在MIPS中,我对如何让mod工作感到困惑。下面是我到目前为止提出的代码。我可能除了mod之外还有更多的错误,但我觉得这些错误是mod误解的结果。我要做的就是在这里获取工作代码(python):
i = 1
k = 0
while i < 9:
if i % 2 != 0:
k = k + i
i += 1
print(k)
正确翻译成MIPS。这是我在汇编时的第一次拍摄,所以在下面的代码中可能有多个mod错误让我感到沮丧:
# Takes the odd integers from 1 to 9, adds them,
# and spits out the result.
# main/driver starts here
.globl main
main:
#data segment
.data
Li: .byte 0x01 # i = 1
Lj: .byte 0x09 # j = 9
Lk: .byte 0x00 # k = 0
Ltwo: .byte 0x02 # 2 for mod usage
# text segment
.text
lb $t0, Li # temp reg for i
lb $t1, Lj # j
lb $t2, Lk # k
lb $t3, Ltwo # 2
L1: beq $t0, $t1, L2 # while i < 9, compute
div $t0, $t3 # i mod 2
mfhi $t6 # temp for the mod
beq $t6, 0, Lmod # if mod == 0, jump over to L1
add $t2, $t2, $t0 # k = k + i
Lmod: add $t0, $t0, 1 # i++
j L1 # repeat the while loop
L2: li $v0, 1 # system call code to print integer
lb $a0, Lk # address of int to print
syscall
li $v0, 10
syscall
答案 0 :(得分:4)
您正在以十六进制查看SPIM寄存器。十六进制10是十进制16。
答案 1 :(得分:3)
完成扭结后,下面的代码就像一个魅力。要在MIPS中正确使用mod运算符,必须使用HI和LO。我需要i%2 == 0作为声明,所以mfhi派上用场了。参考下面的工作结果代码:
# Takes the odd integers from 1 to 9, adds them,
# and spits out the result.
# main/driver starts here
.globl main
main:
#data segment
.data
Li: .byte 0x01 # i = 1
Lj: .byte 0x0A # j = 10
Lk: .byte 0x00 # k = 0
Ltwo: .byte 0x02 # 2 for mod usage
# text segment
.text
lb $t0, Li # temp reg for i
lb $t1, Lj # j
lb $t2, Lk # k
lb $t3, Ltwo # 2
L1: beq $t0, $t1, L2 # while i < 9, compute
div $t0, $t3 # i mod 2
mfhi $t6 # temp for the mod
beq $t6, 0, Lmod # if mod == 0, jump over to Lmod and increment
add $t2, $t2, $t0 # k = k + i
Lmod: add $t0, $t0, 1 # i++
j L1 # repeat the while loop
L2: li $v0, 1 # system call code to print integer
move $a0, $t2 # move integer to be printed into $a0
syscall
li $v0, 10 # close the program
syscall