Question

嘿伙计我需要打印斐波那契序列的值直到第n个值（用户输入）。因此，如果用户输入3，我将不得不打印1,1,2。我必须以递归和迭代方式进行。我得到了迭代部分，但我对递归部分有点麻烦。在程序中我可以让它打印值。任何帮助将不胜感激。

.data

inputMsg: .asciiz "Enter number: "
iterationMsg: .asciiz "\nIterative: "
recursionMsg: .asciiz "\nRecursive: "
series: .asciiz "\nThe series is: "
space: .asciiz " "

.text

main:
    la $a0, inputMsg        #Print out inputMsg
    li $v0, 4
    syscall

    li $v0, 5           #Read in integer into v0
    syscall

    subi $s0, $v0, 1        #s0 = v0-1
    add $s7, $s0, 0         #s7 = s0
    add $s3, $v0, 0     #s3 = v0-1

    addi $s1, $s1, 1        #s1 = 1
    addi $s2, $s2, 1        #s2 = 1

    la $a0, iterationMsg        #Print out text_iteration
    li $v0, 4
    syscall

    la $a0, series          #Print out series
    syscall

    addi $a0, $zero, 1      #a0 = 1
    li $v0, 1
    syscall             #Prints out 1

    jal iterative           #Jump to iterative

    la $a0, recursionMsg        #Print Out Recursion text
    li $v0, 4
    syscall

    la $a0, series          #Print out Series text
    syscall

    add $a0, $s3, $0        #t0 = s3 = inputted val
    add $s0, $0, $0
    jal recursive

exit:       
    li, $v0, 10         #Exit Program
    syscall

iterative:
    beq $s0, 0, iterativeExit   #if s0 = 0 then exit
    subi $s0, $s0, 1        # s0 = s0 

    add $t0, $s2, $zero     # t0 = s2(1) + 0 
    add $s2, $s1, $s2       # s2 = s1(1) + s2(1)
    add $s1, $t0, $zero     # s1 = t0(1) + 0

    la $a0, space
    li $v0, 4
    syscall

    add $a0, $s1, $zero
    li $v0, 1
    syscall

    bnez $s0, iterative

iterativeExit:
    jr $ra

recursive:
    addi $sp, $sp, -4   #create/save stack

    sw $ra, 0($sp)

    sub $a0, $a0, 1     #minus one from inputted val (a0) till 0
    beq $a0, 1, oneRec
    jal recursive

    jal print

        lw  $ra, 0($sp)         #
        addi $sp, $sp, 4      # decrease the stack size

    jr $ra
oneRec:
    li $a0, 1
    li $v0, 1
    syscall

    addi $s0, $a0, 0

    la $a0, space
    li $v0, 4
    syscall

    jr $ra
print:
    add $t0, $0, $a0
    add $a0, $s0, $a0

    li $v0, 1
    syscall

    la $a0, space
    li $v0, 4
    syscall

    add $s0, $t0, $0
    jr $ra

Answer 1

顶部有一个小错误。你有$ s3 =你说的是$ v0 - 1，但实际上它是= $ v0。并且在递归部分中使用$ s3

使用mips中的递归打印Fibonacci序列

1 个答案: