连续算术溢出

Question

我已经尝试更改此代码的大部分内容，但我不断收到无休止的算术溢出警告或“数据/堆栈中的错误地址”错误。我似乎无法正确理解它！欢迎任何见解！我正在尝试实现一个数独板，这些是该函数的说明：给定一个指向数独板的指针，该函数应该返回一个整数，指示板上有多少个单元格已被解决。董事会的状态不应被修改。每个数独单元总共占用十个字节。每个cell占用10字节，整块板占用810字节内存。

有什么见解吗？

# a0: pointer to board
# v0: number of solved cells

count_solved_cells:
    li $v0, 0               # Initialize count to 0
    la $s0, board           # Load the address of the board to $s0
    li $t0, 81              # Initialize loop counter

count_cells_loop:
    jal is_cell_solved      # Check if the current cell is solved
    addi $v0, $v0, 1        # Increment count regardless of cell status

    addi $s0, $s0, 10        # Move to the next cell (each cell is 10 bytes)
    addi $t0, $t0, -1        # Decrement loop counter
    bnez $t0, count_cells_loop  # If not reached the end of the board, continue the loop

    jr $ra

Answer 1

循环中似乎存在问题，无论单元格状态如何，您都会增加计数。仅当单元格已解决时才应增加计数。另外，您需要正确处理 is_cell_solved 函数的返回值。

这是代码的修改版本：

# a0: pointer to board
# v0: number of solved cells

count_solved_cells:
    li $v0, 0               # Initialize count to 0
    la $s0, board           # Load the address of the board to $s0
    li $t0, 81              # Initialize loop counter

count_cells_loop:
    jal is_cell_solved      # Check if the current cell is solved
    beq $v0, $zero, not_solved  # If the cell is not solved, jump to not_solved

    addi $v0, $v0, 1        # Increment count only if the cell is solved

not_solved:
    addi $s0, $s0, 10        # Move to the next cell (each cell is 10 bytes)
    addi $t0, $t0, -1        # Decrement loop counter
    bnez $t0, count_cells_loop  # If not reached the end of the board, continue the loop

    jr $ra

此修改包括一个分支（not_solved），用于在单元格未解决时跳过计数的增量。确保 is_cell_solved 函数正确返回一个值，指示单元格是否已求解。