Question

问题是：请写一个循环来计算当前存储在$ t0中的正数的阶乘，并将结果存储在4个指令中的$ t1中。

这是我到目前为止我很确定它有效，但它有6条指令。

       li $t3, 1
       move $t1, $t0
       move $t2, $t0
 LOOP: addi $t2, $t2, -1
       mul $t1, $t1, $t2
       bne $t2, $t3, LOOP

编辑。这是解决方案

 li $t1 1 
 LOOP: mul $t1 $t1 $t0 
 addi $t0 $t0 -1 
 bgez $t0 LOOP

Answer 1

 li $t1 1 
 LOOP: mul $t1 $t1 $t0 
 addi $t0 $t0 -1 
 bgez $t0 LOOP

Answer 2

const arrayToSort = [
  {
    Country: "Cyprus",
    Date: new Date(2001, 0, 1),
    CreateBy: "William",
  },
  {
    Country: "Belarus",
    Date: new Date(1999, 0, 1),
    CreateBy: "Yuliana",

  },
  {
    Country: "Denmark",
    Date: new Date(2019, 0, 1),
    CreateBy: "Ava",
  },
  {
    Country: "Albania",
    Date: new Date(2000, 0, 1),
    CreateBy: "Zachary",
  }
];

const sortFields = ["Country", "CreateBy", "Date"];
const descending = [true, false, true];

const sortBy = (a,b,field,desc) => {
  if(a[field]<b[field]) return desc ? 1 : -1
  else if(a[field]>b[field]) return desc ? -1 : 1
  else return 0;
};

const result = arrayToSort.sort( (a,b) => {
  for(var i=0;i<sortFields.length;i++){
    var res = sortBy(a,b, sortFields[i], descending[i]);
    if(res != 0) return res;    
  }
  return 0;
})

console.log(result);

Answer 3

在 qtspim 上运行良好。每行都有小注释，以便更好地理解。

请记住，当调用一个函数时，我们应该将它临时存储在堆栈中，这样即使函数更改了寄存器值，我们仍然可以保留以前主函数中的数据。

    .data
    Enter: .asciiz "Enter value of n: \n"

    .text
    .globl main

main:
    lui $s0, 0x1001

    li $v0, 4                   # print message for i/p
    la $a0, Enter
    syscall

    li $v0, 5                   # get n
    syscall

    add $s1, $v0, $zero         # init n = 5
    li $s2, 1                   # s2 = 1 for base case
    
    add $a0, $s1, $zero         # pass fib(a0)
    jal factorial

    li $v0, 1                   # print result
    add $a0, $v1, $zero
    syscall

    li $v0, 10                  # exit
    syscall

factorial:
    addi $sp, $sp, -8           # grow sp below
    sw $a0, 0($sp)              # store arguments and return addr
    sw $ra, 4($sp)

    bne $a0, $s2, recursion     # n != 1
    add $v1, $s2, $zero
    addi $sp, $sp, 8            # restore $sp and remove stack frame
    jr $ra

    recursion: 
        addi $a0, $a0, -1       # fib(a0 - 1)
        jal factorial

        lw $a0, 0($sp)          # load back a0 to use as a0 * fib(a0-1)
        lw $ra, 4($sp)
        addi $sp, $sp, 8        
        mul $v1, $a0, $v1 

        jr $ra

你如何编写代码来完成MIPS中的阶乘？

3 个答案: