问题
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.
回答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
回答2:
It'll make the stack pointer not able to do its job in some cases. for instance, lets say you had a function which pushed a byte onto the stack and then calls another function. The call will end up trying to write a misaligned return address onto the stack, resulting in an error.
回答3:
The stack pointer must be (for some optimalization reasons) 4B aligned -> it should be divisible by four (and, therefore, have last 2 bits zero).
回答4:
what you want to do is use the bit rotation opcodes to rotate through each 32-bit memory location, placing 8 bits at a time into the register until you have rotated back to the starting bit positions. now you should have 4 8-bit quantities lined up side by side in your 32 bit register. now push that onto the stack and you're done.
来源：https://stackoverflow.com/questions/2586591/why-is-it-not-possible-to-push-a-byte-onto-a-stack-on-pentium-ia-32