PERL chmod inconsistent with the LINUX equivalent

[Anonymous Monk]

Hello all. I am trying to run chmod (and check for failures), using PERL. The chmod returns 1, but the permissions reported are different than the ones reported after using LINUX's chmod. I have used perl -e to show the inconcictency. See below:

> perl -e '$m= chmod (777 , "run.debug.2.log")  ; print "$m\n";'
1
> ls -ltr run.debug.2.log
-r----x--x 1 ldagan contract 0 Nov 22 08:59 run.debug.2.log
> chmod 777 run.debug.2.log
> ls -ltr run.debug.2.log
-rwxrwxrwx 1 ldagan contract 0 Nov 22 08:59 run.debug.2.log

As can be seen, chmod 777 in PERL gives permsissions "-r----x--x", which is different than the LINUX equivalent. Any ideas?

[marto]

This is clearly explained in the chmod documentation, I suggest reading and understanding it.

[ColonelPanic]

I disagree.

The problem is that you need a leading zero in order for a number in Perl to be interpreted as octal. That isn't explicitly mentioned in the chmod() documentation, and might not be clear if you don't already know it.

print 0777 . " (octal)\n";
print 777 . " (decimal)\n";

[marto]

IMHO the chmod documenation is fairly comprehensive.

"... The first element of the list must be the numeric mode, which should probably be an octal number, and which definitely should not be a string of octal digits: 0644 is okay, but "0644" is not. Returns the number of files successfully changed. See also oct if all you have is a string."

Note the link to oct, and the fact that the examples clearly show the leading zero. Then follows examples explaining the outcome of several ways of using chmod, including the outcome of using it without the leading zero incorrectly. If you feel this isn't clear enough feel free to contribute to the documentation.

Update: Strike out mistake.

[moritz]

$ perl -wE 'printf "%o\n", 777'
1411