Push on 64-bit intel osx

Question

I want to push a 64-bit address on stack as below,

__asm("pushq $0x1122334455667788");

But I get compilation error and I can only push in following way,

__asm("pushq $0x11223344");

Can someone help me understand my mistake?

I am new to assembly, so please excuse me if my question sounds stupid.

score 18 · Accepted Answer · answered Nov 13 '12 at 00:40

x86-64 has some interesting quirks, which aren't obvious even if you're familiar with 32-bit x86...

Most instructions can only take a 32-bit immediate value, which is sign-extended to 64 bits if used in a 64-bit context. (The instruction encoding stores only 32 bits.)

This means that you can use pushq for immedate values in the range 0x0 - 0x7fffffff (i.e. positive signed 32-bit values which are sign-extended with 0 bits) or 0xffffffff80000000 - 0xffffffffffffffff) (i.e. negative signed 32-bit values which are sign-extended with 1 bits). But you cannot use values outside this range (as they cannot be represented in the instruction encoding).
mov is a special case: there is an encoding which takes a full 64-bit immediate operand. Hence Daniel's answer (which is probably your best bet).
If you really don't want to corrupt a register, you could use multiple pushes of smaller values. However, the obvious thing of pushing two 32-bit values won't work. In the 64-bit world, push will work with a 64 bit operand (subject to point 1 above, if it's an immediate constant), or a 16 bit operand, but not a 32 bit operand (even pushl %eax is not valid). So the best you can do is 4 16-bit pushes:

pushw $0x1122; pushw $0x3344; pushw $0x5566; pushw $0x7788

Comment to point #3: `subq $8, %rsp; movl $0x55667788, (%rsp); movl $11223344, 4(%rsp)` would work too. — Jester, Nov 13 '12 at 01:42
@AkiSuihkonen: No, `pushl` is [not](https://stackoverflow.com/questions/45127993/how-many-bytes-does-the-push-instruction-push-onto-the-stack-when-i-dont-specif) [encodeable](https://stackoverflow.com/questions/40305965/does-each-push-instruction-push-a-multiple-of-8-bytes-on-x64) in 64-bit mode. You want `pushq` (aka `push`) with a sign-extended 32-bit immediate, then `movl $high, 4(%rsp)` to overwrite the high half. — Peter Cordes, Oct 07 '20 at 03:23

score 9 · Answer 2 · answered Nov 12 '12 at 22:04

9

Your best bet would be to do something like this.

movq $0x1122334455667788, %rax
pushq %rax

Replace %rax with any other 64-bit register you find appropriate.

answered Nov 12 '12 at 22:04

score 5 · Answer 3 · answered Nov 12 '12 at 20:56

5

There is no single instruction capable of taking a 64-bit immediate value and pushing that onto the stack.

answered Nov 12 '12 at 20:56

David Schwartz

score 5 · Answer 4 · edited May 23 '17 at 11:45

5

pushq my_const(%rip)
...
my_const: .quad 1122334455667788

edited May 23 '17 at 11:45

Community

answered Nov 13 '12 at 06:51

Aki Suihkonen

4 Answers4