2014-12-01 96 views
1

嗨,我需要打印斐波那契數列的值直到第n個值(用戶輸入)。所以,如果用戶輸入3,我將不得不打印1,1,2。我必須遞歸地和迭代地執行它。我得到了迭代的一部分,但我在遞歸部分遇到了一些麻煩。在程序中,我可以打印它的值。任何幫助,將不勝感激。使用mips遞歸打印Fibonacci序列

.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 

回答

1

上面有一個小錯誤。你有$ s3 =你說的是$ v0 - 1,但你實際上有= = $ v0。你在遞歸部分使用$ s3