ios,swift,binary-operators,unichar

Since unichar is a type aliased to a 16-bit integer, you need to "wrap" it in UnicodeScalar function for the comparison: if UnicodeScalar(character) == "1" { //do this thingy } ...

c#,bitwise-operators,binary-operators,.net-reflector,ulong

How about casting num7 to a ulong? if (!flag && ((ulong)num7 < (((ulong) endLocation) - (((ulong) 4L) + num6)))) This has nothing to do with bitwise operators, by the way....

r = !(~(v>>16)) << 4; Here's what's happening on this line: v >> 16 shifts the value of v (which at this time is the value of x) 16 places to the right. The bottom 16 bits of this expression are the top 16 bits of v, and the top...

matlab,for-loop,binary-operators

arrayfun is your friend for things like this, hopefully you can see how to modify this for your own use: A=randi(5,10,10) B=rand(10) C=arrayfun(@(i) B(A==i),1:5,'UniformOutput',false) C{1} % for example, gives B(A==1) C is a cell array. Beware that the for loop may be faster for larger problems. It would be a...

c,unary-operator,integer-promotion,binary-operators

Usually, yes, unless on your platform uint32_t would be a narrow type. Then it would first be promoted to int and the negation would be made in that type.

c++,operator-overloading,operators,overloading,binary-operators

This program has indeterminate behavior: the compiler is not required to evaluate i[0] and i[2] in a left-to-right order (the C++ language gives this freedom to compilers in order to allow for optimizations). For instance, Clang does it in this instance, while GCC does not. The order of evaluation is...

c#,cil,il,notnull,binary-operators

Short answer: There is no "compare-not-equal" instruction in IL, so the C# != operator has no exact correspondence and cannot be translated literally. There is however a "compare-equal" instruction (ceq, a direct correspondence to the == operator), so in the general case, x != y gets translated like its slightly...

javascript,ternary-operator,logical-operators,binary-operators

Your code has a couple of issues: Don't do that; it's confusing {} is not an empty block; it's an object literal. "" or 0 or 42 would have the same effect. The && operator will do what you want. ...

c#,.net,dynamic-language-runtime,binary-operators

dynamic d1 = d + "add it"; That's not a binary operation, that's string concatenation. Documented in the MSDN article for DynamicObject.TryBinaryOperation(), the Add operation has this description: An addition operation without overflow checking, for numeric operands. The binder already knows how to concatenate strings. All that's required is...