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macosassemblyx86-64

Push on 64-bit intel osx

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.

Solution

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

    1. 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).

    2. 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).

    3. 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