Oct 16, 2017
New from The Xfer (deadmau5 Remixes ) VST & AU. VST & AU FULLY MIXED WITH. A super fast and easy to use midi mapping program.
Sep 23, 2018
Deadmau5 Xfer Samplepack.rar mega
Mar 27, 2019
If you're looking for an original sample pack you can use without getting into. This Drum and Vocal sample pack will help. sound editor vst dr dn masterdg.com/item/xfer-drum-vocal-sample-pack/
Q:
Bitwise Masking for Logical Operators
In a bitwise "and" operation, my understanding is that for an "a & b" it is equivalent to "a & ~b".
However, what is the bitwise meaning of the following?
if (a | b | ~a | ~b)
return false;
Is this simply equivalent to
if (a) return false;
Or should it be equivalent to
if (b) return false;
The second option would make sense to me if "a" and "b" were some non-zero bit values, but here they are zero.
A:
| has higher precedence than &, therefore, b | ~a | ~b is equivalent to a | ~a | ~b.
Your test is equivalent to:
if ((a | b) | ~a | ~b)
return false;
which is equivalent to:
if ((a | ~a) & (b | ~b))
return false;
In particular, it is equivalent to:
if ((a & ~a) & (b & ~b))
return false;
which is equivalent to:
if ((a & a) & (b & b))
return false;
which is equivalent to:
if (a)
return false;
which makes sense.
The only difference is that you have to be careful when you use &&. If the first expression is true, then the whole expression is true (hence the name, logical AND), but if the first expression is false, then the whole expression is false. However, this is only a difference in semantics. The C++ standard does not require the compiler to ac619d1d87
Related links: