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The conditional AND and OR operators (&& and ||, respectively) exhibit short-circuit behavior. That is, the second operand is evaluated only when the result of the conditional operator cannot be deduced solely by evaluating the first operand. Consequently, when the result of the conditional operator can be deduced solely from the result of the first operand, the second operand will remain unevaluated; its side effects, if any, will never occur.

The bitwise AND and OR operators (& and |) do not exhibit this behavior. Like most other Java operators, they evaluate both operands, : first the left operand , and then the right. 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.

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This code can fail as a result of the same errors it is attempting to prevent. When {{array}} is null{{NULL}} or when {{i}} is not a valid index, the reference to {{array\[i\]}} will cause a {{NullPointerException}} or an {{ArrayIndexOutOfBoundsException}} to be thrown. This happens because the {{&}} operator fails to prevent evaluation of its right operand, even when evaluation of its left operand proves that the right operand is invalid.

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This compliant solution mitigates the problem by using &&, which causes the conditional expression to terminate immediately if any of the conditions fail, thereby preventing a potentially - invalid evaluation.

int array[]; // may be null
int i;       // may be a valid index for array
if (array != null &&
    i >= 0 & i < array.length &&
    array[i] >= 0) {
  // handle array
} else {
  // handle error
}

Compliant Solution (Nested

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if Statements)

This compliant solution uses multiple if statements to achieve the proper effect. While Although correct, it is more verbose and could be more difficult to maintain.

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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. It is expected that end1 and end2 will are expected to point to the end of the matching ranges in the two arrays.

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The problem with this code is that when the first condition in the while loop fails, the second condition is not executed. That is, once i1 has reached array1.length, the loop could terminate after i1 is executed. Consequently, the range over array1 is larger than the range over array2, causing the final assertion to fail.

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This compliant solution mitigates the problem by using &, which guarantees that both i1 and i2 are incremented , regardless of the outcome of the first condition.

  while (++i1 < array1.length &     // not &&
         ++i2 < array2.length &&
         array1[i1] == array2[i2])

Compliant Solution (Nested

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if Statements)

This compliant solution uses multiple if statements to achieve the proper effect. While Although correct, it is more verbose.

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CERT C Secure Coding Standard: "EXP02-C. Be aware of the short-circuit behavior of the logical AND and OR operators"
CERT C++ Secure Coding Standard: "EXP02-CPP. Be aware of the short-circuit behavior of the logical AND and OR operators"

Bibliography

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