我试图弄清楚在进行浮点加法时如何规范化 MIPS 中的有效数字。
假设你有 0.001*2^-1。在这种情况下，要规范化，您必须将有效位左移 3 并将指数减 3。
我将如何确定最高有效位回到标准化位置（在我之前的情况下，我怎么知道我在第三次移位后停止移位）？
更新：这是我目前正在使用的代码。给定两个输入（浮点数），程序应该进行半精度浮点加法，这就是我将单精度转换为半精度的原因。输出应该只是两个浮点数的总和
我认为问题在于正常化。
.data
ask_user_a: .asciiz "Enter a decimal number (a): "
ask_user_b: .asciiz "Enter a decimal number (b): "
sum: .asciiz "a+b = "
error_out_of_range: .asciiz "Error: A number you entered is out of range! "
new_line: .asciiz "\n"
.text
main:
#Ask user for the input a
la $a0, ask_user_a # load the addr of ask_user_a into $a0.
li $v0, 4 # 4 is the print_string syscall.
syscall # do the syscall.
#put input a into $s0
li $v0, 6 #6 for getting floating point numbers
syscall # do the syscall
mfc1 $s0, $f0 #move floating point number to $s0
## ask user for the input b
la $a0, ask_user_b # load the addr of ask_user_b into $a0.
li $v0, 4 # 4 is the print_string syscall.
syscall # do the syscall.
#put input b into $s1
li $v0, 6 #6 for getting floating point numbers
syscall # do the syscall
mfc1 $s1, $f0 #move floating point number to $s1
#extract parts for A
srl $t0, $s0, 31 #$t0 = Sign Bit A
sll $t1, $s0, 1
srl $t1, $t1, 24
sub $t1, $t1, 127 #t1 = Exponent Bit A
sll $t2, $s0, 9
srl $t2, $t2, 9 #t2 = Mantissa A
srl $t2, $t2, 13 #Truncate Mantissa
ori $t2, $t2, 0x800000
srl $t2, $t2, 12
#extract parts for B
srl $t5, $s1, 31 #$t5 = Sign Bit B
sll $t6, $s1, 1
srl $t6, $t6, 24
sub $t6, $t6, 127 #t6 = Exponent Bit B
sll $t7, $s1, 9
srl $t7, $t7, 9 #t7 = Mantissa B
srl $t7, $t7, 13 #Truncate Mantissa
ori $t7, $t7, 0x800000
srl $t7, $t7, 12
##Check that exponent A is >16 and <-16, jump to error otherwise
blt $t1, -16, oor_error
bgt $t1, 16, oor_error
##Check that exponent B is >16 and <-16, jump to error otherwise
blt $t6, -16, oor_error
bgt $t6, 16, oor_error
j no_oor_error
#Throw error
oor_error:
la $a0, error_out_of_range # load the addr of ask_user_a into $a0.
li $v0, 4 # 4 is the print_string syscall.
syscall # do the syscall.
j exit
#There was no out of range error
no_oor_error:
#compare exponents
beq $t6, $t1, exponents_match
bgt $t6, $t1 expBgtexpA
sub $t8, $t6, $t1
srlv $t7, $t7, $t8
move $t6, $t1
j exponents_match
expBgtexpA:
sub $t8, $t1, $t6
srlv $t2, $t2, $t8
move $t1, $t6
exponents_match:
#If the signs are the same, add mants
#If the signs are diff, subtract mants
bne $t0, $t5, different_signs
add $t9, $t2, $t7
j significands_added
different_signs:
sub $t9, $t2, $t7
significands_added:
#normalize
#determine size
bne $t0, $t5, normalize_different_sign
normalize_same_sign:
andi $t8, $t9, 0x1000
beq $t8, $zero, done_normalize
srl $t9, $t9, 1
addi $t1, $t1, 1
j normalize_same_sign
normalize_different_sign:
andi $t8, $t9, 0x1000
beq $t8, $zero, done_normalize
sll $t9, $t9, 1
subi $t1, $t1, 1
j normalize_different_sign
done_normalize:
#Check for overflow
##Convert back to float
sll $t0, $t0, 31 ##move sign to 32nd bit
addi $t1, $t1, 127 ##Add bias
sll $t1, $t1, 23 #move exponent to bits 24-31
sll $t9, $t9, 11 #move over mant
andi $t9, $t9, 0x3FFFFF
sll $t9, $t9, 1
or $s2, $t0, $t1
or $s2, $s2, $t9
#Print the float
mtc1 $s2, $f12
li $v0, 2
syscall
exit:
#Exit Program
li $v0, 10 # syscall code 10 is for exit.
syscall # make the syscall.