Translating from c to mips assembly - c

Question

I have spent the greater part of today trying to properly translate this c code to mips and I'm struggling on the formatting and the second half of the pattern. The second part of the pattern breaks in a new way everytime i make a small change to the code. Im lookin to learn so any other tips regarding mips and mistakes made would help a ton.
.data
prompt: .asciiz "Please enter the number of Columns: "
s: .asciiz " "
n: .asciiz "\n"
.text
main:
li $v0, 4
la $a0, prompt
syscall
li $v0,5
syscall
move $s0, $v0 #$s0 holds N
li $s3, 0 #$s3 holds I
li $s4, 0 #$s4 holds J
li $s5, 1 #s5 holds K
add $s2, $zero, $s0 #holds N-i
li $s2, 0
add $s2, $zero, $s0
addi $s2, $s2, 1
sub $s2, $s2, $s3
for1:
blt $s0,$s3, bottom
addi $s3, $s3, 1
li $s4, 0
j for2
for2:
blt $s3, $s4,for3
li $v0, 4
la $a0, s
syscall
addi $s4, $s4, 1
li $s2, 0
add $s2, $zero, $s0
addi $s2, $s2, 1
sub $s2, $s2, $s3
li $s5 , 1
j for2
for3:
blt $s2, $s5,newline1
li $t0, 0
add $t0, $zero, $s5
li $v0, 1
move $a0, $t0
syscall
li $t0, 0
addi $s5, $s5, 1
j for3
bottom:
li $s3, 1 #$s3 holds I
li $s4, 1 #$s4 holds J
li $s5, 1 #s5 holds K
add $s2, $zero, $s0 #holds N-i
addi $s2, $s2, 1
sub $s2, $s2, $s3
for4:
blt $s0,$s3, exit
addi $s3, $s3, 1
li $s4, 0
li $s2, 0
add $s2, $zero, $s0
sub $s2, $s2, $s3
j for5
for5:
blt $s2, $s4,for6
li $v0, 4
la $a0, s
syscall
addi $s4, $s4, 1
li $t2, 1
addi $t2, $s3, 0
li $s5 , 1
j for5
for6:
blt $t2, $s5,newline2
li $t3, 2
addi $t3,$s3, 0
sub $t3,$t3,$s5
li $v0, 1
move $a0, $t3
syscall
addi $s5, $s5, 1
j for6
newline1:
li $v0, 4
la $a0, n
syscall
j for1
newline2:
li $v0, 4
la $a0, n
syscall
j for4
exit:
li $v0, 10
syscall
here is the c code:
#include <stdio.h>
int main(void) {
int n;
printf("Please enter the number of columns:"); // [0.25 mark]
scanf("%d",&n); // [0.25 mark]
//printing the upper part of the pattern.
for(int i=0;i<n;i++){ // [4 mark]
for(int j=0;j<i;j++) { // [4 mark]
printf(" "); // [0.25 mark]
}
for(int k=1;k<=n-i;k++){ // [4 mark]
printf("%d", k); // [0.5 mark]
}
printf("\n"); // [0.25 mark]
}
//printing the lower part of the pattern.
for(int i=1;i<n;i++){ // [4 mark]
for(int j=1;j<n-i;j++){ // [4 mark]
printf(" "); // [0.25 mark]
}
for(int k=1;k<=i+1;k++){ // [4 mark]
printf("%d", (i+2 - k)); // [0.5 mark]
}
printf("\n"); // [0.25 mark]
}
return 0; // [0.25 mark]
}
Any help with this is greatly apprciated!
update working code
.data
prompt: .asciiz "Please enter the number of Columns: "
s: .asciiz " "
n: .asciiz "\n"
.text
main:
li $v0, 4
la $a0, prompt
syscall
li $v0,5
syscall
move $s0, $v0 #$s0 holds N
li $s3, 0 #$s3 holds I
li $s4, 0 #$s4 holds J
li $s5, 1 #s5 holds K
add $s2, $zero, $s0 #holds N-i
li $s2, 0
add $s2, $zero, $s0
addi $s2, $s2, 1
sub $s2, $s2, $s3
for1:
ble $s0,$s3, bottom
li $s4, 0
j for2
for2:
blt $s3, $s4,for3
li $v0, 4
la $a0, s
syscall
addi $s4, $s4, 1
li $s2, 0
add $s2, $s0, 0
sub $s2, $s2, $s3
li $s5 , 1
j for2
for3:
blt $s2, $s5,newline1
li $t0, 0
add $t0, $zero, $s5
li $v0, 1
move $a0, $t0
syscall
li $t0, 0
addi $s5, $s5, 1
j for3
bottom:
li $s3,1 #$s3 holds I
li $s4, 1 #$s4 holds J
li $s5, 1 #s5 holds K
li $s2, 0
add $s2, $s0, 0 #holds N-i
j for4
for4:
blt $s0, $s3, exit
li $s2, 0
add $s2, $s0, 0
sub $s2, $s2, $s3
li $s4, 1
j for5
for5:
blt $s2, $s4,for6
li $v0, 4
la $a0, s
syscall
addi $s4, $s4, 1
li $t1, 1
li $t2, 2
add $t1, $t1, $s3
add $t2, $t2,$s3
li $s5, 1
j for5
for6:
blt $t1, $s5,newline2
li $t3, 0
sub $t3, $t2,$s5
li $v0, 1
move $a0, $t3
syscall
addi $s5, $s5, 1
j for6
newline1:
addi $s3, $s3, 1
li $v0, 4
la $a0, n
syscall
j for1
newline2:
addi $s3, $s3, 1
li $v0, 4
la $a0, n
syscall
j for4
exit:
li $v0, 10
syscall

