问题
Well first I understand (or a I think that I understand) the problems of misaligned stack.
But I know (like a definition) that pushing a 16bit value to 32bit wide stack could cause a stack misaligned.
But the thing I dont understand, is how this could happend...since PUSH and POP check the D flag at the segment descriptor (so is 1 increments/decrements 32bits and is 0 16bits).
Suppose that D flag=1, should PUSH AX do a 32bits decrement? so its like I "miss" 16bits in the stack?
I am not sure that I understand this problem
回答1:
While both push and pop check the D-bit in the segment descriptor to determine the default operand size (i.e. 16 or 32/64 bits), it can be overridden with an operand size override 0x66.
If the D-bit is 0, then:
ff /6
pushes 16 bits onto the stack
and
66 ff /6
pushes 32 bits (or 64 if it's a 64-bit segment).
If the D-bit is 1, then the opposite is true.
In any case, ESP (or RSP or just SP, depending on the address size) is incremented or decremented by 2 (for 16 bit operations), 4 (for 32 bit operations) or 8 (for 64 bit operations).
来源：https://stackoverflow.com/questions/2302661/assembler-stack-alignment-or-better-misaligned-example-with-push