Hi all Im starting to get to grips with some of the bash commands now and was wondering if somebody could let me know if this assumption is correct.....

when using chmod if you specify a 7 (whether for user other users or other groups) it gives read write and execute access

a 6 would give read(4) write(2) access but no execute priveleges

a 5 would give read and execute(1)

a 4 would give read only

a 2 write only

and a 1 execute only

so chmod 751 would give the user full priveleges the other users in the same group read and execute priveleges and other groups execute only priveleges.

Am I correct in my assumptions???

yeah your conclusions are correct ....change the number 7
in binary : 7=111 ....compare it with rwx (Read-Write-eXecute)...
1 at all places means all access....I hope you can now check for
rest.....

many thanks

while on the subject of binary conversion for this command I take it that 7 is the highest decimal number you can use hence only using a 3 bit binary pattern.

Would this therefore also be valid to allow full read write xecute access to all : -

chmod 111111111 or even

chmod 1FF (Dont think this would because its based on the total value of the full string above not three seperate 111)

check it for urself & tell us too
(actually i cant do right now cos in windows)

hmmm I shall do that

it should theoretically be possible I mean Linux is capable of differrentiating between number bases isnt it???

(I cant check now coz Im in work and using Windows if anybody is in Linux and his willing to try it, let us know the outcome please)

Nope dosent work:
[root@localhost amp2000]# chmod 111111111 plugin131.trace
chmod: invalid mode string: `111111111'
[root@localhost amp2000]#

many thanks amp

it may work if you specify that your working in base 2 before hand and also seperate each block of 111. I dont know if you can do that or not does anybody else, Im curious to know???

Don't think about permission being a decimal number - it is octal, and the system understands octal through binaty representation, so a base 8 digit can be represented with three bits binary (2 in the power of three is 8), so this is why you have 7 as 111 (or r=1 w=1 and x=1), and 4 in base 8 you can express as 100 (r=1, no write and executable permissions - READ ONLY), and so forth. So chmod only accepts octal number or letter notation (u+x o-w and so on)

Okay

4 in decimal = 100 in binary

100 in octal is in fact = 63

so in fact the decimal value for 100 (octal) is 63

so your saying the binary encoding used is increasing by 8 for each bit i.e.

64 8 1 <------------- decimal values for 3 bit octal

1 1 1 (r w x)
1 0 0 (r - -)
1 0 1 (r - x)
1 1 0 (r w -)

etc................................................

nope , each digit in the permission pattern is octal (0 through 7) try enter 8 and see chmod complaining about it, now each one of these are represented through 3 bit binary (2 in the power of 3 gives you 8, so it means with three bit binary you can construct an octal digit from 0 to 7 in total of eight) so to say
octal 3-bit binary
0 000
1 001
2 010
3 011
--------------
7 111

So, now each digit in the permission pattern is octal representation of read, write, execution bits for a defined field - owner, group, others.
so if you put it all together
chmod 777 is translated to a binary pattern with each digit represented in 3-bit binary (3-bit - 1 bit for r, 1 bit for w and one bit for x)
chmod 111 111 111
or
chmod rwx rwx rwx

| | |___others
| |______group
|__________owner

is it clear?

ahhhhhh I keep messing up my conversion tables doh!!!!!!!!!!!

I only realised what I was doing when I looked at the hex number F which is equal to 15 in decimal or 1111 binary.

you dont increment 1 8 64 in octal (stupid thing to do)

you increment 1 2 4 8 16 32 64 128 etc.....for all number bases

the only time you do increment 1 8 16 32 etc....is if you convert from binary into octal via decimal i.e.

1111 (binary)

becomes 15 (decimal)

which then becomes 017 (octal)