function in mips that calculates the absolute value [duplicate] - function

Question

This question already has answers here:
Integer absolute value in MIPS?
(5 answers)
Closed 2 years ago.
hello i have an exercise in mips that must call a function to calculates the absolute value of a number
i have write 2 codes but if you have any other solution write it
.data
message:.asciiz "give number: "
.text
main:
li $v0, 4
la $a0, message
syscall
li $v0, 5
syscall
add $t1, $v0, $zero
jal absolute
add $v0,$t2,$zero
li $v0, 1
syscall
li $v0, 10
syscall
absolute: ori $t2,$t1,0 #copy r1 into r2
slt $t3,$t1, $zero #is value < 0 ?
beq $t3,$zero,gg #if r1 is positive, skip next inst
sub $t2,$zero, $t0 #r2 = 0 - r1
jr $ra
gg:
#t2
and the second code is this
.data
question: .asciiz "give number"
.text
main:
li $v0, 4
la $a0, question
syscall
li $v0, 5
syscall
jal absolute
li $v0, 1
syscall
li $v1, 1
syscall
li $v0, 10
syscall
absolute:
slti $t0,$a0,0
bne $t0,$zero,g1
add $v0,$a0,$zero
jr $ra
g1:
sub $t2,$a0,$a0
sub $v1,$t2,$a0
j absolute

Answer 1

Try this:
.data
message: .asciiz "Enter the number: "
.text
.globl main
main:
# print message
li $v0, 4
la $a0, message
syscall
# read integer
li $v0, 5
syscall
slt $t0, $v0, $0 # $t0 = ( $v0 < 0 ? 1 : 0 )
bne $t0, $0, NEGATIVE # if($t0 != 0) goto NEGATIVE
j POSITIVE # goto POSITIVE (and dose nothing)
NEGATIVE:
# ~$v0 + 1
nor $v0, $v0, $0 # NOR with zero = NOT
addi $v0, $v0, 1 # $v0 =+ 1
POSITIVE:
# print $v0
move $a0, $v0
li $v0, 1
syscall
# print new line '\n'
li $v0, 11
addi $a0, $0, 10
syscall
jr $ra
Test:
Enter the number: 10
10
Enter the number: -5
5