$ echo "hello" | od
0000000 062550 066154 005157

I know that the first column represents the byte offset. But I don't see how the other numbers are formed. According to man the above should be "octal bytes". However the option -b is supposed to "select octal bytes" as well and it prints something different:

$echo "hello" | od -b
0000000 150 145 154 154 157 012

EDIT: This is by the way what I would expect to appear i.e. the ascii values of all characters in 'hello\n' as what I would expect to be called "octal bytes".

1 Answer 1


od doesn't show bytes by default, it shows words in octal. This may not quite be intuitive, but don't forget od is a very old command :-) I'll use a somewhat simpler example than you did:

$ echo -en '\01\02' | od
0000000 001001

As Intel uses a little-endian architecture, the bytes \01\02 are interpreted as 00000010 00000001 in binary.

As octal digits each represent 3 bits, we can group that number like this:


So the octal representation of those 2 bytes is:


For day to day use this is pretty useless; perhaps back in the day it was handy for manually debugging memory dumps :-)

Your hello\n example is:

h = 01101000
e = 01100101
l = 01101100
l = 01101100
o = 01101111
\n= 00001010

It's a bit more complicated now, because octal digits represent 3 bits, but bytes are 8 bits; so padding is added :-( The result symbollicaly is:


Remember, each set of 2 bytes is swapped due to the endianness. The P is a padding of 2 bits. The result in octal is (using a slash as separator):


Now in octal groups of 3 bits:

000 110 010 101 101 000 000 110 110 001 101 100 000 000 101 001 101 111

Translated into octal digits:


This matches your result.

In conclusion you've probably learnt that od without options is worse than useless :-)

  • Can you explain to me why this swap of bytes happens in little endianness? Also I don't understand why the padding is added only every two bytes. It would be more intuitive for me if there was one bit of padding for each byte. Nov 18, 2014 at 14:53
  • The padding is added for every word, where a word consists of two bytes. Little endian means that the least significant byte comes first, so you calculate the real value of a word you have to multiply the second byte by 256 and then add the first byte. Multiplying by 256 is the same as shifting the bits to the left 8 positions, so the result is the same as swapping the bytes. I'm sure the wikipedia link can explain it better than I can.
    – wurtel
    Nov 19, 2014 at 8:11
  • Okay I think I understood it a lot better now. Another question: Why does \01\02 equal to two bytes only and 0101 equal to four bytes? (the latter one makes sense to me) Nov 19, 2014 at 16:45
  • The \01 is octal notation for one byte, it can range from \0 to \0377. I don't know why you think 0101 is equal to four bytes though.
    – wurtel
    Nov 24, 2014 at 7:39
  • I thought because of ASCII code? 1 byte = 1 char? Nov 24, 2014 at 9:23

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