我正在尝试完成这个MIPS计算器，超级基本，我的第一个mips程序。它不必处理溢出或类似的东西，只需处理小的正数。
我没有检查我的算法是否有乘法和除法，因为我只是想加入工作。
我不能为我的生活找出为什么int不会读入而且当我调用lb $ a0时，我得到的内存超出界限，op显示运算符输出并且不明白为什么
我是新手，所以任何事情都可能有所帮助。提前谢谢。
.data
# const string for welcome
welc: .asciiz "Welcome to SPIM Calculator 1.0!\n"
p_int: .asciiz "\nPlease give an integer: "
p_op: .asciiz "\nPlease give an operator: "
i_err: .asciiz "\nInput Incorrect, bad operator!\n"
again_str: .asciiz "Another calculation? (y/n)"
rmndr: .asciiz " r: "
new_line: .asciiz "\n"
int1: .word 1 # space to hold int 1
int2: .word 1 # space to hold int 2
raw_in: .space 1 # space to hold raw input
op: .space 1 # space to hold operator char
a_char: .space 1 # space to hold again char
out: .word 1 # space to hold output
remain: .word 1 # space to hold remainder
#operator constants
c_plus: .ascii "+" # const for +
c_min: .asciiz "-" # const for -
c_mult: .asciiz "*" # const for *
c_divi: .asciiz "/" # const for /
c_eq: .asciiz "=" # const for =
c_no: .asciiz "n" # const for n
.text
.globl main
main: li $v0, 4 # syscall 4, print string
la $a0, welc # give argument: string
syscall # actually print string
calc: la $t6, remain # load remainder variable
move $t6, $zero # store 0 in remainder (reset)
li $v0, 4 # syscall 4, print string
la $a0, p_int # give argument: string
syscall # actually print string
li $v0, 5 # tell syscall we want to read int 1
syscall # actually read in int 1
la $s1, int1 # load int1 into $s1
move $s1, $v0 # copy the integer from $v0 to int1
li $v0, 4 # syscall 4, print string
la $a0, p_int # give argument: string
syscall # actually print string
li $v0, 5 # tell syscall we want to read int 2
syscall # actually read in int 2
la $s2, int2 # give $s2 the address to hold int 2
move $s2, $v0 # copy the integer from $v0 to $s2
li $v0, 4 # syscall 4, print string
la $a0, p_op # give argument: string
syscall # actually print string
li $v0, 8 # tell syscall we want to read operator
la $a0, op # give $a0 the address to hold the operator
syscall # actually read in operator
lb $t0, op # load the first byte of op
li $t1, '+' # load const for plus
li $t2, '-' # load const for minus
li $t3, '*' # load const for multiplying
li $t4, '/' # load const for dividing
la $s0, out # load out to $s0
beq $t0, $t1, plus # we're adding
beq $t0, $t2, minus # we're subtracting
beq $t0, $t3, multi # we're multiplying
beq $t0, $t4, divi # we're dividing
# else
j error # incorrect input
plus: add $s0, $s1, $s2 # add our ints, store in $t0
j print
minus: sub $s0, $s1, $s2 # subtract our ints, store in $t0
j print
multi: slt $t1, $t2, $s2 # if our counter is less than int2, set $t1 to 1
beq $t1, $zero, print # if we've reached int2, we're done
add $s0, $s1, $s1 # add int1 and int1, store in out
j multi # loop
divi: la $t0 remain # load remainder into $t0
move $t0, $s1 # set remainder to dividend
add $s0, $zero, $zero # set out to 0, just in case
loop: slt $t1, $t0, $s2 # if remainder is less than divisor, set 1
beq $t1, $zero, print # if we're done branch to done
sub $t0, $t0, $s2 # sub divisor from remainder, store in remainder
addi $s0, $s0, 1 # increment quotient by 1
j loop # loop
print: li $v0, 1 # tell syscall we want to print int
la $a0, int1 # give syscall int1 to print
syscall # actually print int
li $v0, 4 # tell syscall we want to print string
lb $a0, op # tell syscall we want to print operator
syscall # actually print string
li $v0, 1 # tell syscall we want to print int
la $a0, int2 # give syscall int2 to print
syscall # actually print int
li $v0, 4 # tell syscall we want to print string
la $a0, c_eq # tell syscall we want to print operator
syscall # actually print string
li $v0, 1 # tell syscall we want to print integer
la $a0, out # give syscall our output
syscall # actually print int
la $t0, remain # load remainder
beq $t0, $zero, again # if we have no remainder, finish printing
li $v0, 4 # tell syscall we want to print string
la $a0, rmndr # tell syscall we want to print remainder string
syscall # print "r: "
li $v0, 1 # tell syscall we want to print int
la $a0, remain # give syscall our remainder to print
syscall # print remainder
again: li $v0, 4 # tell syscall we want to print string
la $a0, new_line # tell syscall to print new line
syscall
la $a0, again_str # load prompt for again string for syscall
syscall
li $v0, 8 # tell syscall we want to read string
la $a0, a_char # tell syscall to put it in $a0
syscall
lb $t0, a_char
li $t1, 'n' # get n char so we can compare
beq $t0, $t1, exit # if we are done, exit
#else loop
j calc # jump to beginning
error: li $v0, 4 # tell syscall we want to print string
la $a0, i_err # give syscall what to print
syscall # actually print
j again # go to prompt for retry
exit: li $v0, 10 # exit code
syscall #exit!
答案 0 :(得分：3)
问题是您没有使用适当的指令来处理内存。
您应使用move（商店字）代替sw。此代码不会将int存储到int1：
la $s1, int1 # load int1 into $s1
move $s1, $v0 # copy the integer from $v0 to int1
相反，你应该写：
la $s1, int1 # load address of int1 into $s1
sw $v0, 0($s1) # copy the integer from $v0 to int1
与存储一样，从内存加载需要两条指令：
la $s1, p_op # or whatever register you choose to use
lb $a0, 0($s1) # load byte from the address stored in $s0 (in index 0)
如果您要将p_op的地址加载到$a0，则应使用la $a0, p_op，而不是lb