:38

45

In Linux:

/lib/ld-linux.so.2 /bin/chmod +x /bin/chmod

http://www.slideshare.net/cog/chmod-x-chmod

answered Oct 11 '10 at 07:38

bindbn

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bindbn, I was about to post the same link :-) but I was reading myself as I am new to unix / linux stuff !!! – Mutahir Oct 11 '10 at 07:43
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On a 64-bit distro, you may have to use `/lib64/ld-linux-x86-64.so.2` instead. `ldd /bin/chmod` should list exactly which linker to run. – goldPseudo Oct 11 '10 at 07:52
I would write a minimal program that uses chmod(2), but this is cooler – adamo Oct 11 '10 at 07:57
Could someone explain why that works, for the uninitiated? I think I get it, but some documentation would be great. – Stefan Lasiewski Oct 11 '10 at 22:12
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@Stefan: From `man ld-linux.so`: "ld.so [and ld-linux.so] loads the shared libraries needed by a program, prepares the program to run, and then runs it." (`ld.so` is for `a.out` format executables and `ld-linux.so` is for ELF format.) – Dennis Williamson Oct 12 '10 at 00:58
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@Stefan: The kernel doesn't really know how to load and execute dynamically-linked executables, because it is highly complex and follows the glibc, not the kernel. `ld-linux.so` is sort-of executable/library hybrid, sort-of statically linked, runs in userspace, and is responsible of loading dynamically-linked executables and all their dependencies, then running them. – Juliano Nov 05 '10 at 17:09
/lib/ld-linux.so.2: symbolic link to `/lib32/ld-linux.so.2' What if I'm on 64bit machine? – confiq Jul 24 '11 at 15:14

score 18 · Answer 2 · answered Oct 11 '10 at 12:44

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Use python:

# python
Python> import os
Python> os.chmod("/bin/chmod",0755)

answered Oct 11 '10 at 12:44

Douglas Leeder

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And, as a one-liner: `python -c "import os; os.chmod('/bin/chmod', 0755)"` – Thanatos Nov 05 '10 at 16:35
@Thanatos `python -c '__import__("os").chmod("/bin/chmod", 0755)'` does the same without using any semicolons. – Sapphire_Brick Oct 23 '20 at 03:14

score 18 · Answer 3 · answered Oct 11 '10 at 14:55

18

This relies on the fact that permissions of a destination file are preserved rather than the source file when it is being copied over. We're "borrowing" the permissions of ls:

cp /bin/ls /tmp/chmod.tmp
cp /bin/chmod /tmp/chmod.tmp
mv /tmp/chmod.tmp /bin/chmod

answered Oct 11 '10 at 14:55

Dennis Williamson

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I think your first `cp` needs to have the -p flag on – Riking Mar 09 '13 at 21:33
@Riking: It's not necessary for this purpose. – Dennis Williamson Mar 09 '13 at 23:16
This was similar to the solution I came up with, but after `cp`ing the executable, I just `cat`'d the contents of `chmod` into the temp file. – SpellingD Jul 21 '13 at 16:42

score 10 · Answer 4 · answered Oct 11 '10 at 21:54

10

Using Perl:

% perl -e 'chmod 0755, qw[/bin/chmod]'

answered Oct 11 '10 at 21:54

Alnitak

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score 7 · Answer 5 · answered Nov 05 '10 at 17:46

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setfacl -m u::rx /bin/chmod

... will grant the owner execute permissions.

But, the /lib/ld-linux.so.2 trick is neat. :)

answered Nov 05 '10 at 17:46

Kyle Brantley

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score 5 · Answer 6 · answered Oct 11 '10 at 08:08

5

This is weird... I saw something like this a few days ago via someone's tweet...

http://www.slideshare.net/cog/chmod-x-chmod

answered Oct 11 '10 at 08:08

ThatGraemeGuy

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Links to other sites are not helpful as the link will become broken at some point. This is like telling someone to google the answer. – Phil Hannent Oct 11 '10 at 13:19
@Phil Hannent: I've seen that slideshow too, just three days ago, so it was my first thought, if the OP was one of the applicants there. – Boldewyn Oct 11 '10 at 14:04
Same, just saw this on Reddit a few days back... – Dentrasi Oct 11 '10 at 22:31

score 5 · Answer 7 · answered Oct 11 '10 at 08:41

5

Should you be on a system where /bin/chmod can't be loaded by the dynamic linker:

# /bin/mv /bin/chmod /bin/chmod.tmp
# install -p -m 755 /bin/chmod.tmp /bin/chmod

This works on my MacOS X system.

answered Oct 11 '10 at 08:41

Alnitak

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score 2 · Answer 8 · answered Nov 05 '10 at 16:15

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/rescue/chmod 555 /bin/chmod

I think you could also use mtree.

answered Nov 05 '10 at 16:15

Darius

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This would my favorite solution. It doesn't work with linux unfortunately. – Aug 07 '12 at 18:01

score 2 · Answer 9 · answered Nov 17 '10 at 23:58

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I suspect this is not a real question: http://www.slideshare.net/cog/chmod-x-chmod

Reinstall chown: sudo apt-get install --reinstall coreutils
perl -e 'chmod 0755, "chmod"'
more examples in the slides

answered Nov 17 '10 at 23:58

Julien

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score 1 · Answer 10 · answered Oct 11 '10 at 14:42

1

create a new chmod and use that for the original

umask 000
cat chmod > ~/my-chmod
~/my-chmod a+x chmod

answered Oct 11 '10 at 14:42

Kevin M

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Setting a umask of 000 won't give a file execute permission when it's created. It will, at best, get rw-rw-rw- permissions. – Barry Brown Oct 11 '10 at 15:25
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@Barry: It actually depends on the mode passed to creat(2)/open(2)/mkdir(2)/etc. If umask == 0 and the syscall that creates the file is given 0777 for the mode, then the file will have the execute bits turned on. For example, linkers/compilers pass 0777 when writing out an “executable file” (e.g. `(umask 000;gcc -o foo foo.c)` will produce a `foo` with mode 777). However, many (most?) shells pass 0666 when they open/create files for redirection, which means that this answer is not going to work under many shells. – Chris Johnsen Oct 12 '10 at 01:52

I just did a chmod -x chmod

10 Answers10

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