Using bitwise operators for Booleans in C++

Question

Is there any reason not to use the bitwise operators &, |, and ^ for "bool" values in C++?

I sometimes run into situations where I want exactly one of two conditions to be true (XOR), so I just throw the ^ operator into a conditional expression. I also sometimes want all parts of a condition to be evaluated whether the result is true or not (rather than short-circuiting), so I use & and |. I also need to accumulate Boolean values sometimes, and &= and |= can be quite useful.

I've gotten a few raised eyebrows when doing this, but the code is still meaningful and cleaner than it would be otherwise. Is there any reason NOT to use these for bools? Are there any modern compilers that give bad results for this?

Solution 1

|| and && are boolean operators and the built-in ones are guaranteed to return either true or false. Nothing else.

|, & and ^ are bitwise operators. When the domain of numbers you operate on is just 1 and 0, then they are exactly the same, but in cases where your booleans are not strictly 1 and 0 – as is the case with the C language – you may end up with some behavior you didn't want. For instance:

BOOL two = 2;
BOOL one = 1;
BOOL and = two & one;   //and = 0
BOOL cand = two && one; //cand = 1

In C++, however, the bool type is guaranteed to be only either a true or a false (which convert implicitly to respectively 1 and 0), so it's less of a worry from this stance, but the fact that people aren't used to seeing such things in code makes a good argument for not doing it. Just say b = b && x and be done with it.

Solution 2

Two main reasons. In short, consider carefully; there could be a good reason for it, but if there is be VERY explicit in your comments because it can be brittle and, as you say yourself, people aren't generally used to seeing code like this.

Bitwise xor != Logical xor (except for 0 and 1)

Firstly, if you are operating on values other than false and true (or 0 and 1, as integers), the ^ operator can introduce behavior not equivalent to a logical xor. For example:

int one = 1;
int two = 2;

// bitwise xor
if (one ^ two)
{
  // executes because expression = 3 and any non-zero integer evaluates to true
}

// logical xor; more correctly would be coded as
//   if (bool(one) != bool(two))
// but spelled out to be explicit in the context of the problem
if ((one && !two) || (!one && two))
{
  // does not execute b/c expression = ((true && false) || (false && true))
  // which evaluates to false
}

Credit to user @Patrick for expressing this first.

Order of operations

Second, |, &, and ^, as bitwise operators, do not short-circuit. In addition, multiple bitwise operators chained together in a single statement -- even with explicit parentheses -- can be reordered by optimizing compilers, because all 3 operations are normally commutative. This is important if the order of the operations matters.

In other words

bool result = true;
result = result && a() && b();
// will not call a() if result false, will not call b() if result or a() false

will not always give the same result (or end state) as

bool result = true;
result &= (a() & b());
// a() and b() both will be called, but not necessarily in that order in an
// optimizing compiler

This is especially important because you may not control methods a() and b(), or somebody else may come along and change them later not understanding the dependency, and cause a nasty (and often release-build only) bug.

Solution 3

I think

a != b

is what you want

Solution 4

The raised eyebrows should tell you enough to stop doing it. You don't write the code for the compiler, you write it for your fellow programmers first and then for the compiler. Even if the compilers work, surprising other people is not what you want - bitwise operators are for bit operations not for bools.
I suppose you also eat apples with a fork? It works but it surprises people so it's better not to do it.

Solution 5

Disadvantages of the bitlevel operators.

You ask:

“Is there any reason not to use the bitwise operators &, |, and ^ for "bool" values in C++? ”

Yes, the logical operators, that is the built-in high level boolean operators !, && and ||, offer the following advantages:

• Guaranteed conversion of arguments to bool, i.e. to 0 and 1 ordinal value.

• Guaranteed short circuit evaluation where expression evaluation stops as soon as the final result is known.
  This can be interpreted as a tree-value logic, with True, False and Indeterminate.

• Readable textual equivalents not, and and or, even if I don't use them myself.
  As reader Antimony notes in a comment also the bitlevel operators have alternative tokens, namely bitand, bitor, xor and compl, but in my opinion these are less readable than and, or and not.

