目前,我試圖找出如何代碼中發現在MIPS最低的整數算法以下的功能...遞歸以MIPS
int Min(int[] A, int low, int high)
{ if (low== high) return A[low];
int mid = (low+high)/2;
int min1 = Min(int[] A, low, mid);
int min2 =Min(int[] A, mid +1, high);
if(min1>min2) return min2;
return min1;
}
我收到的問題,因爲我嘗試這個代碼在MIPS。這是我目前的MIPS代碼。用戶最多可以輸入6個存儲在數組中的整數。寄存器$ a0,$ a1和$ a2被用作函數的參數。
- $ A0 = INT []甲
- $ A1 = INT低//指數
- $ A2 = INT高//指數
這裏是遞歸函數...
min:
bne $a1, $a2, recur
mul $t4, $a1, 4
add $a0, $a0, $t4
lw $v1, 0($a0)
jr $ra
# recursion start
recur:
addiu $sp, $sp, -12 #reserve 12 bytes on stack
sw $ra, 0($sp) #push return address
# mid = (low+high)/2 t0 = mid t1= min1 t2=min2 t3 = mid+1
add $t0, $a1, $a2 # t0 = low + high
div $t0, $t0, 2 # t0 = (low+high)/2
# mid1 = min(int[]A,low,mid)
min1:
sw $a2, 4($sp) #push high
addi $t3, $t0, 1 # mid+1
sw $t3, 8($sp) #store mid+1
move $a2, $t0 #change high to mid
jal min #compute
# check
move $t1, $v1 #set up the min1 = return value
# mid2 = min(int[]A,mid+1,high)
min2:
lw $a2, 4($sp) #reload high prior call
lw $a1, 8($sp) #change low to mid+1
jal min #compute
move $t2, $v1 #set as the min2 = return value
confirm:
# return mid2 if mid1 > mid2
bgt $t1, $t2, returnMid2
# else return mid1
move $v1, $t1
j minFinal
returnMid2:
move $v1, $t2
minFinal:
lw $ra, 0($sp)
addiu $sp, $sp, 12 #release stack
jr $ra
問題是我在程序中輸入的任何整數組合,我從來沒有得到最小值,而是數字「543976553」。我一直在查看我的代碼和筆記,但我仍然沒有線索。
爲什麼你這樣做這很複雜?簡單的線性搜索至少是一樣的。 – erikkallen 2010-12-05 13:33:57
您是否嘗試過在MARS模擬器中逐步執行? - > http://courses.missouristate.edu/KenVollmar/MARS/download.htm – digEmAll 2010-12-05 13:43:14