如何从MIPS程序集中的堆栈弹出？

Question

我正在尝试学习MIPS程序集，因为我有一些空闲时间，并且我正在尝试编写一个程序，将数字推送到堆栈，然后弹出它们。我希望它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，保持计数弹出了多少负数和正数。

到目前为止，我得到了这个：

 #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"

我把它推向正确，尽管我可以说，但我无法弄清楚正在推动和计数。我该怎么办呢？

Answer 1

流行音乐将与推动相反。因此，如果您使用它来推送$t2：

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

你会弹出它：

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

计算堆栈上否定词的数量将是一个循环，从堆栈中弹出一个单词，如果弹出值>> 0则使用BGEZ退出循环或以其他方式增加一个柜台并重复。