Simply put, each such advantage of the high level operators is a disadvantage of the bitlevel operators.

In particular, since the bitwise operators lack argument conversion to 0/1 you get e.g. 1 & 2 → 0, while 1 && 2 → true. Also ^, bitwise exclusive or, can misbehave in this way. Regarded as boolean values 1 and 2 are the same, namely true, but regarded as bitpatterns they're different.

---

How to express logical either/or in C++.

You then provide a bit of background for the question,

“I sometimes run into situations where I want exactly one of two conditions to be true (XOR), so I just throw the ^ operator into a conditional expression.”

Well, the bitwise operators have higher precedence than the logical operators. This means in particular that in a mixed expression such as

a && b ^ c

you get the perhaps unexpected result a && (b ^ c).

Instead write just

(a && b) != c

expressing more concisely what you mean.

For the multiple argument either/or there is no C++ operator that does the job. For example, if you write a ^ b ^ c than that is not an expression that says “either a, b or c is true“. Instead it says, “An odd number of a, b and c are true“, which might be 1 of them or all 3…

To express the general either/or when a, b and c are of type bool, just write

(a + b + c) == 1

or, with non-bool arguments, convert them to bool:

(!!a + !!b + !!c) == 1

---

Using &= to accumulate boolean results.

You further elaborate,

“I also need to accumulate Boolean values sometimes, and &= and |=? can be quite useful.”

Well, this corresponds to checking whether respectively all or any condition is satisfied, and de Morgan’s law tells you how to go from one to the other. I.e. you only need one of them. You could in principle use *= as a &&=-operator (for as good old George Boole discovered, logical AND can very easily be expressed as multiplication), but I think that that would perplex and perhaps mislead maintainers of the code.

Consider also:

struct Bool
{
    bool    value;

    void operator&=( bool const v ) { value = value && v; }
    operator bool() const { return value; }
};

#include <iostream>

int main()
{
    using namespace std;

    Bool a  = {true};
    a &= true || false;
    a &= 1234;
    cout << boolalpha << a << endl;

    bool b = {true};
    b &= true || false;
    b &= 1234;
    cout << boolalpha << b << endl;
}

Output with Visual C++ 11.0 and g++ 4.7.1:

true
false

The reason for the difference in results is that the bitlevel &= does not provide a conversion to bool of its right hand side argument.

So, which of these results do you desire for your use of &=?

If the former, true, then better define an operator (e.g. as above) or named function, or use an explicit conversion of the right hand side expression, or write the update in full.

View more solutions

Comments

[Greg Rogers]

no such thing as logical xor operator ^^ in C++. For both arguments as bools != does the same thing, but can be dangerous if either argument is not a bool (as would be using ^ for a logical xor). Not sure how this got accepted...

[Greg Rogers]

Not really a fair comparison if you cast to bool in one case and not in the other case... The cast to bool in both cases is what would make it work, and why it is fragile (because you have to remember it).

[Patrick]

In retrospect, that was kinda dumb. Anyway, changed the example to an && to fix the problem. I swear I've used ^^ sometime in the past, but this must not have been the case.

[Johannes Schaub - litb]

C has a bool type that can only be 0 or 1 since 10 years now. _Bool

[Johannes Schaub - litb]

anyway. i think !!two & !!one will fix the first and = 0 case. !! will make any nonzero stuff into 1.

[Exectron]

A ^^ operator would still be useful regardless of the short-circuit consideration. It would a) evaluate the operands in a boolean context and b) guarantee to return a 1 or 0.

[sage]

Patrick, since you have the accepted answer it would be great if you would add something about accumulating with &= and |=. After reading the entire thread and testing something, I checked and learned that gcc gives false after bool passed_tests_ = true; passed_tests_ &= 8;, which means that one of my tests is fragile to later editing... (looking for this is what brought me here)

[sage]

Your post motivated me to test this and gcc did not warn! Bummer, because I was going to use that warning to justify letting me run &= to accumulate bool results, trusting that others would see the warning if they later change my test. My code would fail for even integers, since they evaluate to 0/false - without warnings! Time to refactor...

