在 MIPS 中表示大于 65535 的数字
我正在 MIPS 中工作,并使用超过 65535 的数字,并且出现超出范围的错误。我该如何解决此代码中的问题?
## p2.asm
##
## Andrew Levenson, 2010
## Problem 2 from Project Euler
## In MIPS Assembly, for SPIM
##
## Calculate the sum, s of all
## even valued terms in the
## Fibonacci sequence which
## do not exceed 4,000,000
.text
.globl main
main:
## Registers
ori $t0, $0, 0x0 # $t0 will contain scratch
ori $t1, $0, 0x1 # $t1 will contain initial fib(N-1)
ori $t2, $0, 0x2 # $t2 will contain initial fib(N)
ori $t3, $0, 0x0 # $t3 will be our loop incrementor
ori $t4, $0, 0x0 # $t4 will be our sum
ori $t5, $0, 0x2 # $t5 contains two to test if even
ori $t8, $0, 4000000 # $t8 contains N limit
even_test:
## Test to see if a given number is even
div $t1, $t5 # $t1 / 2
mflo $t6 # $t6 = floor($t1 / 2)
mfhi $t7 # $t7 = $t1 mod 2
bne $t7, $0, inc # if $t7 != 0 then bypass sum
sll $0, $0, $0 # no op
sum:
## Add a given value to the sum
addu $t4, $t4, $t2 # sum = sum + fib(N)
inc:
## Increment fib's via xor swap magic
xor $t1, $t1, $t2 # xor swap magic
xor $t2, $t1, $t2 # xor swap magic
xor $t1, $t1, $t2 # xor swap magic
## Now $t1 = $t2 and $t2 = $t1
## Increment $t2 to next fib
addu $t2, $t1, $t2
## Is $t2 < 4,000,000?
## If so, go to loop
sltu $8, $t2, $t8 # If $t2 < 4,000,000
# then $8 = 1
bne $8, $0, even_test # if $8 == $0 then jump to even_test
sll $0, $0, $0 # no op
print:
li $v0, 0x1 # system call #1 - print int
move $a0, $t4
syscall # execute
li $v0, 0xA # system call #10 - exit
syscall
## End of Program
我该如何解决这个问题?
I am working in MIPS, and using numbers in excess of 65535, and I'm getting an out of range error. How can I work around that in this code?
## p2.asm
##
## Andrew Levenson, 2010
## Problem 2 from Project Euler
## In MIPS Assembly, for SPIM
##
## Calculate the sum, s of all
## even valued terms in the
## Fibonacci sequence which
## do not exceed 4,000,000
.text
.globl main
main:
## Registers
ori $t0, $0, 0x0 # $t0 will contain scratch
ori $t1, $0, 0x1 # $t1 will contain initial fib(N-1)
ori $t2, $0, 0x2 # $t2 will contain initial fib(N)
ori $t3, $0, 0x0 # $t3 will be our loop incrementor
ori $t4, $0, 0x0 # $t4 will be our sum
ori $t5, $0, 0x2 # $t5 contains two to test if even
ori $t8, $0, 4000000 # $t8 contains N limit
even_test:
## Test to see if a given number is even
div $t1, $t5 # $t1 / 2
mflo $t6 # $t6 = floor($t1 / 2)
mfhi $t7 # $t7 = $t1 mod 2
bne $t7, $0, inc # if $t7 != 0 then bypass sum
sll $0, $0, $0 # no op
sum:
## Add a given value to the sum
addu $t4, $t4, $t2 # sum = sum + fib(N)
inc:
## Increment fib's via xor swap magic
xor $t1, $t1, $t2 # xor swap magic
xor $t2, $t1, $t2 # xor swap magic
xor $t1, $t1, $t2 # xor swap magic
## Now $t1 = $t2 and $t2 = $t1
## Increment $t2 to next fib
addu $t2, $t1, $t2
## Is $t2 < 4,000,000?
## If so, go to loop
sltu $8, $t2, $t8 # If $t2 < 4,000,000
# then $8 = 1
bne $8, $0, even_test # if $8 == $0 then jump to even_test
sll $0, $0, $0 # no op
print:
li $v0, 0x1 # system call #1 - print int
move $a0, $t4
syscall # execute
li $v0, 0xA # system call #10 - exit
syscall
## End of Program
How can I fix this?
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(昨天之前我不知道 MIPS 汇编,但我会尝试一下)
带有 0x3D 的 LUI,然后是带有 0x900 的 ORI(4,000,000 是 0x3D0900)?
(I had no idea of MIPS assembly before yesterday, but I'll give a shot)
LUI with 0x3D, followed by ORI with 0x900 (4,000,000 being 0x3D0900)?
我猜这是问题所在?
MIPS 指令只有 16 位常量字段,因此您需要使用更复杂的序列构造大于 65535 的常量,或者从内存加载它们。像这样的东西应该可以工作:
我认为“sll dest,src,count”是在MIPS汇编中左移的方式,但我可能是错的。您还可以使用“li”宏指令,它以某种方式将任何 32 位常量和 finagles 放入寄存器中,如果需要,可以使用多个指令。
I'm guessing this is the problem line?
MIPS instructions have only 16-bit constant fields, so you need to construct constants greater than 65535 with a more complex sequence, or else load them from memory. Something like this should work:
I think "sll dest, src, count" is how you shift left in MIPS assembly, but I could be wrong. You can also use the "li" macro-instruction, which takes any 32-bit constant and finagles that into a register somehow, using more than one instruction if necessary.
我的第一个想法是使用两个寄存器,一个用于高位,一个用于低位。你必须一起跟踪它们,但这正是我想到的。
My first thought would be to use two registers, one for the higher-order bits and one-for the lower-order bits. You'd have to keep track of them together, but that's just what comes to mind for me.