我很难理解如何在 MIPS 中对数字进行平方，然后将 (32,14) 二进制值转换为 (8,5) 十进制值。由于我需要使用 mult 函数，我知道它涉及到一个高低寄存器，然后我将执行一系列移位位，以允许我 OR 一起给我需要的东西，但除此之外，我很迷茫。下面粘贴的是我到目前为止所拥有的，我可能会或可能不会朝着正确的方向前进。
PS 更多上下文：我这样做是为了在 DE2 板上实现。该板有 18 个开关，每个开关代表位位置（如果您“打开”第一个和第二个开关，您将对值 3 进行平方。据我所知，这意味着我乘以两个 (32,14) 二进制值数字在一起，最终需要以 (8,5) 十进制值结束，该值将显示在板上的 LED 显示屏上。
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# HARDCODED $1 <= .1 (429496730)
# HARDCODED $2 <= 10
# HARDCODED $3 <= 100000
# HARDCODED reg30_in ($30) <= 256 for simulation purposes (value of the switch)
add $28, $30, $zero # initialize $28 = value of input
add $29, $zero, $zero
add $30, $zero, $zero
sll $28, $28, 14 # convert the input value into a (32,14) value
add $4, $zero, $zero # initialize x($4) = 0
srl $5, $28, 1 # initialize step($5) = (32,14)input/2
##sll $5, $5, 14 # shift step to convert to (32,14) -- not needed if we already shifted $28??
sqrt: # loop for square root algorithm
mult $4, $4 # {hi,lo} = x^2 in (64,28)
mfhi $6 # move hi part to register 6
srl $6, $6, 18 # shift hi for (32,14) format
mflo $7 # move lo to register 7
sll $7, $7, 14 # shift lo for (32,14) format
or $8, $6, $7 # combine the hi and lo into a converted (32,14) value
sub $9, $8, $28 # val = x^2 - S(input)
bgez $9, gtz # if val >= 0, branch to gtz
add $4, $4, $5 # else x = x + step
srl $5, $5, 1 # step = step/2
bgez $5, sqrt # if step >= 0, go back into loop
j BCD # else continue to BCD for output
gtz: # greater than zero branch
sub $4, $4, $5 # x = x - step
srl $5, $5, 1 # step = step/2
bgez $5, sqrt # if step >= 0, go back into loop
j BCD # else continue to BCD for output
BCD: # function for BCD output to HEX
mult $3, $4 # multiply x value ($4) by 100000
mfhi $5
mflo $6
or $4, $5, $6
srl $4, $4, 13
#Multiply value by 10^5 (100000)
#Shift 13 bits to the right
#If bit 0 is set then add 2
#Shift one bit to the right
# HEX0
mult $4, $1 # {hi,lo} = val*pt_one
mfhi $3 # move hi (whole part) to register 3
mflo $4 # move lo (fractional part) to register 4
multu $4, $2 # {hi,lo} = lo*10
mfhi $5 # $5 (digit) = hi
sll $5, $5, 0 # shift by appropriate amount for digit placement
or $29, $29, $5
# HEX1
mult $3, $1 # {hi,lo} = remaining_val*pt_one
mfhi $3 # move hi (whole part) to register 4
mflo $4 # move lo (fractional part) to register 5
multu $4, $2 # {hi,lo} = lo*10
mfhi $5 # $6 (digit) = hi
sll $5, $5, 4 # shift by appropriate amount for digit placement
or $29, $29, $5
# HEX2
mult $3, $1 # {hi,lo} = remaining_val*pt_one
mfhi $3 # move hi (whole part) to register 4
mflo $4 # move lo (fractional part) to register 5
multu $4, $2 # {hi,lo} = lo*10
mfhi $5 # $6 (digit) = hi
sll $5, $5, 8 # shift by appropriate amount for digit placement
or $29, $29, $5
# HEX3
mult $3, $1 # {hi,lo} = remaining_val*pt_one
mfhi $3 # move hi (whole part) to register 4
mflo $4 # move lo (fractional part) to register 5
multu $4, $2 # {hi,lo} = lo*10
mfhi $5 # $6 (digit) = hi
sll $5, $5, 12 # shift by appropriate amount for digit placement
or $29, $29, $5
# HEX4
mult $3, $1 # {hi,lo} = remaining_val*pt_one
mfhi $3 # move hi (whole part) to register 4
mflo $4 # move lo (fractional part) to register 5
multu $4, $2 # {hi,lo} = lo*10
mfhi $5 # $6 (digit) = hi
sll $5, $5, 16 # shift by appropriate amount for digit placement
or $29, $29, $5
# HEX5
mult $3, $1 # {hi,lo} = remaining_val*pt_one
mfhi $3 # move hi (whole part) to register 4
mflo $4 # move lo (fractional part) to register 5
multu $4, $2 # {hi,lo} = lo*10
mfhi $5 # $6 (digit) = hi
sll $5, $5, 20 # shift by appropriate amount for digit placement
or $29, $29, $5
# HEX6
mult $3, $1 # {hi,lo} = remaining_val*pt_one
mfhi $3 # move hi (whole part) to register 4
mflo $4 # move lo (fractional part) to register 5
multu $4, $2 # {hi,lo} = lo*10
mfhi $5 # $6 (digit) = hi
sll $5, $5, 24 # shift by appropriate amount for digit placement
or $29, $29, $5
# HEX7
mult $3, $1 # {hi,lo} = remaining_val*pt_one
mfhi $3 # move hi (whole part) to register 4
mflo $4 # move lo (fractional part) to register 5
multu $4, $2 # {hi,lo} = lo*10
mfhi $5 # $6 (digit) = hi
sll $5, $5, 28 # shift by appropriate amount for digit placement
or $30, $29, $5