短路计算应该比按位计算更快，但事实并非如此.为什么

Question

比较短路求值逻辑运算符(例如&&(与按位求值逻辑运算符(即 &，我写了这个例子并运行了它：

package examples1;
import java.util.Scanner;
public class ShortCircuitOperatorsVSBitwiseOperators {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        System.out.print("Enter A: "); boolean a = scanner.nextBoolean();
        System.out.print("Enter B: "); boolean b = scanner.nextBoolean();
        long startTimeShortCircuited = System.currentTimeMillis();
        boolean resultShortCircuited = true;
        for(long i = 0; i < 10000000000L; i++) {
            resultShortCircuited = a && b;
        }
        long endTimeShortCircuited = System.currentTimeMillis();
        System.out.println(resultShortCircuited + " in " + (endTimeShortCircuited - startTimeShortCircuited) + " milliseconds, short-circuited");
        long startTimeBitwise = System.currentTimeMillis();
        boolean resultBitwise = true;
        for(long i = 0; i < 10000000000L; i++) {
            resultBitwise = a & b;
        }
        long endTimeBitwise = System.currentTimeMillis();
        System.out.println(resultBitwise + " in " + (endTimeBitwise - startTimeBitwise) + " milliseconds, bitwise");
        scanner.close();
    }
}

示例运行显示以下内容：

java examples1/ShortCircuitOperatorsVSBitwiseOperators
Enter A: false
Enter B: true
false in 4829 milliseconds, short-circuited
false in 3276 milliseconds, bitwise

这说不通。我希望短路评估会更快，因为在这种情况下，如果左侧false，它不会评估&&的右侧。反直觉结果的原因是什么？

Answer 1

短路操作很复杂。

public static boolean shortCircuitedAnd(boolean a, boolean b) {
    return a && b;
}
public static boolean bitwiseAnd(boolean a, boolean b) {
    return a & b;
}

它们被编译为

public static boolean shortCircuitedAnd(boolean, boolean);
  Code:
     0: iload_0
     1: ifeq          10
     4: iload_1
     5: ifeq          10
     8: iconst_1
     9: ireturn
    10: iconst_0
    11: ireturn
public static boolean bitwiseAnd(boolean, boolean);
  Code:
     0: iload_0
     1: iload_1
     2: iand
     3: ireturn