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463

answers:

4
  1. Why is operator '&' defined for bool?, and operator '&&' is not?
  2. How exactly does this 1) bool? & bool? and 2) bool? and bool work?

Any other "interesting" operator semantics on Nullable? Any overloaded operators for generic T?

+1  A: 

Operator "&" is bitwise operator, ¨whereas "&&" is logical operator. The logcial "AND" operator can not be aplied to the threestate value, only for twostate.

TcKs
A: 

-- In addition, the logical && operator along with the || operator cannot be overloaded.

2) The reason that the bitwise operator works is because it operates on single bits, which is a simple 1 or 0. This is what boolean operators essentially are, true and false, 1 and 0.

Craig Wilson
+2  A: 

There are no short-circuiting operators (&& ||) defined for bool?

Only are the logical AND, inclusive OR, operators and they behave like this:

x        y      x & y   x | y   
true    true true true
true    false false true
true    null null true
false   true false true
false   false false false
false   null false null
null    true null true
null    false false null
null    null null null
CMS
badbadboy
They are "short-circuiting" because they only evaluate multiple conditions in a logical statement if necessary. See http://developer.mozilla.org/en/Core_JavaScript_1.5_Reference/Operators/Logical_Operators for more information.
Scott Dorman
+6  A: 

Operators on Nullable<T> are "lifted" operators. What this means is: if T has the operator, T? will have the "lifted" counterpart.

&& and || aren't really operators in the same sense as & and | - for example, they can't be overloaded - from the ECMA spec 14.2.2 Operator overloading:

The overloadable binary operators are: + - * / % & | ^ << >> == != > < >= <= Only the operators listed above can be overloaded. In particular, it is not possible to overload member access, method invocation, or the =, &&, ||, ??, ?:, checked, unchecked, new, typeof, as, and is operators.

Likewise, from the ECMA spec, 14.2.7 Lifted operators, the lifted operators are:

For the unary operators + ++ - -- ! ~

For the binary operators + - * / % & | ^ << >>

For the equality operators == !=

For the relational operators < > <= >=

So basically, the short-circuiting operators aren't defined as lifted operators.

[edit: added crib sheet]

  • Lifted operator: a compiler provided operator on Nullable<T>, based on the operators of T - for example: the int "+" operator gets "lifted" onto int?, defined as:

    (int? x, int? y) => (x.HasValue && y.HasValue) ? (x.Value + y.Value) : (int?) null;

  • Operator overloading: the act of providing a custom operator implementation for a given type; for example decimal and DateTime provide various operator overloads

  • Short-circuiting: the normal behavior of && and || (in many languages, including C++ and C#) - i.e. the second operand might not be evaluated - i.e.

    (expression1, expression2) => expression1() ? expression2() : false;

Or perhaps a simpler example:

bool someFlag = Method1() && Method2();

if Method1() returns false, then Method2() isn't executed (since the compiler already knows that the overall answer is false). This is important if Method2() has side-effects, since as saving to the database...

Marc Gravell
badbadboy
I'll add a crib sheet ;-p
Marc Gravell
@Marc Gravell: thanks!!
badbadboy