Sort a list of user-entered numbers in MIPS
Question
I'm writing MIPS program that will examine a list of 10 numbers to be input from the terminal. And these numbers will output on the terminal in an ascending order. And below are my MIPS program...please can someone help me look into it, because i m running and isn't working proply....
.data
array: .space 40
prompt: .asciiz "Enter a number: "
spacee: .asciiz " "
output: .asciiz "The numbers are: "
.text
main:
li $t1,10
la $a1,array
loop:
addi $t1,$t1,-1
li $v0,4
la $a0,prompt
syscall
li $v0,5
syscall
sw $v0,0($a1)
addi $a1,$a1,4
bnez $t1,loop
li $t1,9
li $t2,9
la $a1,array
loop1:
beqz $t2,here
addi $t2,$t2,-1
lw $t5,0($a1)
lw $t6,4($a1)
add $a1,$a1,4
ble $t5,$t6,loop1
sw $t5,0($a1)
sw $t6,-4($a1)
bnez $t2,loop1
here:
la $a1,array
addi $t1,$t1,-1
add $t2,$t2,$t1
bnez $t1,loop1
li $v0,4
la $a0,output
syscall
la $a1,array
li $t1,10
loop2:
li $v0,1
lw $a0,0($a1)
syscall
li $v0,4
la $a0,spacee
syscall
add $a1,$a1,4
addi $t1,$t1,-1
bnez $t1,loop2
li $v0,10 #exit
syscall
Solution
If you use addi instead of add, does it work? Also with assembly, it sometimes helps to comment a lot since it doesn't read anywhere close to a natural language.
.data
array: .space 40
prompt: .asciiz "Enter a number: "
spacee: .asciiz " "
output: .asciiz "The numbers are: "
.text
main:
li $t1,10 #load 10 into $t1
la $a1,array #load a pointer to array into $a1
loop:
addi $t1,$t1,-1 #subtract 1 from $t1, save to $t1
li $v0,4 #load 4 into $v0 (print string)
la $a0,prompt #load prompt text into $a
syscall #display prompt
li $v0,5 #load 5 into $v0 (read integer)
syscall #prompt for input
sw $v0,0($a1) #store input int to array
addi $a1,$a1,4 #add 4 to $a1, save to $a1
bnez $t1,loop #if $t1 isn't zero,goto loop
li $t1,9 #if $t1 is zero, load 9 into $t1
li $t2,9 #and load 9 into $t2
la $a1,array #load array pointer into $a1
loop1:
beqz $t2,here #if $t2 is zero, goto here
addi $t2,$t2,-1 #subtract 1 from $t2, save to $t2
lw $t5,0($a1) #load an input int into $t5
lw $t6,4($a1) #load the next one into $t6
addi $a1,$a1,4 #add 4 to $a1, save to $a1
ble $t5,$t6,loop1 #if $t5 <= $t6, goto loop1
sw $t5,0($a1) #else, store $t5 in $a1
sw $t6,-4($a1) #and store $t6 in $a1-4 (swapping them)
bnez $t2,loop1 #if $t2 is not zero, to go loop1
here:
la $a1,array #load array into $a1
addi $t1,$t1,-1 #subtract 1 from $t1, save to $t1
add $t2,$t2,$t1 #add $t2 to $t1, save to $t2
bnez $t1,loop1 #if $t1 isn't zero, goto loop1
li $v0,4 #load 4 into $v0 (print string)
la $a0,output #load 'the numbers are' into $a0
syscall #display message to screen
la $a1,array #load array pointer into $a1
li $t1,10 #load 10 into $t1
loop2:
li $v0,1 #load 1 into $v0 (print int)
lw $a0,0($a1) #load $a1 into $a0
syscall #print first number to screen
li $v0,4 #load 4 into $v1 (print string)
la $a0,spacee #load ' ' into $a0
syscall #print ' ' to screen
addi $a1,$a1,4 #add 4 to $a1, save to $a1
addi $t1,$t1,-1 #subtract 1 from $t1, save to $t1
bnez $t1,loop2 #if $t1 isn't zero, goto loop2
li $v0,10 #exit
syscall
I don't have a MIPS processor, but this worked in C: #include "stdafx.h"
int _tmain(int argc, _TCHAR* argv[])
{
int t1;
int t2;
int* a1;
int t5;
int t6;
int arr[10] = {10,9,8,7,6,5,4,3,2,1};
t1 = 9;
t2 = 9;
a1 = arr;
loop1:
if(t2 == 0)
goto here;
t2 = t2 - 1;
t5 = *a1;
t6 = *(a1 + 1);
a1 = a1 + 1;
if(t5 <= t6)
goto loop1;
*a1 = t5;
*(a1-1) = t6;
if(t2 != 0)
goto loop1;
here:
a1 = arr;
t1 = t1 - 1;
t2 = t2 + t1;
if(t1 != 0)
goto loop1;
printf("the numbers are\n");
a1 = arr;
t1 = 10;
loop2:
printf("%i ", *a1);
a1 = a1 + 1;
t1 = t1 - 1;
if(t1 != 0)
goto loop2;
return 0;
}
OTHER TIPS
To begin with, you are using some instructions the wrong way.
add $a1,$a1,4
should be
addiu $a1,$a1,4
Because you are adding an inmediate,not two registers.
In addition to that, you should take a look to your comparing logic. It's quite confusing and error prone.