Question

I’ve come to learn that you cannot push a byte directly onto the Intel Pentium’s stack, can anyone explain this to me please?

The reason that I’ve been given is because the esp register is word-addressable (or, that is the assumption in our model) and it must be an “even address”. I would have assumed decrementing the value of some 32-bit binary number wouldn’t mess with the alignment of the register, but apparently I don’t understand enough.

I have tried some NASM tests and come up that if I declare a variable (bite db 123) and push it on to the stack, esp is decremented by 4 (indicating that it pushed 32-bits?). But, “push byte bite” (sorry for my choice of variable names) will result in a kind error:

test.asm:10: error: Unsupported non-32-bit ELF relocation

Any words of wisdom would be greatly appreciated during this troubled time. I am first year undergraduate so sorry for my naivety in any of this.

Answer 1

Its based on how the stack was created:

The address-size attribute of the
stack segment determines the stack
pointer size (16, 32 or 64 bits). The
operand-size attribute of the current
code segment determines the amount the
stack pointer is decremented (2, 4 or
8 bytes).

In non-64-bit modes: if the
address-size and operand-size
attributes are 32, the 32-bit ESP
register (stack pointer) is
decremented by 4. If both attributes
are 16, the 16-bit SP register (stack
pointer) is decremented by 2.

Source: http://www.intel.com/Assets/PDF/manual/253667.pdf

pg. 4-320 Vol. 2B

Edit

Just wanted to point out also that an interesting read is the section on stacks in the manual, it will explain creating a stack segment further.

http://www.intel.com/Assets/PDF/manual/253665.pdf

Chapter 6.2