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I did this MIPS problem that prompts the user to enter at least 4 numbers and print them in ascending order. I was wondering if someone can take a look at it and tell me what you think about it? If I wanted to get it to print in descending order too, how would I get that started? This program is just for fun before I have a real program to do!!!! 

.data
array: .space 100
input: .asciiz "Enter at least 4 integers: Enter the number 1000 to exit \n"
output: .asciiz "The array in ascending order: \n"
commas: .asciiz ","

.text
.globl main
main:

la $a1, array   #loads a pointer to array into $a1
li $a2,9        #loads 9 into $a2
li $t0,0    
li $t1,1000     

loops:

la $a0, input   #loads input text into $a
li $v0, 4       #loads 4 into $v0 (prints string) 
syscall         
li $v0, 5       #loads 5 into $v0 (read interger)
syscall         
beq $v0,$t1,swap
addi $t0,$t0,4   #add 4 to $t0, save to $t0
sw $v0, ($a1)    #stores input into array
addi $a1, $a1,4  #add 4 to $a1, save to $a1
j loops

swap:

la $t4, array    #loads array to $t4
la $t1, array    #loads array to $t1
addi $t1,$t1,4   #add 4 to $t1, save to $t1 
la $t8,array     #loads array to $t8
add $t8,$t0,$t8  #add $t8 to $t0, save to $t8
la $t9,array
add $t9,$t0,$t9  #add $t9 to $t0, save to $t9
addi $t9,$t9,-4  #subtracts 4 from $t9, save to $t9

loop:

lw $t2,($t4)     #load input into $t2
lw $t3,($t1)     #load input into $t3
blt $t2,$t3,loop1 #if $t2 < $t3, go to loops
sw $t3,($t4)     #store $t3 in $t4
sw $t2,($t1)     #store $t2 in $t1

loop1:

addi $t1,$t1,4    #add 4 to $t1, save to $t1
blt $t1,$t8,loop  #if $t1<$t8, go to loop
addi $t4,$t4,4    #add 4 to $t4, save to $t4
move $t1,$t4
addi $t1,$t1,4    #add 4 to $t1, save to $t1
blt $t4,$t9,loop  #if $t4<$t9, to go loop

print:

la $a1,array      #loads array to $a1
la $a0, output    #loads output to $a0
li $v0, 4         #loads 4 into #v0
syscall

loop2:

blez $t0, done   #if $t0<=0, go to done
li $v0, 1        #loads 1 into $v0
lw $a0, 0($a1)   #load an inout into $a0
syscall
la $a0, commas    #loads commas into $a0
li $v0, 4        #loads 4 into $v0
syscall
addi $a1, $a1, 4  #add 4 to $a1, save to $a1
addi $t0, $t0, -4 #subtracts 4 from #t0, save to $t0
j loop2

done:
j done
4

1 回答 1

1

方法1：

只需按降序对数组进行排序。这应该是推荐的。
与大多数常见的排序一样，排序是基于比较和交换方案。在代码中，它发生在 labelloop和loop1. 所以你可以改变第 44 行的比较部分：

blt $t2,$t3,loop1 #if $t2 < $t3, go to loops

至：

bgt $t2,$t3,loop1 #if $t2 > $t3, go to loops

然后结果应该是递减的。

方法2：

仍然按升序对数组进行排序，同时反向打印数组。
打印部分位于 label loop2，您需要在其中更改它，例如替换第 68 行：

lw $a0, 0($a1)   #load an inout into $a0

和：

add $t1, $a1, $t0
addi $t1, $t1, -4
lw $a0, 0($t1)   #load an inout into $a0

以及第 73 行评论：

addi $a1, $a1, 4  #add 4 to $a1, save to $a1
于 2014-04-16T11:46:27.290 回答