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Is there any difference between and with bool (s)? 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
Using bitwise operators for Booleans in C++ - Stack Overflow However might I suggest reducing 'if' statements to readable english wherever possible by using well-named boolean vars For example, and this is using boolean operators but you could equally use bitwise and name the bools appropriately:
How do I create a numpy array of all True or all False? It seems more natural to fill an array with bools, than to fill it with numbers to cast them to bools The ones and zeros answers do not construct an array of integers They build an array of bools directly Is numpy full((2,2), True) an equivalent? It is in numpy 1 12+ I dont remember whether it applies to former versions either
c++ - How is a bool represented in memory? - Stack Overflow The standard doesn't mandate anything for the binary representation of bools; it just says that, when converting to other integral types, a true bool will become 1 and a false bool will become 0 This of course suggests an implementation similar in spirit to the one you said, where such conversions would become essentially no-ops or plain integer widening (but remember that bool is mandated to
Why use flags+bitmasks rather than a series of booleans? It uses considerably less memory Putting all 4 of your example values in a bitmask would use half a byte Using an array of bools, most likely would use a few bytes for the array object plus a long word for each bool If you have to store a million values, you'll see exactly why a bitmask version is superior
How to create a byte out of 8 bool values (and vice versa)? How they work Suppose we have 8 bools b[0] to b[7] whose least significant bits are named a-h respectively that we want to pack into a single byte Treating those 8 consecutive bool s as one 64-bit word and load them we'll get the bits in reversed order in a little-endian machine Now we'll do a multiplication (here dots are zero bits)