在沒有FPU的MIPS彙編中將單精度（IEEE 754）分爲雙精度

Question

我必須在MIPS彙編中編寫一個程序，它將雙精度浮點數除以單精度數（使用標準IEEE 754），而不使用浮點指針單元。我唯一不能理解的是如何處理這兩個數字的尾數來計算分割。如何遍歷個別位？

.data 

res: .asciiz "Result is: " 
neg_res:.asciiz "Result is: - " 
choice: .asciiz "\nMake your choice:\n1. Execute\n2. Exit\n" 
wrong: .asciiz "Inserted value is wrong! Please try one more time." 
first: .asciiz "Insert the first number: " 
n:  .asciiz "\n" 
second: .asciiz "Insert the second number: " 
divz: .asciiz "Can't divide for 0! Please try one more time." 

.text 

main: 
    la $a0, choice 
    li $v0, 4 
    syscall     # Print menu 

    li $v0, 5 
    syscall     # Input choice 

    move $t7, $v0  
    beq  $t7, 2, end  # Control=2 
    blt  $t7, 1, err  # Control<1 
    bgt  $t7, 2, err  # Control>2 

getdoub: 
    la $a0, first 
    li $v0, 4 
    syscall    # Message for input numerator 

    li $v0, 7 
    syscall 
    mov.d $f4, $f0  # Input double numerator 

    c.eq.d $f4, $f30 
    bc1t print_dou  # Print result if numerator is = 0 

getfloat: 
    la $a0, second 
    li $v0, 4 
    syscall    # Message for input denominator 

    li $v0, 6 
    syscall 
    mov.s $f6, $f0  # Input float denominator 

    c.eq.s $f6, $f30 
    bc1t errz   # Print error if denominator is = 0 

    sign_d: 
    cvt.d.s $f6, $f6  # Convert from single to double to avoid errors in compare 

    c.lt.d $f4, $f6   # Control sign of double 
    bc1t minus_d   # If c1 is true (that means $f4<$f8) jump to minus_d 
    b plus_d    # Otherwise plus_d 

minus_d: 
    li $t0, 1   # Negative sign 

plus_d: 
    li $t0, 0   # Positive sign 

extr_double: 
    mfc1.d $t2, $f4   # Iin $t2 I've sign, exponent and 20 bit of mantissa in t3, 32 bit of mantissa 
    li $t6, 2146435072 # Mask for 11 bit of exponent 
    and  $t1, $t2, $t6  # In $t1 Exponent 
    li $t6, 1048575 # Mask for 20 bit of mantissa 
    and  $t2, $t2,$t6 
    add  $t2, $t2, 2097152 # In t2 I've mantissa 1 and in t3 ho la mantissa 2 automatically 


sign_s: 
    cvt.s.d $f8, $f8  # Convert $f8 from double to single 
    c.lt.s $f6, $f8  # Same control made for numerator 
    bc1t minus_s 
    b plus_s 

minus_s: 
    li $t4, 1 
    b extr_sing 

plus_s: 
    li $t4,0   # t4 sign 

extr_sing: 
    mfc1 $t6, $f6 
    li $t7, 2139095040 # 8 bit of exponent a 1 
    and  $t5, $t6, $t7  # In $t5 I've exponent 

    li $t7, 8388607 # Mask for 23 bit 
    and  $t6, $t6, $t7 
    add  $t6, $t6, 16777216 # In $t6 I've mantissa 

sign: # Sign of result 
    xor  $t0, $t0, $t4  
    beqz $t0, print_res 
    b p_neg_res 

exp: 
    sub  $t5, $t5, 1023 # Taking real exponent 
    sub  $t1, $t1, 127 # Taking real exponent 
    sub  $t1, $t1, $t5  # In $t1 I've the exponent of the final number 
    add  $t1, $t1, 1023 # Adding bias (the final number will be a double) 
    sll  $t1, $t1, 20  # Shifting so when I add $t2 (the first part of mantissa) I have the number in the right position 

divis: 
    div  $t3, $t3, $t6  # First part of division (32 bit of mantissa) 
    div  $t2, $t2, $t6  # Second part of division (20 bit of mantissa) 

     #HUGE QUESTION MARK 
    add $t2, $t1, $t2  # Adding exponent and mantissa, in this way in $t2 I have all the information 
    mtc1.d $t2, $f4   # $t2 & $t3 are moved in $f4 & $f5 so I have a double number in $f4...I Hope 
    b print_dou 

print_res: # Print positive result 
    la $a0, res 
    li $v0, 4 
    syscall 
    b divis 

p_neg_res: # Print negative result 
    la $a0, neg_res 
    li $v0, 4 
    syscall 
    b divis 

err: # Error in the choice 
    la $a0, wrong 
    li $v0, 4 
    syscall 
    b main 

end: # End of program 
    li $v0, 10 
    syscall 

errz: # Error dividing by 0 
    la $a0, divz 
    li $v0, 4 
    syscall 
    b main 

print_dou: # Print the number!! 
    mov.d $f12, $f30 # bylo #f4 
    li $v0, 3 
    syscall 
    b main 

Answer 1

當計算結果$t1的指數，你有正確的想法，但雖然1.0/1.5f和1.5/1.0f各自的操作數相同，則結果的指數相差一個。在第一種情況下，您可以做的是，在減少$t1的同時，將分子左側的有效位移動1。你需要在提取兩個有效位之後設置各自的隱含有效位（顯然，在第一段中建議的調整之前設置先前隱含的位，否則將變得複雜）。

之後，您將實施一個部門。以下說明做不實現53位除法：

div  $t3, $t3, $t6  # First part of division (32 bit of mantissa) 
div  $t2, $t2, $t6  # Second part of division (20 bit of mantissa) 

首先，他們忘了隱位，和第二，這只是不是它是如何工作。

您以1xxxxxxxxx（52或53位）xxxxxxxxxxx和1yyyyy（23位）yyyyy開頭。在第一段中，您確定分子的有效位數大於分母的位數，因此第一位數字計算爲1（它將隱式存儲在最終結果中）。接下來你需要從前者中減去後者。如果你還沒有在步驟1中轉移，減法是這樣的：

 
1xxxxxxxxxxx 
-1yyyyyyy 
_____________ 
zzzzzzzzzzzz 

如果你在步驟1中已經改變，它看起來像這樣：

 
1xxxxxxxxxxx 
- 1yyyyyyy 
_____________ 
zzzzzzzzzzzz 

移由一個離開的結果，並將新值與1yyyyyyyy進行比較。如果新值大於或等於1yyyyyyy，則下一個數字爲1，必須再次減去1yyyyyyy。如果它較低，則下一個數字是0（並且不減）。

在這兩種情況下，再次左移並計算下一個數字。 等等......