Is there any difference between && and & with bool(s)?

Question

In C++, is there any difference between doing && (logical) and & (bitwise) between bool(s)?

bool val1 = foo();
bool val2 = bar();

bool case1 = val1 & val2;
bool case2 = val1 && val2;

Are case1 and case2 identical or if not how exactly do they vary and why would one choose one over the other? Is a bitwise and of bools portable?

Answer 1

The standard guarantees that false converts to zero and true converts to one as integers:

4.7 Integral conversions

...

If the destination type is bool, see 4.12. If the source type is bool, the value false is converted to zero and the value true is converted to one.

So the effect in the example you give is guaranteed to be the same and is 100% portable.

For the case you give, any decent compiler is likely to generate identical (optimal) code.

However, for Boolean expressions expr1 and expr2, it is not true in general that expr1 && expr2 is the same as expr1 & expr2 because && performs "short-circuit" evaluation. That is, if expr1 evaluates to false, expr2 will not even be evaluated. This can affect performance (if expr2 is complicated) and behavior (if expr2 has side-effects). (But note that the & form can actually be faster if it avoids a conditional branch... Toying with this sort of thing for performance reasons is almost always a bad idea.)

So, for the specific example you give, where you load the values into local variables and then operate on them, the behavior is identical and the performance is very likely to be.

In my opinion, unless you are specifically relying on the "short-circuit" behavior, you should choose the formulation that most clearly expresses your intention. So use && for logical AND and & for bit-twiddling AND, and any experienced C++ programmer will find your code easy to follow.