Just want to know about a function in MIPS that can take under root of any number in my MIPS program.
4 Answers
You can use simplified version of Newton Method to find roots of integer
x=N
iterate 20 times:
x'=(x+N/x) /2
x=x'
Mips implementation
.data
.text
main:
li $t0,25 #N
move $t1,$t0 #x
li $t4,0 #loop variable
sqrLoop:
#Newton Formula
div $t3, $t0, $t1 # N/x
add $t1, $t3, $t1 # x + N/x
div $t1, $t1, 2 # (x + N/x)/2
#loop
add $t4, $t4, 1
blt $t4, 20, sqrLoop
end:
li $v0,10
syscall
try this out.
#DATA
.data
square: .asciiz "Enter the number you wish to find the square root for: "
answer: .asciiz "The answer is: "
newline: .asciiz "\n"
#Text
.text
.globl main
main:
li $v0, 4 #Prompt user for input
la $a0, square
syscall
li $v0, 5 #Receive said input
syscall
move $a0, $v0
move $t4, $zero #Move variables to t registers
move $t1, $a0
addi $t0, $zero, 1 #Set $t0 to 1
sll $t0, $t0, 30 #Bit Shift $t0 left by 30
#For loop
loop1:
slt $t2, $t1, $t0
beq $t2, $zero, loop2
nop
srl $t0, $t0, 2 #Shift $t0 right by 2
j loop1
loop2:
beq $t0, $zero, return
nop
add $t3, $t4, $t0 #if $t0 != zero add t0 and t4 into t3
slt $t2, $t1, $t3
beq $t2, $zero, else1
nop
srl $t4, $t4, 1 #shift $t4 right by 1
j loopEnd
else1:
sub $t1, $t1, $t3 #Decrement $t1 by $t3
srl $t4, $t4, 1 #Shift $t4 right by 1
add $t4, $t4, $t0 #then add $t0 to that
loopEnd:
srl $t0, $t0, 2 #shift $t0 to the right
j loop2
return:
li $v0, 4 #print out the answer then exit
la $a0, answer
syscall
li $v0, 1
move $a0, $t4
syscall
li $v0, 10
syscall
.data
.text
.globl main
.ent main
Sqrt:
move $v1, $a1 # $v0 = x = N li $t0, 0 # counter
sqrLoop:
div $t8, $a1, $v1 # N/x add $v1, $t8, $v1 # x + N/x div $v1, $v1, 2 # (x + N/x)/2 add $t0, $t0, 1 blt $t0, 20, sqrLoop
jr $ra
.end Sqrt
This is a function when called will take out the square root of the number . Just call the function by writing "jal Sqrt" where necessary or needed .
You can try this algorithm, which gives the integer smaller than or equal to the square root of your number.
Suppose you want the square root of n. Then keep repeating the following calculations:
x = (x + n/x) / 2
Choose x = n to start and keep repeating until x stops changing.
Here is the following MIPS program library you can add in your program.
#SquareRoot.s
#DATA
.data
square: .asciiz "Enter the number you wish to find the square root for: "
answer: .asciiz "The answer is: "
newline: .asciiz "\n"
#Text
.text
.globl main
main:
li $v0, 4 #Prompt user for input
la $a0, square
syscall
li $v0, 5 #Receive said input
syscall
move $a0, $v0
move $t4, $zero #Move variables to t registers
move $t1, $a0
addi $t0, $zero, 1 #Set $t0 to 1
sll $t0, $t0, 30 #Bit Shift $t0 left by 30
#For loop
loop1:
slt $t2, $t1, $t0
beq $t2, $zero, loop2
nop
srl $t0, $t0, 2 #Shift $t0 right by 2
j loop1
loop2:
beq $t0, $zero, return
nop
add $t3, $t4, $t0 #if $t0 != zero add t0 and t4 into t3
slt $t2, $t1, $t3
beq $t2, $zero, else1
nop
srl $t4, $t4, 1 #shift $t4 right by 1
j loopEnd
else1:
sub $t1, $t1, $t3 #Decrement $t1 by $t3
srl $t4, $t4, 1 #Shift $t4 right by 1
add $t4, $t4, $t0 #then add $t0 to that
loopEnd:
srl $t0, $t0, 2 #shift $t0 to the right
j loop2
return:
li $v0, 4 #print out the answer then exit
la $a0, answer
syscall
li $v0, 1
move $a0, $t4
syscall
li $v0, 10
syscall
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