Question

所以我确定我在代码中的某个地方出了问题但是我试图将用户输入程序的整数转换为Hex。我的第一个问题是，我似乎无法让程序打印输入的整数。相反，它似乎打印整数的地址。这是代码：

.data # Data declaration
  # A+B -(C-D x E)

    string1: .asciiz "Please enter an integer from range 0 to 32768: "
    string2: .asciiz "Next integer: "
    string3: .asciiz "Invalid input. Start over: "
    string4: .asciiz "The hexadecimal value of the decimal integer "
    string5: .asciiz "is: "
    newline: .asciiz "\n"
    .align 2
    userinput: .space 100
    var6: .asciiz "The result of A+B -(C-D x E) is: "
.text

main:
    la $a0, string1 #load string one and print
    li $v0, 4
    syscall

    li $s0, 1
    la $a1, userinput 
    li $t1, 5 #set temporary variable to 5
    li $t0, 0 #start of counter

input:  
    beq $t0, $t1, exit
    li $v0, 5 # read integer
    syscall
    blt $v0, $zero, invalid # if input is less than zero
    bgt $v0, 32768, invalid # if input is greater than 32768
    beq $t0, 0, case1
    beq $t0, 1, case2
    beq $t0, 2, case3
    beq $t0, 3, case4
    beq $t0, 4, case5

case1: 
    sw $v0, 0($a1)
    addi $t0, 1
    la $a0, string2
    li $v0, 4
    syscall
    j input
case2: 
    sw $v0, 4($a1)
    addi $t0, 1
    la $a0, string2
    li $v0, 4
    syscall
    j input 
case3: 
    sw $v0, 8($a1)
    addi $t0, 1
    la $a0, string2
    li $v0, 4
    syscall
    j input
case4: 
    sw $v0, 12($a1)
    addi $t0, 1
    la $a0, string2
    li $v0, 4
    syscall
    j input
case5: 
    sw $v0, 16($a1)
    addi $t0, 1
    la $a0, string2
    li $v0, 4
    syscall
    j input
invalid: 
    la $a0, string3 #load string one and print
    li $v0, 4
    syscall
    j input

exit:
    la $t3, userinput # stores base address of user input array into $t3
    li $t1, 1
    li $t0, 0

    la $a0, string4 # load string4
    li $v0, 4 # print
    syscall

    move $a0, $t3 # move first number into a0
    li $v0, 1 # print
    syscall

    la $a0, string5
    li $v0, 4
    syscall

    jal convert

    li $t1, 1
    jal hexconvert

    addi $s0, $s0, 1
    slti $t7, $s0, 6
    bne $t2, $zero, exit

    j close

convert:
    andi $t4, $t3, 15
    sll $s4, $s4, 8
    slti $t5, $t4, 10
    bne $t5, $zero, else
    addi $t4, $t4, 65
    jal next
else:   
    addi $t4, $t4, 48

next:   
    add $s4, $s4, $t3
    srl $t3, $t3, 4
    addi $t1, $t1, 1
    slti $t5, $t1, 5
    bne $t5, $zero, convert
    jr $ra
hexconvert:
    andi $t4, $s4, 255
    move $a0, $t4
    li $v0, 11
    syscall
    srl $s4, $s4, 8
    addi $s3, $s3, 1
    slti $t1, $s3, 5
    bne $t2, $zero, hexconvert
    la $a0, newline
    li $v0, 4
    syscall
    jr $ra

close:  
    li $v0, 10 # exit instruction
    syscall

我希望有人能指出我正确的方向。我对MIPS仍然很陌生。谢谢！

Answer 1

我似乎无法让程序打印输入的整数。相反，它似乎打印整数的地址。

打印地址似乎正是您在这里所做的：

la $t3, userinput # stores base address of user input array into $t3

...

move $a0, $t3 # move first number into a0
li $v0, 1 # print
syscall

move $a0, $t3只是将$t3的值（即userinput的地址）复制到$a0。如果你真正想要的是从$t3所指向的记忆中读取你应该写的lw $a0,($t3)

MIPS打印错误

1 个答案: