java - Converting Boolean to Integer in Java without If-Statements

Question

I'm wondering if there's a way to convert a boolean to an int without using if statements (as not to break the pipeline). For example, I could write

int boolToInt( boolean b ){
  if ( b )
    return 1
  return 0

But I'm wondering if there's a way to do it without the if statement, like Python's

bool = True 
num = 1 * ( bool )

I also figure you could do

boolean bool = True;
int myint = Boolean.valueOf( bool ).compareTo( false );

This creates an extra object, though, so it's really wasteful and I found it to be even slower than the if-statement way (which isn't necessarily inefficient, just has the one weakness).

Answer 1

除了 if 之外，您不能使用布尔值。但是，这并不意味着在装配级别会有一个分支。

如果您检查该方法的编译代码（顺便说一下，使用return b ? 1 : 0;编译到完全相同的指令），您将看到它不使用跳转：

0x0000000002672580: sub    $0x18,%rsp
0x0000000002672587: mov    %rbp,0x10(%rsp)    ;*synchronization entry
0x000000000267258c: mov    %edx,%eax
0x000000000267258e: add    $0x10,%rsp
0x0000000002672592: pop    %rbp
0x0000000002672593: test   %eax,-0x2542599(%rip)        # 0x0000000000130000
                                                ;   {poll_return}
0x00000000025b2599: retq  

注意：这是在热点服务器 7 上 - 您可能会在不同的 VM 上得到不同的结果。

Answer 2

使用 ?: 运算符：( b ? 1 : 0 )

Answer 3

You can use the ternary operator:

return b ? 1 : 0;

If this is considered an "if", and given this is a "puzzle", you could use a map like this:

return new HashMap<Boolean, Integer>() {{
    put(true, 1);
    put(false, 0);
}}.get(b);

Although theoretically the implementation of HashMap doesn't need to use an if, it actually does. Nevertheless, the "if" is not in your code.

Of course to improve performance, you would:

private static Map<Boolean, Integer> map = new HashMap<Boolean, Integer>() {{
    put(true, 1);
    put(false, 0);
}};

Then in the method:

return map.get(b);

Answer 4

否则，您可以使用Apache Commons BooleanUtils.toInteger方法，它就像一个魅力......

// Converts a boolean to an int specifying the conversion values.    
static int  toInteger(boolean bool, int trueValue, int falseValue)

// Converts a Boolean to an int specifying the conversion values.
static int  toInteger(Boolean bool, int trueValue, int falseValue, int nullValue)

Answer 5

I found a solution by framework. Use compare for Boolean.

// b = Your boolean result
// v will be 1 if b equals true, otherwise 0
int v = Boolean.compare(b, false);

Answer 6

I can't say I recommend this. It's both slower than the ternary operator by itself, and it's too clever to be called good programming, but there's this:

-Boolean.FALSE.compareTo(value)

It uses the ternary under the covers (a couple of method calls later), but it's not in your code. To be fair, I would be willing to bet that there's a branch somewhere in the Python execution as well (though I probably only bet a nickel ;) ).

Answer 7

这不是直接可能的，无论如何在Java中都不是。如果您确实需要避免分支，您可以考虑直接使用intorbyte代替 a 。boolean

在这种情况下， VM 也有可能足够聪明以消除分支（if或?:）本身，因为无论如何，它boolean的内部表示很可能是文字 1 或 0。这是一篇关于如何检查为 Oracle JDK 生成的本机机器代码的文章，如果您需要速度，请确保您使用的是“服务器”JVM ，因为它比“客户端”执行更积极的优化。

Answer 8

由于您不需要 if / else 解决方案，因此您的表达方式很完美，尽管我会稍微更改一下

int myint = Boolean.valueOf( bool ).compareTo( Boolean.FALSE );

不涉及对象创建， Boolean.valueOf(boolean b) 返回 Boolean.TRUE 或 Boolean.FALSE，请参阅 API

Answer 9

You can try using ternary operator like this

int value = flag ? 1 : 0;

Answer 10

一个合理的替代方案来避免“如果”：

private static Boolean[] array = {false, true};

int boolToInt( boolean b ){
    return Arrays.binarySearch(array, b);
}

请注意，我认为这是一个“谜题”问题，所以如果我自己编码，我会使用三元......

Answer 11

如今，jdk 提供了一个有用的 Utils 方法：BooleanUtils.toInteger()

源码中jdk实现的方法一定是高效的：

public static int toInteger(boolean bool) {
    return bool ? 1 : 0;
}

所以，我觉得票数最多的答案非常棒，return bool ? 1 : 0是最佳实践。

要使用的示例代码BooleanUtils如下：

BooleanUtils.toInteger(false);

Answer 12

int ansInt = givenBoolean ? 1 : 0;