使用附带的代码，在JDK 15上，@Mecki测试用例得到了完全不同的结果。这基本上是在5个循环中运行代码，第一个循环稍微短一些，给VM一些时间来热身。

结果:

Loop 1 10000 cycles
method1 took 1 ms, result was 2
method2 took 0 ms, result was 2
method3 took 22 ms, result was 2
method4 took 22 ms, result was 2
method5 took 24 ms, result was 2
Loop 2 10000000 cycles
method1 took 39 ms, result was 2
method2 took 39 ms, result was 2
method3 took 1558 ms, result was 2
method4 took 1640 ms, result was 2
method5 took 1717 ms, result was 2
Loop 3 10000000 cycles
method1 took 49 ms, result was 2
method2 took 48 ms, result was 2
method3 took 126 ms, result was 2
method4 took 88 ms, result was 2
method5 took 87 ms, result was 2
Loop 4 10000000 cycles
method1 took 34 ms, result was 2
method2 took 34 ms, result was 2
method3 took 33 ms, result was 2
method4 took 98 ms, result was 2
method5 took 58 ms, result was 2
Loop 5 10000000 cycles
method1 took 34 ms, result was 2
method2 took 33 ms, result was 2
method3 took 33 ms, result was 2
method4 took 48 ms, result was 2
method5 took 49 ms, result was 2

package hs.jfx.eventstream.api;

public class Snippet {
  int value;


  public int getValue() {
      return value;
  }

  public void reset() {
      value = 0;
  }

  // Calculates without exception
  public void method1(int i) {
      value = ((value + i) / i) << 1;
      // Will never be true
      if ((i & 0xFFFFFFF) == 1000000000) {
          System.out.println("You'll never see this!");
      }
  }

  // Could in theory throw one, but never will
  public void method2(int i) throws Exception {
      value = ((value + i) / i) << 1;
      // Will never be true
      if ((i & 0xFFFFFFF) == 1000000000) {
          throw new Exception();
      }
  }

  private static final NoStackTraceRuntimeException E = new NoStackTraceRuntimeException();

  // This one will regularly throw one
  public void method3(int i) throws NoStackTraceRuntimeException {
      value = ((value + i) / i) << 1;
      // i & 1 is equally fast to calculate as i & 0xFFFFFFF; it is both
      // an AND operation between two integers. The size of the number plays
      // no role. AND on 32 BIT always ANDs all 32 bits
      if ((i & 0x1) == 1) {
          throw E;
      }
  }

  // This one will regularly throw one
  public void method4(int i) throws NoStackTraceThrowable {
      value = ((value + i) / i) << 1;
      // i & 1 is equally fast to calculate as i & 0xFFFFFFF; it is both
      // an AND operation between two integers. The size of the number plays
      // no role. AND on 32 BIT always ANDs all 32 bits
      if ((i & 0x1) == 1) {
          throw new NoStackTraceThrowable();
      }
  }

  // This one will regularly throw one
  public void method5(int i) throws NoStackTraceRuntimeException {
      value = ((value + i) / i) << 1;
      // i & 1 is equally fast to calculate as i & 0xFFFFFFF; it is both
      // an AND operation between two integers. The size of the number plays
      // no role. AND on 32 BIT always ANDs all 32 bits
      if ((i & 0x1) == 1) {
          throw new NoStackTraceRuntimeException();
      }
  }

  public static void main(String[] args) {
    for(int k = 0; k < 5; k++) {
      int cycles = 10000000;
      if(k == 0) {
        cycles = 10000;
        try {
          Thread.sleep(500);
        }
        catch(InterruptedException e) {
          // TODO Auto-generated catch block
          e.printStackTrace();
        }
      }
      System.out.println("Loop " + (k + 1) + " " + cycles + " cycles");
      int i;
      long l;
      Snippet t = new Snippet();

      l = System.currentTimeMillis();
      t.reset();
      for (i = 1; i < cycles; i++) {
          t.method1(i);
      }
      l = System.currentTimeMillis() - l;
      System.out.println(
          "method1 took " + l + " ms, result was " + t.getValue()
      );

      l = System.currentTimeMillis();
      t.reset();
      for (i = 1; i < cycles; i++) {
          try {
              t.method2(i);
          } catch (Exception e) {
              System.out.println("You'll never see this!");
          }
      }
      l = System.currentTimeMillis() - l;
      System.out.println(
          "method2 took " + l + " ms, result was " + t.getValue()
      );

      l = System.currentTimeMillis();
      t.reset();
      for (i = 1; i < cycles; i++) {
          try {
              t.method3(i);
          } catch (NoStackTraceRuntimeException e) {
            // always comes here
          }
      }
      l = System.currentTimeMillis() - l;
      System.out.println(
          "method3 took " + l + " ms, result was " + t.getValue()
      );


      l = System.currentTimeMillis();
      t.reset();
      for (i = 1; i < cycles; i++) {
          try {
              t.method4(i);
          } catch (NoStackTraceThrowable e) {
            // always comes here
          }
      }
      l = System.currentTimeMillis() - l;
      System.out.println( "method4 took " + l + " ms, result was " + t.getValue() );


      l = System.currentTimeMillis();
      t.reset();
      for (i = 1; i < cycles; i++) {
          try {
              t.method5(i);
          } catch (RuntimeException e) {
            // always comes here
          }
      }
      l = System.currentTimeMillis() - l;
      System.out.println( "method5 took " + l + " ms, result was " + t.getValue() );
    }
  }

  public static class NoStackTraceRuntimeException extends RuntimeException {
    public NoStackTraceRuntimeException() {
        super("my special throwable", null, false, false);
    }
  }

  public static class NoStackTraceThrowable extends Throwable {
    public NoStackTraceThrowable() {
        super("my special throwable", null, false, false);
    }
  }
}

我用JVM 1.5做了一些性能测试，使用异常至少慢了两倍。平均:一个非常小的方法的执行时间超过3倍(3倍)。一个必须捕获异常的小循环的自时间增加了2倍。

我在产品代码和微基准测试中也看到过类似的数字。

异常绝对不应该用于任何频繁调用的东西。每秒抛出数千个异常将导致巨大的瓶颈。

例如，使用“Integer.ParseInt(…)”在一个非常大的文本文件中找到所有错误的值——非常糟糕的想法。(我曾在产品代码上看到过这种实用方法的性能下降)

使用异常在用户GUI表单上报告错误的值，从性能的角度来看可能并不是那么糟糕。

无论这是否是一个好的设计实践，我都会遵循这样的规则:如果错误是正常的/预期的，那么就使用返回值。如果不正常，请使用异常。例如:读取用户输入，错误值是正常的—使用错误代码。将值传递给内部实用程序函数时，应该通过调用代码来过滤坏值——使用异常。

HotSpot非常能够删除系统生成的异常代码，只要它是内联的。但是，显式创建的异常和其他未删除的异常要花费大量时间来创建堆栈跟踪。重写fillInStackTrace以查看这会如何影响性能。

使用附带的代码，在JDK 15上，@Mecki测试用例得到了完全不同的结果。这基本上是在5个循环中运行代码，第一个循环稍微短一些，给VM一些时间来热身。

结果:

Loop 1 10000 cycles
method1 took 1 ms, result was 2
method2 took 0 ms, result was 2
method3 took 22 ms, result was 2
method4 took 22 ms, result was 2
method5 took 24 ms, result was 2
Loop 2 10000000 cycles
method1 took 39 ms, result was 2
method2 took 39 ms, result was 2
method3 took 1558 ms, result was 2
method4 took 1640 ms, result was 2
method5 took 1717 ms, result was 2
Loop 3 10000000 cycles
method1 took 49 ms, result was 2
method2 took 48 ms, result was 2
method3 took 126 ms, result was 2
method4 took 88 ms, result was 2
method5 took 87 ms, result was 2
Loop 4 10000000 cycles
method1 took 34 ms, result was 2
method2 took 34 ms, result was 2
method3 took 33 ms, result was 2
method4 took 98 ms, result was 2
method5 took 58 ms, result was 2
Loop 5 10000000 cycles
method1 took 34 ms, result was 2
method2 took 33 ms, result was 2
method3 took 33 ms, result was 2
method4 took 48 ms, result was 2
method5 took 49 ms, result was 2

package hs.jfx.eventstream.api;

public class Snippet {
  int value;


  public int getValue() {
      return value;
  }

  public void reset() {
      value = 0;
  }

  // Calculates without exception
  public void method1(int i) {
      value = ((value + i) / i) << 1;
      // Will never be true
      if ((i & 0xFFFFFFF) == 1000000000) {
          System.out.println("You'll never see this!");
      }
  }

