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The logical operators && (“and”) and || (“or”) combine conditional expressions; the ! (“not”) operator negates them. The ! has the highest precedence, then &&, then ||; you will need parentheses to force a different order.
(Beware of the bitwise “and” and “or” operators & and | – these should never be used to combine conditions; they are for set-and-mask bit fiddling.)
The operators for comparing numeric values are == (equal), != (not equal), > (greater), < (less), >= (greater or equal), and <= (less or equal). These all have higher precedence than &&, but lower than ! or any arithmetic operators.
if ((a<b && x>a && x<b) || (a>b && x<a && x>b)) write, "x is between a and b" |
Here, the expression for the right operand to && will execute only if its left operand is actually true. Similarly, the right operand to || executes only if its left proves false. Therefore, it is important to order the operands of && and || to put the most computationally expensive expression on the right — even though the logical “and” and “or” functions are commutative, the order of the operands to && and || can be critical.
In the example, if a>b, the x>a and x<b subexpressions will not actually execute since a<b proved false. Since the left operand to || was false, its right operand will be evaluated.
Despite the cleverness of the && and || operators in not executing the expression for their right operands unless absolutely necessary, the example has obvious inefficiencies: First, if a>=b, then both a<b and a>b are checked. Second, if a<b, but x is not between a and b, the right operand to || is evaluated anyway. Yorick has a ternary operator to avoid this type of inefficiency:
expr_A_or_B= (condition? expr_A_if_true : expr_B_if_false); |
The ?: operator evaluates the middle expression if the condition is true, the right expression otherwise. Is that so? Yes : No. The efficient betweeness test reads:
if (a<b? (x>a && x<b) : (x<a && x>b)) write, "x is between a and b"; |
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