Questions about the stack segment SS register.

qqion1988

There can only be one stack in the stack segment of the program. SS is located at the starting address (base address, low address) of the stack segment, and the bottom of the stack is at the end (high address) of the effective address of the stack segment. SP is initialized to the size of the stack segment, and SS:SP always points to the top of the stack. During initialization, SS:SP points to the highest address of the stack segment (at this time, both the bottom and top of the stack point to this memory address). As the element is pushed, SP keeps getting smaller, and then the top address of the stack represented by SS:SP gets smaller, not equal to the bottom address of the stack, but gradually approaching the starting address of the stack segment. When SP is 0, SS:SP represents The top address of the stack is equal to the start address of the stack segment represented by SS:0000, and then the stack is full. I guess the CPU also checks whether the stack overflows by judging SP<0.

qqion1988

I found a question and its answer on the Internet, please refer to it.

1. If a stack starts from address 3250H:0000H, the offset address of its last word is 0100H, and the content of SP is 0052H, ask:
1 What is the top address of the stack?
2 What is the bottom address of the stack?
3 What is the content of SP after saving data 1234H and 5678H?

answer:
1 The highest address of the stack is called the bottom of the stack, and the stack indicator SP always points to the top of the stack.
Stack top address=(SS)×10H+(SP)=3250H×10H + 0052H=32552H
2 Stack bottom address = (SS)×10H + 0100H=32600H
3 The stack access in 8086CPU must be based on words. After the data 1234H and 5678H are stored, that is, after the two words are pushed onto the stack, SP is reduced by 4. The content of SP is: (SP)-number of words×2=0052H-2×2=004EH. The content of SS remains unchanged.

大核桃

In the next beginners, the microcomputer principle supportsqqion1988. Recently, my classmates have asked the same question. I agree with Mr. M. Then, first open up the 64k maximum space in the memory and give the stack ss point to the low address. Sp always points to the top of the stack. Initialization should be put into the stack at the high address. Is in line with sp-2

xjzdd3534296

The above is so complicated, but the 21# example is easier to understand! ! ! ! ! !

deng2020

Owner, help you top the post

deng2020

The original poster, help you top the post,,,,