  // Could in theory throw one, but never will
  public void method2(int i) throws Exception {
      value = ((value + i) / i) << 1;
      // Will never be true
      if ((i & 0xFFFFFFF) == 1000000000) {
          throw new Exception();
      }
  }

  private static final NoStackTraceRuntimeException E = new NoStackTraceRuntimeException();

  // This one will regularly throw one
  public void method3(int i) throws NoStackTraceRuntimeException {
      value = ((value + i) / i) << 1;
      // i & 1 is equally fast to calculate as i & 0xFFFFFFF; it is both
      // an AND operation between two integers. The size of the number plays
      // no role. AND on 32 BIT always ANDs all 32 bits
      if ((i & 0x1) == 1) {
          throw E;
      }
  }

  // This one will regularly throw one
  public void method4(int i) throws NoStackTraceThrowable {
      value = ((value + i) / i) << 1;
      // i & 1 is equally fast to calculate as i & 0xFFFFFFF; it is both
      // an AND operation between two integers. The size of the number plays
      // no role. AND on 32 BIT always ANDs all 32 bits
      if ((i & 0x1) == 1) {
          throw new NoStackTraceThrowable();
      }
  }

  // This one will regularly throw one
  public void method5(int i) throws NoStackTraceRuntimeException {
      value = ((value + i) / i) << 1;
      // i & 1 is equally fast to calculate as i & 0xFFFFFFF; it is both
      // an AND operation between two integers. The size of the number plays
      // no role. AND on 32 BIT always ANDs all 32 bits
      if ((i & 0x1) == 1) {
          throw new NoStackTraceRuntimeException();
      }
  }

  public static void main(String[] args) {
    for(int k = 0; k < 5; k++) {
      int cycles = 10000000;
      if(k == 0) {
        cycles = 10000;
        try {
          Thread.sleep(500);
        }
        catch(InterruptedException e) {
          // TODO Auto-generated catch block
          e.printStackTrace();
        }
      }
      System.out.println("Loop " + (k + 1) + " " + cycles + " cycles");
      int i;
      long l;
      Snippet t = new Snippet();

      l = System.currentTimeMillis();
      t.reset();
      for (i = 1; i < cycles; i++) {
          t.method1(i);
      }
      l = System.currentTimeMillis() - l;
      System.out.println(
          "method1 took " + l + " ms, result was " + t.getValue()
      );

      l = System.currentTimeMillis();
      t.reset();
      for (i = 1; i < cycles; i++) {
          try {
              t.method2(i);
          } catch (Exception e) {
              System.out.println("You'll never see this!");
          }
      }
      l = System.currentTimeMillis() - l;
      System.out.println(
          "method2 took " + l + " ms, result was " + t.getValue()
      );

      l = System.currentTimeMillis();
      t.reset();
      for (i = 1; i < cycles; i++) {
          try {
              t.method3(i);
          } catch (NoStackTraceRuntimeException e) {
            // always comes here
          }
      }
      l = System.currentTimeMillis() - l;
      System.out.println(
          "method3 took " + l + " ms, result was " + t.getValue()
      );


      l = System.currentTimeMillis();
      t.reset();
      for (i = 1; i < cycles; i++) {
          try {
              t.method4(i);
          } catch (NoStackTraceThrowable e) {
            // always comes here
          }
      }
      l = System.currentTimeMillis() - l;
      System.out.println( "method4 took " + l + " ms, result was " + t.getValue() );


      l = System.currentTimeMillis();
      t.reset();
      for (i = 1; i < cycles; i++) {
          try {
              t.method5(i);
          } catch (RuntimeException e) {
            // always comes here
          }
      }
      l = System.currentTimeMillis() - l;
      System.out.println( "method5 took " + l + " ms, result was " + t.getValue() );
    }
  }

  public static class NoStackTraceRuntimeException extends RuntimeException {
    public NoStackTraceRuntimeException() {
        super("my special throwable", null, false, false);
    }
  }

  public static class NoStackTraceThrowable extends Throwable {
    public NoStackTraceThrowable() {
        super("my special throwable", null, false, false);
    }
  }
}

