Quick question. How the following code will perform order of check :
if ((NULL != ObjectPtr) && (ObjectPtr->isValid()) )
{
}
Is the order on if-statement depend on compiler used? Could that code crash if ObjectPtr is NULL?
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1Too many parentheses.
– Pete Becker Jul 25 '17 at 21:58 -
[Yoda Conditions](https://en.wikipedia.org/wiki/Yoda_conditions) were relevant to C++ 20 years ago but that's no longer the case. – Captain Obvlious Jul 25 '17 at 22:02
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1This is an obvious duplicate, come on 120k and 200k users. – user229044 Jul 25 '17 at 22:05
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2@meagar: You can't blame anyone for not finding that question, its title is 100% wrong (now fixed). – Ben Voigt Jul 25 '17 at 22:07
2 Answers
3
Is the order depend on
ifstatement depend on compiler used?
No.
Could that code crash if
ObjectPtrisNULL?
No.
The language guarantees that.
R Sahu
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In C++, the && operator is guaranteed to be short-circuiting. This means that the left-hand operand is checked first, and if it is false, none of the right-hand operand is evaluated.
So your code is safe and will not perform member access through the NULL pointer.
Similarly, || is also short-circuiting and will not evaluate any of its right-hand operand if the left operand is true.
With boolean operands, the bitwise operators & and | give the same result as the logical operators && and ||, but the bitwise operators do not short-circuit and their right-hand operand is always evaluated (possibly before the left-hand one).
Also as Quentin mentions in a comment, user-provided overloads of these operators do not short-circuit, because they are actually function calls and have the evaluation order of function calls (all arguments evaluated before the call).
Ben Voigt
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Note: this only works with the built-in operators (which may or may not be obvious). – Quentin Jul 25 '17 at 22:03