One thing I haven't seen others point out is that a standard thing to do in the "please write me some code" section of the job interview is to say "Could you improve that?" or "Are you completely happy with that" or "is that as optimized as possible?" when you say you are done. It's possible you heard "how would you improve that" as "this might be improved; how?". In this case changing the if(x) return true; else return false; idiom to just return x is an improvement - but be aware that there are times they just want to see how you react to the question. I have heard that some interviewers will insist there is a flaw in perfect code just to see how you cope with it.

我们可以将bool转换为整数，并执行以下简单的检查:

(int(a) + int(b) + int(c)) >= 2

三元运算符让人觉得很无聊，但它们也会令人困惑(使代码更难维护，从而增加了注入错误的可能性)。Jeff Attwood说得很好:

这是一个权衡取舍的完美例子 毫无意义的一次 写的时候节省了几十个 阅读时间理解惩罚——它 让我思考。

为了避免三元操作符，我创建了以下函数:

function atLeastTwoTrue($a, $b, $c) {
        $count = 0;

        if ($a) { $count++; }
        if ($b) { $count++; }
        if ($c) { $count++; }

        if ($count >= 2) {
                return true;
        } else {
                return false;
        }
}

这个和其他解一样酷吗?不。这样更容易理解吗?是的。这是否会使代码更具可维护性、bug更少?是的。

我不喜欢三元(return a ?(b || c):(b && c);从最上面的答案)，我想我没有看到任何人提到过它。它是这样写的:

boolean atLeastTwo(boolean a, boolean b, boolean c) {
    if (a) {
        return b||c;
    } 
    else {
        return b&&C;
    }

以下是目前为止的答案:

public class X
{
    static boolean a(final boolean a, final boolean b, final boolean c)
    {
    return ((a && b) || (b && c) || (a && c));
    }

    static boolean b(final boolean a, final boolean b, final boolean c)
    {
    return a ? (b || c) : (b && c);
    }

    static boolean c(final boolean a, final boolean b, final boolean c)
    {
    return ((a & b) | (b & c) | (c & a));
    }

    static boolean d(final boolean a, final boolean b, final boolean c)
    {
    return ((a?1:0)+(b?1:0)+(c?1:0) >= 2);
    }
}

并通过反编译器运行它们(javap -c X > results.txt):

Compiled from "X.java"
public class X extends java.lang.Object{
public X();
  Code:
   0:   aload_0
   1:   invokespecial   #1; //Method java/lang/Object."<init>":()V
   4:   return

static boolean a(boolean, boolean, boolean);
  Code:
   0:   iload_0
   1:   ifeq    8
   4:   iload_1
   5:   ifne    24
   8:   iload_1
   9:   ifeq    16
   12:  iload_2
   13:  ifne    24
   16:  iload_0
   17:  ifeq    28
   20:  iload_2
   21:  ifeq    28
   24:  iconst_1
   25:  goto    29
   28:  iconst_0
   29:  ireturn

static boolean b(boolean, boolean, boolean);
  Code:
   0:   iload_0
   1:   ifeq    20
   4:   iload_1
   5:   ifne    12
   8:   iload_2
   9:   ifeq    16
   12:  iconst_1
   13:  goto    33
   16:  iconst_0
   17:  goto    33
   20:  iload_1
   21:  ifeq    32
   24:  iload_2
   25:  ifeq    32
   28:  iconst_1
   29:  goto    33
   32:  iconst_0
   33:  ireturn

static boolean c(boolean, boolean, boolean);
  Code:
   0:   iload_0
   1:   iload_1
   2:   iand
   3:   iload_1
   4:   iload_2
   5:   iand
   6:   ior
   7:   iload_2
   8:   iload_0
   9:   iand
   10:  ior
   11:  ireturn

static boolean d(boolean, boolean, boolean);
  Code:
   0:   iload_0
   1:   ifeq    8
   4:   iconst_1
   5:   goto    9
   8:   iconst_0
   9:   iload_1
   10:  ifeq    17
   13:  iconst_1
   14:  goto    18
   17:  iconst_0
   18:  iadd
   19:  iload_2
   20:  ifeq    27
   23:  iconst_1
   24:  goto    28
   27:  iconst_0
   28:  iadd
   29:  iconst_2
   30:  if_icmplt   37
   33:  iconst_1
   34:  goto    38
   37:  iconst_0
   38:  ireturn
}

你可以看到?:那些比你原来的修复版本稍好。最好的方法是完全避免分支的方法。从更少的指令(在大多数情况下)的角度来看，这很好，对于CPU的分支预测部分来说也更好，因为分支预测中的错误猜测可能会导致CPU失速。

我觉得最有效的是《月影》里的那个。它平均使用最少的指令，减少了CPU中管道停顿的机会。

为了100%确定，您需要找出每条指令的成本(以CPU周期为单位)，不幸的是，这是不容易获得的(您必须查看热点的源代码，然后查看CPU供应商的规格，以确定每条生成指令所花费的时间)。

请参阅Rotsor对代码的运行时分析的更新答案。

可读性应该是目标。阅读代码的人必须立即理解您的意图。这就是我的解。

int howManyBooleansAreTrue =
      (a ? 1 : 0)
    + (b ? 1 : 0)
    + (c ? 1 : 0);

return howManyBooleansAreTrue >= 2;