我认为第一篇文章提到遍历调用堆栈和创建堆栈跟踪是最昂贵的部分，虽然第二篇文章没有这样说，但我认为这是对象创建中最昂贵的部分。John Rose在一篇文章中描述了加速异常的不同技术。(预分配和重用异常，没有堆栈跟踪的异常，等等)

但我仍然认为这应该被认为是一种必要的邪恶，一种最后的手段。John这样做的原因是为了模拟JVM中(还)没有的其他语言的特性。你不应该养成对控制流使用异常的习惯。尤其是因为性能原因!正如您自己在第2条中提到的，这样做可能会掩盖代码中的严重错误，而且对于新程序员来说，维护起来会更加困难。

Java中的微基准测试出奇地难以正确(有人告诉过我)，特别是在进入JIT领域时，因此我真的怀疑在现实生活中使用异常是否比“返回”更快。例如，我怀疑您在测试中有2到5个堆栈帧?现在假设您的代码将由JBoss部署的JSF组件调用。现在您可能有一个数页长的堆栈跟踪。

也许您可以发布您的测试代码?

Java和c#中的异常性能还有待改进。

作为程序员，这迫使我们遵循“异常应该很少引起”的规则，仅仅是出于实际性能的考虑。

However, as computer scientists, we should rebel against this problematic state. The person authoring a function often has no idea how often it will be called, or whether success or failure is more likely. Only the caller has this information. Trying to avoid exceptions leads to unclear API idoms where in some cases we have only clean-but-slow exception versions, and in other cases we have fast-but-clunky return-value errors, and in still other cases we end up with both. The library implementor may have to write and maintain two versions of APIs, and the caller has to decide which of two versions to use in each situation.

这里有点乱。如果异常具有更好的性能，我们就可以避免这些笨拙的习惯用法，并按照它们应该使用的方式使用异常……作为结构化错误返回工具。

我真的希望看到异常机制使用更接近返回值的技术来实现，这样我们的性能就能更接近返回值。因为这是我们在性能敏感代码中恢复的内容。

下面是一个比较异常性能和错误返回值性能的代码示例。

公共类test {

int value;


public int getValue() {
    return value;
}

public void reset() {
    value = 0;
}

public boolean baseline_null(boolean shouldfail, int recurse_depth) {
    if (recurse_depth <= 0) {
        return shouldfail;
    } else {
        return baseline_null(shouldfail,recurse_depth-1);
    }
}

public boolean retval_error(boolean shouldfail, int recurse_depth) {
    if (recurse_depth <= 0) {
        if (shouldfail) {
            return false;
        } else {
            return true;
        }
    } else {
        boolean nested_error = retval_error(shouldfail,recurse_depth-1);
        if (nested_error) {
            return true;
        } else {
            return false;
        }
    }
}

public void exception_error(boolean shouldfail, int recurse_depth) throws Exception {
    if (recurse_depth <= 0) {
        if (shouldfail) {
            throw new Exception();
        }
    } else {
        exception_error(shouldfail,recurse_depth-1);
    }

}

public static void main(String[] args) {
    int i;
    long l;
    TestIt t = new TestIt();
    int failures;

    int ITERATION_COUNT = 100000000;


