大端至小端采用MIPS没有逻辑操作？ [英] Big Endian to Little Endian using MIPS without logical operations?

问题描述

我使用MIPS（QtSpim）隐蔽从大端至小端32位字。我下面展示检查无误。不过，我想知道还有什么其他的方法可以让我做转换。我虽然只用旋转和变化，但我没能做到这一点没有逻辑运算。

所以我的问题是，它可以没有逻辑运算来完成？

 转换李$＃t0,0x12345678数量应该在$ T0
ROL $ T1，$ t0,8
李$ t2,0x00FF00FF＃$ t2包含面罩0x00FF00FF
及$ T3，$ T1，$ T2＃字节0和2有效
ROR $ T1，$ t0,8
不是$ T2，T2 $＃$ t2包含面罩0xFF00FF00
及$ T1，$ T1，$ T2＃字节1和3的有效
或$ T3，T3 $，$＃T1在小T3 $尾数号
 

解决方案

下面去，不使用逻辑运算符的解决方案。然而，这仅仅是一个黑客：

 转换李$＃t0,0x12345678数量应该在$ T0  SWL $ T0，划痕+ 3
  在全方位学习$ T1，从头＃负载MSB LSB
  LWR $ T1，划痕+ 3＃加载LSB在MSB
  SWL $ T0，划痕+ 2
  LWR $ T2，从头＃交换第二和
  LWL $ T2，划痕+ 1＃第三个字节  的sw $零，从头开始
  LWL $ T2，从头＃休假MSB和LSB零
  LWR $ T2，划痕+ 3
  阿杜$ T3，T1 $，$＃T2添加部分结果得到最终结果 。数据为0x2000＃在哪里可以找到暂存空间（4字节）
刮：
 。空间4
 

输入为 $ T0 ，部分结果是 $ T1 和 $ T2 和最终的结果是 $ T3 。它还使用4个字节的内存（位于从头）

I am using MIPS (QtSpim) to covert a 32bit word from Big Endian to Little Endian. What I show below is checked and correct. However I would like to know what other ways will let me do the conversion. I though only by using rotates and shifts, but I didn't manage to do it without logical operations.

So my question is, can it be done without logical operations?

li $t0,0x12345678     # number to be converted supposed to be in $t0
rol $t1,$t0,8         
li $t2,0x00FF00FF     # $t2 contains mask 0x00FF00FF
and $t3,$t1,$t2     # byte 0 and 2 valid
ror $t1,$t0,8       
not $t2,$t2        # $t2 contains mask 0xFF00FF00
and $t1,$t1,$t2     # byte 1 and 3 valid
or $t3,$t3,$t1      # little endian-number in $t3

解决方案

Here goes a solution that does not use logical operators. However, it is just a hack:

  li $t0,0x12345678   # number to be converted supposed to be in $t0

  swl $t0, scratch+3
  lwl $t1, scratch    # Load MSB in LSB
  lwr $t1, scratch+3  # Load LSB in MSB


  swl $t0, scratch+2
  lwr $t2, scratch    # Swap second and
  lwl $t2, scratch+1  # third bytes

  sw $zero, scratch 
  lwl $t2, scratch    # Leave MSB and LSB in zero
  lwr $t2, scratch+3
  addu $t3, $t1, $t2  # Add partial results to get final result

 .data 0x2000  # Where to locate scratch space (4 bytes)
scratch:
 .space 4

Input is $t0, partial results are in $t1 and $t2 and final result is in $t3. It also uses 4 bytes of memory (located at scratch)

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