如何建议 gcc 编译器更可能的分支

发布于 2024-08-19 02:58:58 字数 194 浏览 12 评论 0原文

示例：

if (almost_always_false_condition) {
       // do something
}

有没有办法建议编译器在 99% 的情况下将为 false。条件计算需要约 60 个周期进行检查，并且编译器本身无法在编译时计算。

（海湾合作委员会4.3）

原文

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)

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可爱咩 2024-08-26 02:58:58

如果你想向 GCC 建议某个条件可能有给定值作为安排代码流的提示，你应该使用 __builtin_expect( ) ：

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

但是，听起来你想找到一种方法以避免评估条件，而 __builtin_expect( ) 不会这样做。有没有一种方法可以快速近似条件，并且仅在近似为真时才进行全面检查：

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
    }
}

您能告诉我们更多有关条件的信息吗？

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?

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梦途 2024-08-26 02:58:58

如果单次运行期间结果可能有所不同，您可以使用布尔运算符的惰性求值将您的条件分为便宜部分和昂贵部分，并首先运行便宜部分。

if (a == 5 && somethingexpensive())
{
   ...
}

由于计算 a == 5 比 somethingexppressive() 更便宜，并且如果它几乎总是 false，您应该首先运行它，这样可以避免评估somethingexpcious 子句。

另一方面，如果程序运行的结果是恒定的，则可以通过将计算结果存储在静态或全局变量中来优化它。

static int result = doevalfunctiononlyonce();

if (result)
{
   ....    
}

通过这种方式，您可以将 if 的成本降低为简单的内存查找。

如果条件仅因响应另一个过程中的操作而变化，则可以从该过程更新全局：

int condition;

void appendToList(int a)
{
   list.append(a);
   if (list.somethingexpensive())
   {
     condition = true;
   } else
   {
     condition = false;
   }
}

void someotherfunction()
{
  // if (list.somethingexpensive())
  if (condition)
  {
    ...
  }
}

如果 someotherfunction 的调用频率比 appendtolist 的调用频率高，这非常有用功能。

If the result might vary during a single run, you may be able to use lazy evaluation of boolean operators to split your condition into a cheap part and an expensive part, and run the cheap part first.

if (a == 5 && somethingexpensive())
{
   ...
}

Since calculating a == 5 is cheaper than somethingexpensive(), and if it is almost always false you should run it first, which avoids evaluating the somethingexpensive clause.

If on the other hand the result is constant for a run of the program, you can optimise it by storing the result of the calculation in a static or global variable.

static int result = doevalfunctiononlyonce();

if (result)
{
   ....    
}

This way you have reduced the cost of the if to a simple memory lookup.

If the condition only changes in response to an action in another procedure, you can update the global from that procedure:

int condition;

void appendToList(int a)
{
   list.append(a);
   if (list.somethingexpensive())
   {
     condition = true;
   } else
   {
     condition = false;
   }
}

void someotherfunction()
{
  // if (list.somethingexpensive())
  if (condition)
  {
    ...
  }
}

This is useful if someotherfunction is called lots more often than the appendtolist function.

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一江春梦 2024-08-26 02:58:58

首先，在 else 子句或程序的其他地方花费了多少个周期？如果您进行简介或拍摄stackshots，您是否至少花费了你用 10% 的时间来做那个测试？如果没有，您可能应该首先考虑更大的问题。

其次，如果您在该测试中花费超过 10% 的时间，您应该查看是否可以调整算法以使决策点更接近 50-50 的概率。 50-50 决策点在执行时会产生 1 位信息，而 99-1 决策点仅产生约 0.07 位*（即它不会告诉您太多信息，因此 CPU 周期的使用效率较低。）这种现象的一个例子是将线性搜索与二分搜索进行比较。

*如果您有一个二元决策点，并且结果的概率为 a 和 b，则以位为单位的信息产量（熵）为 -(a*log (a) + b*log(b))/log(2)。