    // (0) baseline null workload
    for (int recurse_depth = 2; recurse_depth <= 10; recurse_depth+=3) {
        for (float exception_freq = 0.0f; exception_freq <= 1.0f; exception_freq += 0.25f) {            
            int EXCEPTION_MOD = (exception_freq == 0.0f) ? ITERATION_COUNT+1 : (int)(1.0f / exception_freq);            

            failures = 0;
            long start_time = System.currentTimeMillis();
            t.reset();              
            for (i = 1; i < ITERATION_COUNT; i++) {
                boolean shoulderror = (i % EXCEPTION_MOD) == 0;
                t.baseline_null(shoulderror,recurse_depth);
            }
            long elapsed_time = System.currentTimeMillis() - start_time;
            System.out.format("baseline: recurse_depth %s, exception_freqeuncy %s (%s), time elapsed %s ms\n",
                    recurse_depth, exception_freq, failures,elapsed_time);
        }
    }


    // (1) retval_error
    for (int recurse_depth = 2; recurse_depth <= 10; recurse_depth+=3) {
        for (float exception_freq = 0.0f; exception_freq <= 1.0f; exception_freq += 0.25f) {            
            int EXCEPTION_MOD = (exception_freq == 0.0f) ? ITERATION_COUNT+1 : (int)(1.0f / exception_freq);            

            failures = 0;
            long start_time = System.currentTimeMillis();
            t.reset();              
            for (i = 1; i < ITERATION_COUNT; i++) {
                boolean shoulderror = (i % EXCEPTION_MOD) == 0;
                if (!t.retval_error(shoulderror,recurse_depth)) {
                    failures++;
                }
            }
            long elapsed_time = System.currentTimeMillis() - start_time;
            System.out.format("retval_error: recurse_depth %s, exception_freqeuncy %s (%s), time elapsed %s ms\n",
                    recurse_depth, exception_freq, failures,elapsed_time);
        }
    }

    // (2) exception_error
    for (int recurse_depth = 2; recurse_depth <= 10; recurse_depth+=3) {
        for (float exception_freq = 0.0f; exception_freq <= 1.0f; exception_freq += 0.25f) {            
            int EXCEPTION_MOD = (exception_freq == 0.0f) ? ITERATION_COUNT+1 : (int)(1.0f / exception_freq);            

            failures = 0;
            long start_time = System.currentTimeMillis();
            t.reset();              
            for (i = 1; i < ITERATION_COUNT; i++) {
                boolean shoulderror = (i % EXCEPTION_MOD) == 0;
                try {
                    t.exception_error(shoulderror,recurse_depth);
                } catch (Exception e) {
                    failures++;
                }
            }
            long elapsed_time = System.currentTimeMillis() - start_time;
            System.out.format("exception_error: recurse_depth %s, exception_freqeuncy %s (%s), time elapsed %s ms\n",
                    recurse_depth, exception_freq, failures,elapsed_time);              
        }
    }
}

}

结果如下:

