blocks|key|89135|text|使用?：运算符：(+b+?+1+:+0+)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|89136|entityMap^0|8|D|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

Use the ?: operator: <code>( b ? 1 : 0 )</code>

blocks|key|306163|text|可以尝试使用三元运算符，如下所示|type|unstyled|depth|inlineStyleRanges|entityRanges|data|306164|int+value+=+flag+?+1+:+0;|code-block|syntax|javascript|306165|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

You can try using ternary operator like this

<pre><code>int value = flag ? 1 : 0;
</code></pre>

blocks|key|89245|text|这是不可能的，至少在Java中是不可能的。如果确实需要避免分支，可以考虑直接使用int或byte而不是boolean。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|89246|在这种情况下，VM也有可能足够智能地消除分支(+if或?:)本身，因为boolean的内部表示很可能是文字1或0。Here+is+an+article介绍如何检查为Oracle+JDK生成的本机代码，如果您需要速度，请确保使用“服务器”as+it+performs+more+aggressive+optimization而不是“客户端”。|89247|entityMap|0|LINK|mutability|MUTABLE|url|https://wikis.oracle.com/display/HotSpotInternals/PrintAssembly|1|https://stackoverflow.com/questions/198577/real-differences-between-java-server-and-java-client^0|14|3|18|4|1F|7|0|O|2|R|2|Z|7|1L|I|0|3A|17|1|0^^$0|@$1|2|3|4|5|6|7|R|8|@$9|S|A|T|B|C]|$9|U|A|V|B|C]|$9|W|A|X|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]|$9|13|A|14|B|C]]|D|@$9|15|A|16|1|17]|$9|18|A|19|1|1A]]|E|$]]|$1|H|3|-4|5|6|7|1B|8|@]|D|@]|E|$]]]|I|$J|$5|K|L|M|E|$N|O]]|P|$5|K|L|M|E|$N|Q]]]]

This is not directly possible, not in Java anyway. You could consider directly using an <code>int</code> or <code>byte</code> instead of a <code>boolean</code> if you really need to avoid the branch.

It's also possible that the VM is smart enough to eliminate the branch (the <code>if</code> or <code>?:</code>) itself in this case, as the <code>boolean</code>'s internal representation is quite likely to be the literal 1 or 0 anyway. <a href="https://wikis.oracle.com/display/HotSpotInternals/PrintAssembly" rel="nofollow noreferrer">Here is an article</a> on how to examine the generated native machine code for the Oracle JDK, and if you need speed, make sure you're using the "server" JVM <a href="https://stackoverflow.com/questions/198577/real-differences-between-java-server-and-java-client">as it performs more aggressive optimization</a> than the "client" one.

blocks|key|306298|text|除非在if中，否则不能使用布尔值。但是，这并不意味着在程序集级别上会有一个分支。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|306299|如果您检查该方法的编译代码(顺便说一句，使用return+b+?+1+:+0;编译成完全相同的指令)，您将看到它没有使用跳转：|offset|length|style|CODE|306300|0x0000000002672580:+sub++++$0x18,%25rsp
0x0000000002672587:+mov++++%25rbp,0x10(%25rsp)++++;*synchronization+entry
0x000000000267258c:+mov++++%25edx,%25eax
0x000000000267258e:+add++++$0x10,%25rsp
0x0000000002672592:+pop++++%25rbp
0x0000000002672593:+test+++%25eax,-0x2542599(%25rip)++++++++#+0x0000000000130000
++++++++++++++++++++++++++++++++++++++++++++++++;+++{poll_return}
0x00000000025b2599:+retq++|code-block|syntax|javascript|306301|注意:这是在hotspot服务器7上-你可能会在不同的虚拟机上得到不同的结果。|306302|entityMap^0|0|M|H|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@$D|S|E|T|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|U|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

You can't use a boolean other than in a if. However it does not mean that there will be a branch at the assembly level.

If you check the compiled code of that method (by the way, using <code>return b ? 1 : 0;</code> compiles to the exact same instructions), you will see that it does not use a jump:

<pre><code>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 
</code></pre>

Note: this is on hotspot server 7 - you might get different results on a different VM.

blocks|key|306332|text|因为您不需要if+/+else解决方案，所以您的表达式是完美的，尽管我会稍微更改它|type|unstyled|depth|inlineStyleRanges|entityRanges|data|306333|int+myint+=+Boolean.valueOf(+bool+).compareTo(+Boolean.FALSE+);|code-block|syntax|javascript|306334|不涉及对象创建，Boolean.valueOf(boolean+b)返回Boolean.TRUE或Boolean.FALSE，见接口|306335|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Since you want no if / else solution your expression is perfect, though I would slightly change it

<pre><code>int myint = Boolean.valueOf( bool ).compareTo( Boolean.FALSE );
</code></pre>

