如何处理存储在 MIPS 程序集堆栈指针中的三个或更多值？

Question

该程序旨在允许用户使用破折号（“-”和“|”）形成框，连接 3x3 网格的顶点。

因此，如果您愿意，可以将此称为初级汇编程序员问题，但我在将三个或更多值存储在堆栈指针上时遇到了问题。我想通过从堆栈指针中弹出 return 地址，从 loop1 return 到 main；但是，在 checkMove 中，堆栈指针被添加到两次。

如何解决我的问题，以便我可以从 loop1 return 进入 main？

    .data #Data declaration segment
board: .ascii   "\n\n   . . .          . 0 . 1 ."
       .ascii     "\n                  2   3   4"
       .ascii     "\n   . . .          . 5 . 6 ."
       .ascii     "\n                  7   8   9"
       .asciiz    "\n   . . .          . a . b .\n"
offset: .byte     6,   8,  33,  35,  37,  62,  64,  89,  91,  93,  118, 120
marker: .byte    '-', '-', '|', '|', '|', '-', '-', '|', '|', '|', '-', '-'

    .text #Text declaration statement

main:
    li $t0, 6 #Number of loops/moves to make
    jal loop1
    
    #Print game board
    li $v0, 4
    la $a0, board
    syscall
    
    #Exit Program
    li $v0, 10
    syscall
    
loop1:
    #Load return address onto stack pointer
    subu $sp, $sp, 4 
    sw $ra, ($sp) #Store $ra into $sp
    
    #Read integer for marking excluding a and b
    li $v0, 5
    syscall
    
    move $a0, $v0 #Set $v0 to $a0
    jal markMove #Jump to markMove; return address is set in $ra
    sub $t0, $t0, 1 #Decrement counter by 1
    bnez $t0, loop1 #Return to L1 if counter not 0
    
    #Once counter is 0, pop return address off stack pointer
    lw $ra, ($sp)
    addu $sp, $sp, 4
    
    jr $ra
    
#Mark a move on the game board
#Input : $a0 (Integer move 0-9)
markMove:
    ##Prepare to pass necessary variables ($v0, $ra) into function.
    
    #Push return address ($ra) onto stack pointer
    subu $sp, $sp, 4 
    sw $ra, ($sp) #Store $ra into $sp
    #Push $t0 onto stack pointer
    subu $sp, $sp, 4

    sw $t0, ($sp) #Store $t0 into $sp
    
    lb $t0, offset($a0)
    
    #Actually mark now
    lb $t1, marker($a0) #Find the marker
    sb $t1, board($t0) #Place the marker at spot in board

#Clear stack pointer ($sp) and return to $ra
loop2:
    ##Prepare to take the variables ($v0, $ra) from above.
    lw $t0, ($sp) #Pop $t0 off of stack pointer
    addu $sp, $sp, 4
    lw $ra, ($sp) #Pop $t0 off of stack pointer
    addu $sp, $sp, 4
    
    jr $ra #Return to return address; jump register

Answer 1

你的问题是你在每次从用户那里读取数字的迭代中推动 $ra（根据你评论的代码，6 loops/moves）。

每次输入新的 function/procedure 时，您只应存储一次 return 地址，并在从中 return 之前恢复它。

因此，在保存上下文并在控制循环中使用它之后，您将添加另一个标签。在这个例子中我使用了 inner_loop:

...
loop1:
    #Load return address onto stack pointer
    subu $sp, $sp, 4 
    sw $ra, ($sp) #Store $ra into $sp
    
inner_loop:
    #Read integer for marking excluding a and b
    li $v0, 5
    syscall
    
    move $a0, $v0 #Set $v0 to $a0
    jal markMove #Jump to markMove; return address is set in $ra
    sub $t0, $t0, 1 #Decrement counter by 1
    bnez $t0, inner_loop #Return to L1 if counter not 0
    
    #Once counter is 0, pop return address off stack pointer
...

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