baseline: recurse_depth 2, exception_freqeuncy 0.0 (0), time elapsed 683 ms
baseline: recurse_depth 2, exception_freqeuncy 0.25 (0), time elapsed 790 ms
baseline: recurse_depth 2, exception_freqeuncy 0.5 (0), time elapsed 768 ms
baseline: recurse_depth 2, exception_freqeuncy 0.75 (0), time elapsed 749 ms
baseline: recurse_depth 2, exception_freqeuncy 1.0 (0), time elapsed 731 ms
baseline: recurse_depth 5, exception_freqeuncy 0.0 (0), time elapsed 923 ms
baseline: recurse_depth 5, exception_freqeuncy 0.25 (0), time elapsed 971 ms
baseline: recurse_depth 5, exception_freqeuncy 0.5 (0), time elapsed 982 ms
baseline: recurse_depth 5, exception_freqeuncy 0.75 (0), time elapsed 947 ms
baseline: recurse_depth 5, exception_freqeuncy 1.0 (0), time elapsed 937 ms
baseline: recurse_depth 8, exception_freqeuncy 0.0 (0), time elapsed 1154 ms
baseline: recurse_depth 8, exception_freqeuncy 0.25 (0), time elapsed 1149 ms
baseline: recurse_depth 8, exception_freqeuncy 0.5 (0), time elapsed 1133 ms
baseline: recurse_depth 8, exception_freqeuncy 0.75 (0), time elapsed 1117 ms
baseline: recurse_depth 8, exception_freqeuncy 1.0 (0), time elapsed 1116 ms
retval_error: recurse_depth 2, exception_freqeuncy 0.0 (0), time elapsed 742 ms
retval_error: recurse_depth 2, exception_freqeuncy 0.25 (24999999), time elapsed 743 ms
retval_error: recurse_depth 2, exception_freqeuncy 0.5 (49999999), time elapsed 734 ms
retval_error: recurse_depth 2, exception_freqeuncy 0.75 (99999999), time elapsed 723 ms
retval_error: recurse_depth 2, exception_freqeuncy 1.0 (99999999), time elapsed 728 ms
retval_error: recurse_depth 5, exception_freqeuncy 0.0 (0), time elapsed 920 ms
retval_error: recurse_depth 5, exception_freqeuncy 0.25 (24999999), time elapsed 1121   ms
retval_error: recurse_depth 5, exception_freqeuncy 0.5 (49999999), time elapsed 1037 ms
retval_error: recurse_depth 5, exception_freqeuncy 0.75 (99999999), time elapsed 1141   ms
retval_error: recurse_depth 5, exception_freqeuncy 1.0 (99999999), time elapsed 1130 ms
retval_error: recurse_depth 8, exception_freqeuncy 0.0 (0), time elapsed 1218 ms
retval_error: recurse_depth 8, exception_freqeuncy 0.25 (24999999), time elapsed 1334  ms
retval_error: recurse_depth 8, exception_freqeuncy 0.5 (49999999), time elapsed 1478 ms
retval_error: recurse_depth 8, exception_freqeuncy 0.75 (99999999), time elapsed 1637 ms
retval_error: recurse_depth 8, exception_freqeuncy 1.0 (99999999), time elapsed 1655 ms
exception_error: recurse_depth 2, exception_freqeuncy 0.0 (0), time elapsed 726 ms
exception_error: recurse_depth 2, exception_freqeuncy 0.25 (24999999), time elapsed 17487   ms
exception_error: recurse_depth 2, exception_freqeuncy 0.5 (49999999), time elapsed 33763   ms
exception_error: recurse_depth 2, exception_freqeuncy 0.75 (99999999), time elapsed 67367   ms
exception_error: recurse_depth 2, exception_freqeuncy 1.0 (99999999), time elapsed 66990 ms
exception_error: recurse_depth 5, exception_freqeuncy 0.0 (0), time elapsed 924 ms
exception_error: recurse_depth 5, exception_freqeuncy 0.25 (24999999), time elapsed 23775  ms
exception_error: recurse_depth 5, exception_freqeuncy 0.5 (49999999), time elapsed 46326 ms
exception_error: recurse_depth 5, exception_freqeuncy 0.75 (99999999), time elapsed 91707 ms
exception_error: recurse_depth 5, exception_freqeuncy 1.0 (99999999), time elapsed 91580 ms
exception_error: recurse_depth 8, exception_freqeuncy 0.0 (0), time elapsed 1144 ms
exception_error: recurse_depth 8, exception_freqeuncy 0.25 (24999999), time elapsed 30440 ms
exception_error: recurse_depth 8, exception_freqeuncy 0.5 (49999999), time elapsed 59116   ms
exception_error: recurse_depth 8, exception_freqeuncy 0.75 (99999999), time elapsed 116678 ms
exception_error: recurse_depth 8, exception_freqeuncy 1.0 (99999999), time elapsed 116477 ms

检查和传播返回值与基线空调用相比确实增加了一些成本，而该成本与调用深度成正比。在调用链深度为8时，错误返回值检查版本比不检查返回值的基线版本慢了约27%。

相比之下，异常性能不是调用深度的函数，而是异常频率的函数。然而，随着异常频率的增加，这种退化更为显著。当错误频率只有25%时，代码运行速度变慢了24倍。当错误频率为100%时，异常版本几乎要慢100倍。

这在我看来可能是在我们的异常实现中做出了错误的权衡。异常可以更快，可以避免代价高昂的跟踪遍历，也可以直接将异常转换为编译器支持的返回值检查。在此之前，当我们希望代码运行得更快时，我们不得不避免它们。