如何在MIPS程序集中从堆栈中弹出？

Question

我试图去学习MIPS程序集，因为我有一些空闲时间，而且我正在尝试编写一个程序，将数字压入堆栈，然后将它们弹出。我希望它能够计算在达到负数之前弹出的数字的数量，然后每次得到负数时都会记录下弹出多少个负数和正数。

到目前为止，我得到这个：

#count the number of negative words on the stock by poping the stack until a non- negative word is found 
#and print out the number of words found 

.text 
.globl main 
#this code reads the numbers from the data area and stores in them in a stack 
#the numbers are located in the test area and the number of numbers in the num area 

main: la $t0, test 
lw $t1, num 
loop: lw $t2,($t0) 
sub $sp, $sp, 4 
sw $t2($sp) 
add $t0, $t0, 4 
add $t1, $t1, -1 
bnez $t1, loop 


#pop from the stack and print the number of numbers in the stack before a nonnegative number is reached 
#then keep count of how many negative and positive ones there are total 

#code I cannot come up with would go here 

.data 
test: .word 
2, 0xfffabfff,2,-4,-9,0x99999999,0x90000000,-2147479536,0x80000000 
num: .word 10 
ans: .asciiz "Number is = " 
endl: .asciiz "\n" 

我把它推吧，据我所知道的，但我想不通的推动和计数的权利。我必须从这里做什么？

Answer 1

流行将是推动的对立面。所以，如果你用它来推动$t2：

lw $t2,($sp) 
addiu $sp,$sp,4 

计数的堆栈上否定词的数量将有一个循环弹出一个字的问题：

sub $sp,$sp,4 
sw $t2,($sp) 

你会用弹出它如果弹出的值大于等于0，则使用BGEZ退出循环，否则增加计数器并重复。