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The bitwise AND and OR operators (&
and |
) do not exhibit this lack short-circuit behavior. Similar to most Java operators, they evaluate both operands. They return the same Boolean result as &&
and ||
respectively but can have different overall effects depending on the presence or absence of side effects in the second operand.
Consequently, either the &
or the &&
operator can be used when performing Boolean logic. However, there are times when the short-circuiting behavior is preferred and other times when the short-circuiting behavior causes subtle bugs.
Noncompliant Code Example (Improper &
)
This noncompliant code example, derived from Flanagan [Flanagan 2005], has two variables , with no guarantees regarding their current valueswith unknown variables. The code must validate its data and then check whether array[i]
is a valid index.
Code Block | ||
---|---|---|
| ||
int array[]; // mayMay be null int i; // mayMay be an invalid index for array if (array != null & i >= 0 & i < array.length & array[i] >= 0) { // useUse array } else { // handleHandle error } |
This code can fail as a result of the same errors it is attempting trying to prevent. When array
is NULL
or when or i
is not a valid index, the reference to to array
and array[i]
will cause either a NullPointerException
or an ArrayIndexOutOfBoundsException
to be thrown. This happens The exception occurs because the &
operator fails to prevent evaluation of its right operand even when evaluation of its left operand proves that the right operand is invalidinconsequential.
Compliant Solution (Use &&
)
This compliant solution mitigates the problem by using &&
, which causes the evaluation of the conditional expression to terminate immediately if any of the conditions fail, thereby preventing a runtime exception.:
Code Block | ||
---|---|---|
| ||
int array[]; // mayMay be null int i; // mayMay be an invalid index for array if (array != null && i >= 0 && i i < array.length && array[i] >= 0) { // handleHandle array } else { // handleHandle error } |
Compliant Solution (Nested if
Statements)
This compliant solution uses multiple if
statements to achieve the proper effect. Although correct, it is more verbose and could be more difficult to maintain.
Code Block | ||
---|---|---|
| ||
int array[]; // mayMay be null int i; // mayMay be a valid index for array if (array != null) { if (i >= 0 && i < array.length) { if (array[i] !>= -10) { // useUse array } else { // handleHandle error } } else { // handleHandle error } } else { // handleHandle error } |
Although correct, this solution is more verbose and could be more difficult to maintain. Nevertheless, this solution is preferable if when the error-handling code for each potential failure condition failure were is different.
Noncompliant Code Example (Improper &&
)
This noncompliant code example demonstrates code that compares two arrays for ranges of members that match. Here i1
and i2
are valid array indices in array1
and array2
respectively. Variables end1
and end2
are expected to point to the end the indices of the ends of the matching ranges in the two arrays.
Code Block | ||
---|---|---|
| ||
if (end1 >= 0 & i2 >= 0) { int begin1 = i1; int begin2 = i2; while (++i1 < array1.length && ++i2 < array2.length && array1[i1] == array2[i2]) { // arraysArrays match so far } int end1 = i1; int end2 = i2; assert end1 - begin1 == end2 - begin2; } |
The problem with this code is that when the first condition in the while
loop fails, the second condition is does not executedexecute. That is, once i1
has reached array1.length
, the loop will terminate terminates after i1
is incremented. Consequently, the apparent range over array1
is larger than the apparent range over array2
, causing the final assertion to fail.
Compliant Solution (Use &
)
This compliant solution mitigates the problem by judiciously using &
, which guarantees that both i1
and i2
are incremented regardless of the outcome of the first condition.:
Code Block | ||
---|---|---|
| ||
public void exampleFunction() { while (++i1 < array1.length & // notNot && ++i2 < array2.length && array1[i1] == array2[i2]){ //doSomething } } |
Applicability
Failure to understand the behavior of the bitwise and conditional operators can cause unintended program behavior.
Bibliography
2§2.5.6., "Boolean Operators" | |||
|
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