如何提示gcc编译器更可能的分支 [英] how to suggest gcc compiler more probable branch

问题描述

  if（almost_always_false_condition）{
 //做某事
} 
  

有没有一种方法可以建议编译器在99％的条件下是错误的。
条件计算需要60个周期来检查，编译器本身无法在编译时计算。

（gcc 4.3）

解决方案

如果您想向GCC建议某个条件可能具有给定值作为安排代码流的提示，则应该使用 __ builtin_expect（）：

  if（__builtin_expect（almost_always_false_condition，0））{ 
 //做某事
} 
  

然而，这听起来像是你想找到避免评估条件的方法，其中 __ builtin_expect（）不会执行。有没有一种方法可以快速逼近条件，并且只有在近似值为真时才进行全面检查：

  if（ __builtin_expect（fastCheckThatIsTrueIfFullConditionIsTrue，0））{
 //大部分时间，我们甚至都没有到达这里，所以你不需要
 //来评估条件
 if（ almostAlwaysFalseCondition）{
 //做某事
} 
} 
  

你能告诉我们更多关于什么情况吗？

Example:

if (almost_always_false_condition) {
       // do something
}

Is there a way to suggest compiler that in 99% condition will be false. The condition calculation takes ~60 cycles to be checked, and it cannot be calculated at compile time by compiler itself.

(gcc 4.3)

解决方案

If you want to suggest to GCC that a condition is likely to have a given value as a hint for arranging code flow, you should use __builtin_expect( ):

if (__builtin_expect(almost_always_false_condition,0)) {
   // do something
}

However, it sounds like you want to find a way to avoid evaluating the condition, which __builtin_expect( ) won't do. Is there a way that you can quickly approximate the condition, and only do the full check when the approximation is true:

if (__builtin_expect(fastCheckThatIsTrueIfFullConditionIsTrue,0)) {
    // most of the time, we don't even get to here, so you don't need
    // to evaluate the condition
    if (almostAlwaysFalseCondition) {
        // do something
    }
}

Can you tell us more about what the condition is?

这篇关于如何提示gcc编译器更可能的分支的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！