我有mips的这个程序，我不是要把它改为双精度。看起来单精度和双精度浮动指令具有相同的指令但不是.s它是.d如果任何人有共识或帮助它会帮助我很多。
#Babylonian method: The babylonian method is one of the many method to approximate Roots.
#The method uses the following C code:
# float x = n;
# float y = 1;
# float e = 0.000001; /* e decides the accuracy level*/
# while(x - y > e)
# {
# x = (x + y)/2;
# y = n/x;
# }
.data
StartingStatment: .asciiz "Please enter a number that you would like to find the square root of: \n"
Answer: .asciiz "Your answer is: \n"
NumberofIterations: .asciiz "Please enter the number of Iteration. Example: 100\n"
ErrorQ: .asciiz "Pleae enter a number which is greater than zero!"
Ended: .asciiz "\n--Ended--"
Y: .float 1.0
SmallValue: .float 0.000001
Two: .float 2.0
.text
.globl main
main:
#printing message for StartingStatment
li $v0, 4
la $a0, StartingStatment
syscall
li $v0, 6 #load a six into vo for syscall user input
syscall
mov.d $f2, $f0 #$f2 have value
mov.d $f8, $f2 ##### convert float to int
cvt.w.d $f8, $f8
mfc1 $t1, $f8 #converted int stored in t1
blt $t1,0,ErrorG #if int is less than one then throw an error
l.d $f4, Y #loading float into $f4
l.d $f12, SmallValue #load value into f12
l.d $f8, Two #load value into f8
j FindSqaureRoot #now find the square root
FindSqaureRoot:
#$f2 #x
#f4 #y
#f12 #SmallValue
mov.d $f14,$f2 #n
#while(x - y > e)
# {
# x = (x + y)/2;
# y = n/x;
# }
# return x;
SubtractionChecking:
sub.d $f2,$f2,$f4 #x-y
c.lt.d $f2,$f12 # is (x-y) < e?
bc1t Print #yes: print(Move towards answer)
c.lt.d $f12,$f2 # is (x-y) > e?
bc1t Loop # yes: Continue with the #loop!
Loop:
add.d $f6,$f2,$f4 #x+y
div.d $f2,$f6,$f8 #divding it by 2
div.d $f4,$f14,$f2 #n/x
j SubtractionChecking
#Printing the answer!
Print:
mov.d $f16,$f2
li $v0,2
syscall
j EndProgram
#Error that is generated when a number is less than 0!
ErrorG:
li $v0, 4
la $a0, ErrorQ
syscall
j EndProgram
EndProgram:
li $v0, 4 # printing message for xvalue
la $a0, Ended
syscall
li $v0,10
syscall