计算结果和多路复用与否(compute results and mux or not)

 在这里使用伪代码。 这些风格是否有利有弊： 
 
 假设你有一个可以添加的alu，或者和xor。 是否有更好的代码可以一直计算可能的答案，然后根据操作码选择答案（在这种情况下是一个热门）： 
 
alu_add = a + b;
alu_and = a & b;
alu_or  = a | b;
alu_xor = a ^ b;

...

if(opcode[0])      alu_out = alu_add;
else if(opcode[1]) alu_out = alu_and;
else if(opcode[2]) alu_out = alu_or;
else if(opcode[3]) alu_out = alu_xor;
 
 另一种方法是像这样编码它： 
 
if(opcode[0])      alu_out = a + b;
else if(opcode[1]) alu_out = a & b;
else if(opcode[2]) alu_out = a | b;
else if(opcode[3]) alu_out = a ^ b;
 
 我也看到它： 
 
alu_add = a + b;
alu_and = a & b;
alu_or  = a | b;
alu_xor = a ^ b;

...

alu_out = 
  ( 8{opcode[0]} & alu_add ) |
  ( 8{opcode[1]} & alu_and ) | 
  ( 8{opcode[2]} & alu_or ) |
  ( 8{opcode[3]} & alu_xor );
 
 这两种方法都有利弊，或者它们最终会出现同样的问题吗？ 

Using pseudo code here. Are there pros and cons to these styles:
 
Say you have an alu that can do add, and, or and xor. Is it better to have code that computes the possible answers all the time then select the answer based on the opcode (in this case a one hot):
 
alu_add = a + b;
alu_and = a & b;
alu_or  = a | b;
alu_xor = a ^ b;

...

if(opcode[0])      alu_out = alu_add;
else if(opcode[1]) alu_out = alu_and;
else if(opcode[2]) alu_out = alu_or;
else if(opcode[3]) alu_out = alu_xor;
 
An alternative would be to code it like this:
 
if(opcode[0])      alu_out = a + b;
else if(opcode[1]) alu_out = a & b;
else if(opcode[2]) alu_out = a | b;
else if(opcode[3]) alu_out = a ^ b;
 
I have also seen it as:
 
alu_add = a + b;
alu_and = a & b;
alu_or  = a | b;
alu_xor = a ^ b;

...

alu_out = 
  ( 8{opcode[0]} & alu_add ) |
  ( 8{opcode[1]} & alu_and ) | 
  ( 8{opcode[2]} & alu_or ) |
  ( 8{opcode[3]} & alu_xor );
 
Are there pros and cons to either method or do they come out about the same in the end?

原文：https://stackoverflow.com/questions/1395308