184.108.40.206 Combining conditions with && and ||
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
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
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
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";