CODEWITHSUNDEEP
x86gdbendianness
ultimate cause
ultimate cause

Reputation: 2304

Little Endian in gdb

        0x0      0x6      0x0       0x1

BIG - 00000000 00000110 00000000 00000001

LIT - 00000001 00000000 00000110 00000000

So I was expecting 0xcfb01420 + 2 to be 0x6, but that is not the case. Am I missing something?

(gdb) p &tx_q_sem->mlock.u.mtxa_owner
$41 = (volatile __uintptr_t *) 0xcfb01420
(gdb) p/x tx_q_sem->mlock.u.mtxa_owner
$34 = 0x601
(gdb) p (char)*(char*)(0xcfb01420+0)
$36 = 0x1
(gdb) p (char)*(char*)(0xcfb01420+1)
$37 = 0x6
(gdb) p (char)*(char*)(0xcfb01420+2)
$38 = 0x0
(gdb) p (char)*(char*)(0xcfb01420+3)
$39 = 0x0

My assumption is that byte sequence for 0x601 in little endian format should match the way I represented it above. Therefore second byte should have been all zeros. There I have put all the nibbles correctly I guess.

Upvotes: 0

Views: 1421

Answers (1)

Tom Tromey
Tom Tromey

Reputation: 22589

Your initial hex values are all missing digits, relative to what you're actually examining.

00000001 is 0x01 -- it is 8 binary digits, so a full byte.

So 0x601 == 0x0601 == 00000110 00000001.

You can see this in gdb with:

(gdb) print /t 0x601
$1 = 11000000001

Upvotes: 2

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