There is no object creation involved, Boolean.valueOf(boolean b) returns either Boolean.TRUE or Boolean.FALSE, see API

blocks|key|89387|text|我不能说我推荐这样做。它比三元运算符本身更慢，而且它太聪明了，不能被称为好的编程，但有以下几点：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|89388|-Boolean.FALSE.compareTo(value)|code-block|syntax|javascript|89389|它在幕后使用三元组(稍后会调用几个方法)，但它不在您的代码中。公平地说，我敢打赌在Python执行中的某个地方也有一个分支(尽管我可能只下了五分钱;)。|89390|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

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:

<pre><code>-Boolean.FALSE.compareTo(value)
</code></pre>

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 ;) ).

blocks|key|89423|text|对于避免“if”的三元组，有一个合理的替代方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|89424|private+static+Boolean[]+array+=+{false,+true};

int+boolToInt(+boolean+b+){
++++return+Arrays.binarySearch(array,+b);
}|code-block|syntax|javascript|89425|请注意，我认为这是一个“难题”问题，因此如果我自己编写它，我将使用三元..|89426|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

A reasonable alternative to ising to the ternary to avoid an "if":

<pre><code>private static Boolean[] array = {false, true};

int boolToInt( boolean b ){
 return Arrays.binarySearch(array, b);
}
</code></pre>

Note that I consider this s "puzzle" question, so if coding it myself i would use the ternary..

blocks|key|306403|text|否则，您可以使用Apache+Commons+BooleanUtils.toInteger方法，它的工作方式非常出色……|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|306404|//+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)|code-block|syntax|javascript|306405|entityMap|0|LINK|mutability|MUTABLE|url|http://commons.apache.org/proper/commons-lang/apidocs/org/apache/commons/lang3/BooleanUtils.html^0|8|11|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

Otherwise, you could use the <a href="http://commons.apache.org/proper/commons-lang/apidocs/org/apache/commons/lang3/BooleanUtils.html" rel="nofollow">Apache Commons BooleanUtils.toInteger</a> method which works like a charm...

<pre><code>// 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)
</code></pre>

blocks|key|306407|text|我通过框架找到了一个解决方案。对布尔值使用比较。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|306408|//+b+=+Your+boolean+result
//+v+will+be+1+if+b+equals+true,+otherwise+0
int+v+=+Boolean.compare(b,+false);|code-block|syntax|javascript|306409|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

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

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

blocks|key|89464|text|int+ansInt+=+givenBoolean+?+1+:+0;|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|89465|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>int ansInt = givenBoolean ? 1 : 0;
</code></pre>

blocks|key|306427|text|目前，jdk已经提供了一个有用的工具方法：BooleanUtils.toInteger()。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|306428|在源代码中，jdk实现它的方法必须是高效的：|306429|public+static+int+toInteger(boolean+bool)+{
++++return+bool+?+1+:+0;
}|code-block|syntax|javascript|306430|因此，我认为最多票数的答案是非常好的，return+bool+?+1+:+0是最佳实践。|306431|使用BooleanUtils的示例代码如下：|306432|BooleanUtils.toInteger(false);|306433|entityMap^0|L|O|0|0|0|J|J|0|2|C|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|X|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|Y|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|Z|8|@$9|10|A|11|B|C]]|D|@]|E|$]]|$1|O|3|P|5|6|7|12|8|@$9|13|A|14|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|J|7|15|8|@]|D|@]|E|$K|L]]|$1|S|3|-4|5|6|7|16|8|@]|D|@]|E|$]]]|T|$]]

Nowadays, jdk has delivered a useful Utils method: <code>BooleanUtils.toInteger()</code>
In the source code, the method that jdk realize it must be efficient:
<pre class="lang-java prettyprint-override"><code>public static int toInteger(boolean bool) {
 return bool ? 1 : 0;
}
</code></pre>
So, I think the most votes answer is very great, <code>return bool ? 1 : 0</code> is the best practice.
Example Code to use <code>BooleanUtils</code> as followed:
<pre class="lang-java prettyprint-override"><code>BooleanUtils.toInteger(false);
</code></pre>

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

<pre><code>int boolToInt( boolean b ){
 if ( b )
 return 1
 return 0
</code></pre>

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

<pre><code>bool = True 
num = 1 * ( bool )
</code></pre>

I also figure you could do 

<pre><code>boolean bool = True;
int myint = Boolean.valueOf( bool ).compareTo( false );
</code></pre>

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).

Converting Boolean to Integer in Java without If-Statements

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我想知道是否有一种方法可以在不使用if语句的情况下将布尔值转换为int (因为不会中断管道)。例如，我可以这样写int boolToInt( boolean b ){  if ( b )    return 1  return 0但是我想知道是否有一种方法可以不使用if语句，就像Python一样bool = True ...

问在没有If语句的Java中将Boolean转换为Integer
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