Question

我需要在MIPS组装方面提供一些认真的帮助。这个程序应该做两件事。这两件事是：

计算C（n，k）= n！/（（n-k）！k！）

输入：spim -f 90.s 4 6

输出：
6 4
720 24
15

2.显示Pascal三角形的行n，其中k = 0到n
输入：与之前相同（spim -f 90.s 4 6）

输出： 6 4 720 24 15 1 6 15 20 15 6 1 （这是最终输入应该是什么。）

我基本上需要编写一个函数displayRow，它显示Pascal三角形的行n，其中k = 0到n。忽略传入的k的值。由于displayRow函数调用Cnk，我必须构造一个堆栈帧。在displayRow中，我需要编写一个从k = 0到n的循环。在循环体内显示Cnk（n，k）的返回值。我卡住了，我的输出看起来像这样：

spim :(解析器）解析文件90.s的第174行的错误 bgtz $ v0 以下符号未定义：的atoi

6 4 720 24 24

我不知道从哪里开始。一些认真的帮助将不胜感激！这是我的代码......

     # Display Row n of Pascal's Triangle for k=0 to n
     # usage: spim -f 90.s 4 6

    .data
    newline: .asciiz "\n"
    space: .asciiz " "

    .text
    .globl main

    main:
    # hard coded
    li $t0 6
    li $t1 4

    # grab command line stuff - a0 is arg count and a1 points to list of args
    move $s0 $a0
    move $s1 $a1

    # zero out these registers just to be safe
    move $s2 $zero
    move $s3 $zero
    move $s4 $zero

    # check if less than three arguments are given
    li $t3 2
    blt $a0 $t3 default
    bgt $a0 $t3 default

    # parse the first number
    lw $a0 4($s1)
    jal atoi
    move $s2 $v0

    # parse the second number
    lw $a0 8($s1)
    jal atoi
    move $s3 $v0

    # load a0 and a1 with two integers
    move $a0 $s2
    move $a1 $s3

    move $t0, $a0
    move $t1, $a1


    # swap if n < k
    slt $t3 $t0 $t1
    beq $t3 $0 done
    move $t2 $t0
    move $t0 $t1
    move $t1 $t2

    done:
    # move the result from t0 to v0 to print it
    move $a0 $t0
    li $v0 1
    syscall

    la $a0 space
    li $v0 4
    syscall

    move $a0 $t1
    li $v0 1
    syscall

    la $a0 newline
    li $v0 4
    syscall

    move $t8 $t0
    jal factorial

    move $a0 $t9
    li $v0 1
    syscall
    move $s2 $t9

    la $a0 space
    li $v0 4
    syscall

    move $t8 $t1
    jal factorial

    move $a0 $t9
    li $v0 1
    syscall
    move $s3 $t9

    la $a0 newline
    li $v0 4
    syscall

    move $a0 $s3
    li $v0 1
    syscall

    la $a0 newline
    li $v0 4
    syscall

    li $v0 10  # 10=exit
    syscall

    default:
    move $a0 $t0
    move $a1 $t1
    j done

    pascal:
    addi $sp, $sp, -16             # create stack frame
    sw $ra, 12($sp)                # save return address
    sw $s0, 8($sp)                 # save three registers
    sw $s1, 4($sp)
    sw $s2, 0($sp)

    beq $a1, $zero, one            # branch if col == 0
    beq $a1, $a0, one              # branch if col == row

    add $s0, $a0, $zero            # save argument values
    add $s1, $a1, $zero
    addi $a0, $a0, -1              # compute args for first recursion
    addi $a1, $a1, -1
    jal pascal                     # pascal(row-1, col-1)
    add $s2, $v0, $zero            # save returned value

    addi $a0, $s0, -1              # compute args for second recursion
    add $a1, $s1, $zero
    jal pascal                     # pascal(row-1, col)
    add $v0, $v0, $s2              # add the two values

    #done:
    lw $ra, 12($sp)                # restore registers
    lw $s0, 8($sp)
    lw $s1, 4($sp)
    lw $s2, 0($sp)
    addi $sp, $sp, 16
    jr $ra                         # return

    one:
    addi $v0, $zero, 1 # return constant 1
    j done

    #---------------------Factorial
    factorial:
    li $t9 1

    fac:
    beqz $t8 return
    mul $t9 $t9 $t8
    addi $t8 $t8 -1
    j fac

    return:
    move $v0 $t8
    jr $ra

    #---------------
    Cnk:
    addi $sp, $sp, -32  # Stack frame is 32 bytes long
    sw   $ra, 20($sp)   # Save return address
    sw   $fp, 16($sp)   # Save frame pointer
    addi $fp, $sp, 28   # Set up frame pointer
    sw   $a0, 0($fp)    # Save argument (n)

    lw $v0 0($fp)
    bgtz $v0
    # --------- ATOI FUNCTION
    atoi:
    move $v0, $zero

    # detect sign
    li $t0, 1
    lbu $t1, 0($a0)
    bne $t1, 45, digit
    li $t0, -1
    addu $a0, $a0, 1

    digit:
    # read character
    lbu $t1, 0($a0)

    # finish when non-digit encountered
    bltu $t1, 48, finish
    bgtu $t1, 57, finish

    # translate character into digit
    subu $t1, $t1, 48

    # multiply the accumulator by ten
    li $t2, 10
    mult $v0, $t2
    mflo $v0

    # add digit to the accumulator
    add $v0, $v0, $t1

    # next character
    addu $a0, $a0, 1
    b digit

    finish:
    mult $v0, $t0
    mflo $v0
    jr $ra
    #----------------------------------------

Answer 1

答案在错误消息中：

spim: (parser) parse error on line 174 of file 90.s 
          bgtz $v0
                  ^ 
The following symbols are undefined: 
atoi

btgz需要一个目标而你没有给出任何目标。弄清楚你希望它分支到哪里并分配目标。

MIPS中Pascal的三角形

1 个答案: