Question

我试图创建一个函数，将数字排列在正方形网格中，其中每行，每列中的数字以及前向和后向主对角线中的数字都加起来相同的数字。虽然我对mips相对较新，但这是我最近的尝试：

square:  lw $t0,$zero #i = 0
     lw $t2,1 # k = 1
     div $t1,$a1,2 # j = n/2
     mul $t3,$a1,$a1 #  n*n

while:  ble $t2,$t3,else1 # k <= n*n
    bgt $t0,-1,else2 #
    bgt $t1,-1,else2 #
    ble $t1,$a1,else3 #
    ble $t0,$a1,else4 #
    bne $,$,else5#a[i,j] == 0

else1:  #a[i,j] = k
    addi $t0,$t0,-1 #
    addi $t1,$t1,1 #
    addi $t2,$t2,1 #

else2:  lw $t0,1 #
    addi $t1,$a1,-1 #

else3:  lw $t1,$zero #


else4:  addi $t0,$a1,-1 #

else5:  addi $t0,$t0,2 #
    addi $t1,$t1,-1 #

end:    jr $ra

另外这是寄存器的结构：

$ a0 =数组的基地址（矩阵），

$ a1 = n，矩阵的大小（行数和列数）

$ t0 = i

$ t1 = j

$ t2 = k

$ t3 = n * n

$ t4 = arraya index - 分步

$ t5 =数组值

我遇到的麻烦是创建[i，j]它是如何完成的？这里是一个算法：

 i = 0, k = 1 and j = n/2
 while (k <= n*n)
   if (i > -1 and j > -1 and j < n and i < n and a[i,j] == 0)
     a[i,j] = k
     i = i - 1, j = j +1 and k = k + 1
   else if (i < 0 and j == n) move out of upper right square
     i = 1 and j = n - 1
   else if (j == n)           move out of right side of square
     j = 0
   else if (i < 0)            move above top row
     i = n - 1
   else                       move to an already filled square
     i = i + 2 and j = j - 1
   end if - else
 end while loop

Answer 1

您可以通过地址计算连续存储2D数组：a[i, j]存储在base address + i + j * n。

为了优化，行/列遍历以+/-1或+/-n为增量。

或者，您可以为行分配不同的区域，并提供行开始表。

Mips函数数组

1 个答案: