问题
so if I have x = 0x01 and y = 0xff and I do x & y, I would get 0x01. If I do x && y do I still get 0x01, but the computer just says its true if its anything than 0x00? My question is are the bits the same after the operation regardless of bit-wise or logical AND/OR, but the computer just interprets the final result differently? In other words are the hex values of the result of & and && (likewise | and ||) operations the same?
edit: talking about C here
回答1:
The answer depends on the precise language in use.
In some weakly-typed languages (e.g. Perl, JavaScript) a && b will evaluate to the actual value of b if both a and b are "truthy", or a otherwise.
In C and C++ the && operator will just return 0 or 1, i.e. the int equivalent of true or false.
In Java and C# it's not even legal to supply an int to && - the operands must already be booleans.
I don't know of any language off the top of my head where a && b == a & b for all legal values of a and b.
回答2:
In C#, the operators behave differently: && is only allowed for boolean types and & for any integer type as well.
a && b returns true, if both a and b are true. a & b returns the result of the bitwise AND operation.
来源：https://stackoverflow.com/questions/28282002/bitwise-and-logical-and-or-in-terms-of-hex-result