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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).

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8
  • what do you mean by "as not to break the pipeline"? Commented Jun 1, 2013 at 1:35
  • And your second piece of code does the exact same thing (i.e. it uses an if). Commented Jun 1, 2013 at 1:36
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    stackoverflow.com/questions/315306/is-if-expensive I think addresses the pipeline issue well Commented Jun 1, 2013 at 1:39
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    The question is moot. At the absolute lowest level, whatever the language, the truth remains that some branching will occur. And this kind of branching is so fast that it can happen 10^n times, n <= 10, without your having the time to blink. So, ultimately: why does it matter at all? Commented Jun 1, 2013 at 1:47
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    I won't put this as an answer as it's not faster, but here's another way to do it: return (Boolean.valueOf(b).hashCode() >> 1) & 1; The return value of Boolean.valueOf is always one of the class constants so there's no object creation overhead, but it still involves two (hidden) ifs. Commented Jun 1, 2013 at 2:07

12 Answers 12

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31

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 return b ? 1 : 0; compiles to the exact same instructions), you will see that it does not use a jump:

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

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

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1 Comment

I'm pretty convinced. Thanks!
15

Use the ?: operator: ( b ? 1 : 0 )

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9 Comments

I believe that the ternary operator is just sugar for an if-statemnent. Is that incorrect?
@mjCSguy it is essentially an inline if statement <-- over simplification
That does not answer the question of how a branch can be avoided.
@mjCSguy whether you use if or the ?: operator, it is highly likely that the JVM, due to JIT, will optimize to the same runtime bytecode eventually, so...
@assylias the question is, at the lowest level, can a branch be avoided? Answer: no.
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5

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);
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5 Comments

I think the HashMap wins the award for the slowest possible way to do this.
Ya, I appreciate that you DID probably answer the question, but I would be creating objects for primitives, to start. I actually tried that, just to see, but it's predictably really slow.
+1 The map solution did answer the question. Actually it's the only answer without a logical/imperative if, it uses the functional approach. If the HashMap is static, this solution is 6x slower than ternary operator (1000000 iterations: 6 vs 36ms). So in this case it is slower, but generally speaking, starting at a certain limit (regarding the number of values you have to map), the map/hash solution will be faster than a sequence of if statements.
@Boann i consider this to be a "puzzle" question, so performance is irrelevant.
@beryllium Hiding the if statement behind a method call can't possibly be the answer. And in HasMap code, it will be more than one if probably.
4

Otherwise, you could use the Apache Commons BooleanUtils.toInteger method which works like a charm...

// 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)
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2

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);
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How is it different? compare implementation is return (x == y) ? 0 : (x ? 1 : -1);
1

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

It's also possible that the VM is smart enough to eliminate the branch (the if or ?:) itself in this case, as the boolean's internal representation is quite likely to be the literal 1 or 0 anyway. Here is an article 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 as it performs more aggressive optimization than the "client" one.

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1

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 ;) ).

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0

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

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

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

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1 Comment

That definitely uses the ternary a few frames later. OP seems to be concerned with performance, so you might want to mention that. Also, you can do that with one less method call. Not that it matters, because you shouldn't be doing this in your code at all ;)
0

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

private static Boolean[] array = {false, true}; int boolToInt( boolean b ){ return Arrays.binarySearch(array, b); }

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

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1 Comment

Another question that just calls a method containing if statements.
0

You can try using ternary operator like this

int value = flag ? 1 : 0;
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0

Nowadays, jdk has delivered a useful Utils method: BooleanUtils.toInteger()

In the source code, the method that jdk realize it must be efficient:

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

So, I think the most votes answer is very great, return bool ? 1 : 0 is the best practice.

Example Code to use BooleanUtils as followed:

BooleanUtils.toInteger(false);
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-1 
int ansInt = givenBoolean ? 1 : 0;
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1 Comment

@Hermant, three other people already answered this question with the same answer you gave. PLS read for unique answers before adding the same answer again. :-)

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