nested loop problem of mips:Printing a pattern and gettin an error in my code - mips

Question

the following mips code is of printing a number pattern like this: 1 12 123 1234 12345 i cannot find the error in my code.I don't understand what is missing.
main:
li $t0,1
li $t1,6
li $t2,1
li $t3,6
out:
beq $t0, $t1, exit
in:
beq $t2,$t3, exit
move $a0,$t2
li $v0, 1
syscall
addi, $t2, $t2, 1 #counter
j in
lineloop:
addi $t0,$t0,1
li $v0, 4 # print new line
la $a0, newline
syscall
j out
exit:

Answer 1

you have not changed $t2 to 1 in the last loop so that it starts printing from 1 everytime the line changes.
Also in the "in" loop you have not added the last loop instead you wrote exit which will make it leave the outer loop as well so you have to write it like this:
beq $t2,$t3,lineloop
SO the final code looks something like this
main:
li $t0,1
li $t1,6
li $t2,1
li $t3,6
out:
beq $t0, $t1, exit
in:
beq $t2,$t3, lineloop
move $a0,$t2
li $v0, 1
syscall
addi, $t2, $t2, 1 #counter
j in
lineloop:
addi $t0,$t0,1
li $v0, 4 # print new line
la $a0, newline
syscall
li $t2,1 #changes value in $t2 back to 1
j out
exit:

Answer 2

Here's some C code equivalent to what you're working on:
for ( int i = 1; i != 6; i++ ) {
   for ( int j = 11; j != 62; j++ ) {
     printf ( "%d", j );
   }
   printf ( "\n" );
}
1  The inner loop should be fully nested within the outer loop — that includes its initialization part.  Your assembly code has relocated inner loop initialization out of its proper context, and placed it with the (initialization section of the) outer loop:
for ( int i = 1, int j = 11; ; i != 6; i++ ) {
   for ( ; j != 62; j++ ) {
     printf ( "%d", j );
   }
   printf ( "\n" );
}
Without the inner loop's initialization being repeated it won't function equivalent to the above C code.  You only need move your li $t2, 1 to the proper location:
#li $t2,1 # improper placement of inner loop initialzation
li $t3,6
out:
beq $t0, $t1, exit
li $t2, 1 # here's where inner loop's initialization belongs
# properly nested within the body of the outer loop
# and immediately preceding the inner loop repeating body
in:
beq $t2,$t3, lineloop
2  In order to change the behavior of the inner loop each iteration, the inner loop should be dependent upon the outer loop's control variable, but you don't have it that way.  Without that, it will function independently, yielding 12345 each time (so: 12345 12345 12345 12345 for output).  The inner loop should take a dependence on the outer loop's control variable to decide when to stop as follows:
for ( int i = 1; i != 6; i++ ) {
   for ( int j = 11; j != i+12; j++ ) {
     printf ( "%d", j );
   }
   printf ( "\n" );
}
So, compute i+1 and use that computation to stop the inner loop.
Let's note that you can remove the li $t3,6 since that isn't actually when to stop the inner loop.  Also i+1 needs to be computed for the inner loop to use, and could be done within in the inner loop itself, but it is invariant to the inner loop (i only changes in the outer loop), so that computation can be done inside the body of the outer loop instead of inside the body of the inner loop.
#li $t2,1 # improper placement of inner loop initialzation
#li $t3,6 # wrong termination value & computation location (its not constant)
out:
beq $t0, $t1, exit
li $t2, 1 # here's where inner loop's initialization belongs
# properly nested within the body of the outer loop
addi $t3, $t0, 1 # compute i+1 for the inner loop to use as when to stop
in:
#addi $t3, $t0, 1 # alternate location to compute i+1
# but would (re)compute the same value each time
# so can do this outside the inner loop instead as above
beq $t2,$t3, lineloop