How to copy the value at a certain address in memory to a register in gcc AT&T style

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I want to copy the value at a certain address in memory to a register using AT&T style assembly. I know this shouldn't be hard, and I think in Intel style it's something like:

mov rdi, [0xdeadbeef]

But I don't know much about the AT&T style (or assembly in general). I searched about it but all the examples about mov that I got didn't include this one.

So can anyone tell me how that instruction looks like?

Also, where can I find a complete list of x86_64 assembly instructions in AT&T style?

assembly
x86-64
att
asked on Stack Overflow Oct 21, 2013 by Gnijuohz • edited Apr 24, 2019 by phuclv

1 Answer

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To copy the value at a certain address in memory to a register in 32-bit mode we use

mov edi, [0xdeadbeef] ; Intel
movl 0xdeadbeef, %edi ; AT&T

In AT&T any literal that is not prefixed by $ is an address

But in x86_64 64-bit absolute addressing is not allowed, so you can't use movq 0xdeadbeef, %rdi like above. The only instruction that has 64-bit immediate is mov (movabs in gas), which can assign a 64-bit constant to any registers, or move value at a 64-bit absolute address to Areg

mov rax, [0xdeadbeef]   ; Intel
movabs 0xdeadbeef, %rax ; AT&T

If you really need to move the value from a 64-bit absolute address to a register different from Areg you must use indirect addressing instead

mov rdi, 0xdeadbeef     ; Intel
mov rdi, [rdi]

movq $0xdeadbeef, %rdi  ; AT&T
movq (%rdi), %rdi

or if you want the value to be copied to both rax and rdi then

mov rax, [0xdeadbeef]   ; Intel
mov rdi, rax

movabs 0xdeadbeef, %rax ; AT&T
movq %rax, %rdi

Here the q suffix means quadword (64-bit) registers

In AT&T syntax the size of memory operands is determined from the last character of the instruction mnemonic. Mnemonic suffixes of b, w, l and q specify byte (8-bit), word (16-bit), long (32-bit) and quadruple word (64-bit) memory references. Intel syntax accomplishes this by prefixing memory operands (not the instruction mnemonics) with byte ptr, word ptr, dword ptr and qword ptr. Thus, Intel mov al, byte ptr foo is movb foo, %al in AT&T syntax.

In 64-bit code, movabs can be used to encode the mov instruction with the 64-bit displacement or immediate operand.

https://sourceware.org/binutils/docs/as/i386_002dVariations.html

More information about 64-bit mov instruction here: Difference between movq and movabsq in x86-64. As you can see there's no version for moving from a 32-bit absolute address to a 64-bit register, so even in rare cases when the address fits in 32 bits like 0xdeadbeef, you still have to use movabs Areg, moffs64

answered on Stack Overflow Oct 21, 2013 by phuclv • edited Jul 4, 2018 by phuclv

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