Answer 1

This is the proper solution to the question. The mark comments are for grading purposes as this is an assignment. The issue I ran to was I was incrementing I within the first for loop moving it to the newline function helped fix the issue I was having. The pattern was only meant to reach 1-9. If there are any new questions ill try my best to answer them.
.data
prompt: .asciiz "Please enter the number of Columns: "
s: .asciiz " "
n: .asciiz "\n"
.text
main:
li $v0, 4
la $a0, prompt # asks column amount
syscall
li $v0,5
syscall
move $s0, $v0 #$s0 holds N
li $s3, 0 #$s3 holds I
li $s4, 0 #$s4 holds J #declares locations for variables
li $s5, 1 #s5 holds K
add $s2, $zero, $s0 #holds N-i
li $s2, 0
add $s2, $zero, $s0
addi $s2, $s2, 1 # sets up n-i
sub $s2, $s2, $s3
for1: # first for loop
ble $s0,$s3, bottom # break when less than to bottom func
li $s4, 0 # resets j to 0 eveery iteration
j for2 #jumps to second loop
for2:
blt $s3, $s4,for3 # breaks when less than to for3
li $v0, 4 #prints spaces
la $a0, s
syscall
addi $s4, $s4, 1 # increments J
li $s2, 0
add $s2, $s0, 0
sub $s2, $s2, $s3 #updates n-i
li $s5 , 1
j for2 #repeats loop
for3:
blt $s2, $s5,newline1 #break when less than to newline1
li $t0, 0
add $t0, $zero, $s5
li $v0, 1
move $a0, $t0 #prints numbers (K)
syscall
li $t0, 0
addi $s5, $s5, 1
j for3 # repeats loop
bottom:
li $s3,1 #$s3 holds I
li $s4, 1 #$s4 holds J #sets variables for second 3 loops
li $s5, 1 #s5 holds K
li $s2, 0
add $s2, $s0, 0 #holds N-i
j for4
for4:
blt $s0, $s3, exit #break to exit when less than
li $s2, 0
add $s2, $s0, 0
sub $s2, $s2, $s3 #sets up n-i
li $s4, 1 #sets j to 1
j for5
for5:
blt $s2, $s4,for6 #break to sixth loop
li $v0, 4
la $a0, s #prints spaces
syscall
addi $s4, $s4, 1 #increments j
li $t1, 1
li $t2, 2 #setting up i+1 and i+2
add $t1, $t1, $s3
add $t2, $t2,$s3
li $s5, 1 #sets k to 1
j for5 #repeats loop
for6:
blt $t1, $s5,newline2 #breaks to newline 2 func
li $t3, 0
sub $t3, $t2,$s5 #prints numbers
li $v0, 1
move $a0, $t3
syscall
addi $s5, $s5, 1
j for6 #repeats loops
newline1:
addi $s3, $s3, 1 #increments i
li $v0, 4
la $a0, n #prints new line
syscall
j for1 # jumps to for loop 1
newline2:
addi $s3, $s3, 1 #increments i
li $v0, 4 #prints new line
la $a0, n
syscall
j for4 #jumps to for loop 4
exit: #ends program
li $v0, 10
syscall