[aaronman]

They are still no the same even for bools because they do not short circut

[Eloff]

Yes, this should be the top answer. Follow the principle of least surprise. Code that does unusual things is harder to read and understand, and code is read far more often than it is written. Don't use cutesy tricks like this.

[Antimony]

bitwise operators also have text equivalents

[Cheers and hth. - Alf]

@Antimony: I didn't understand that comment at first but yes, the bitlevel operations have alternative tokens bitand, bitor, xor and compl. I think that's why I used the qualification "readable". Of course, the readability of e.g. compl 42 is subjective. ;-)

[Marc van Leeuwen]

This is true, +1. But there is no assigning version of this operator (!== so to speak) so if you are computing the XOR of a sequence of bool values you would need to write acc = acc!= condition(i); in a loop body. The compiler probably can handle this as efficiently as if !== existed, but some might find that it does not look nice, and prefer the alternative of adding the booleans as integers, and then testing whether the sum is odd.

[Marc van Leeuwen]

@CraigMcQueen. It is simple to define inline bool XOR(bool a,bool b) { return a!=b; } and get what you want, except that it is a function rather than an (infix) operator. Or you could directly use != or overload some other operator with this meaning, but then of course you need to be very careful to not accidentally end up using an unintended overload of the same name. And incidentally || and && return true or false, not 1 or 0.

[Marc van Leeuwen]

And it seems that the fact that !, || and && evaluate their arguments in a Boolean context is only because these operators are rarely overloaded to accept other types; as far as I can tell the language does allow such overloads, and therefore does not guarantee evaluation of arguments in a Boolean context.

[Marc van Leeuwen]

As far as I can tell logical operators can be overloaded with other argument types (though they usually are not), so the first point "guaranteed conversion of arguments to bool" is just a consequence of convention, not a guarantee that the C++ language actually gives you.

[Cheers and hth. - Alf]

@MarcvanLeeuwen: Possibly your visual subsystem failed to notice “the built-in high level boolean operators !, && and ||”. You're right that these operators can be overloaded, and then a main problem is that the semantics change subtly, no longer short circuit evaluation. But that's a problem with overloads, not with the built-in operators.

[Marc van Leeuwen]

That is right (I did miss that). But then the statement is still misleading, because you are just saying that those operators take boolean arguments; if (and only if) the built-in versions of the operators get selected, then the compiler will have inserted a conversion of the arguments. Similarly the built-in (high level) unsigned integer addition operator + guarantees conversion of its arguments to unsigned, but that does not prevent unsigned int n=-1; n+=3.14; from actually using the double (or is it float?) addition operation. (Compare your &= example.)

[Cheers and hth. - Alf]

@MarcvanLeeuwen: Sorry, now you're into what appears to be a nonsense defense. But possibly you're unaware that operators can't be overloaded for built-in types. Guarantees about the built-in operators are guarantees about the built-in operators.

[codetaku]

Yeah, I strongly believe this should not be the accepted answer until you change "When the domain of numbers you operate on is just 1 and 0, then they are exactly the same" to "When the domain of numbers you operate on is just 1 and 0, the only difference is that the bitwise operators do not short-circuit". The former statement is flatly wrong, so in the meanwhile I have downvoted this answer since it contains that sentence.

[codetaku]

The explanation of short-circuiting is why this should absolutely be the accepted answer; Patrick's answer (err... the other Patrick, who is just named Patrick) is completely wrong in saying "When the domain of numbers you operate on is just 1 and 0, then they are exactly the same"

[anton_rh]

I'm here because my colleague used bitwise "and" operator for bool. And now I'm spending my time trying to understand if this is correct or not. If he didn't do this I would be doing something more useful now -_-. If you value time of your colleagues PLEASE don't use bitwise operators for bool!