This started from a joke:
Interviewer: What is the difference between C and C++?
Candidate: ONE
My question is whether the expressions abs(C++ - C) and abs(C - C++) invokes undefined behavior or not?
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Quixotic
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This is not only undefined but also unspecified behavior since the order of evaluation of sub-expressions is not specified. – Shafik Yaghmour Feb 13 '14 at 15:54
3 Answers
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It depends on the type of C, but at the best (a user defined
type, where ++ is a function), it is unspecified whether the
second C is evaluated before or after the evaluation of
C.operator++.
Of course, for a built-in type, the expression is undefined
behavior, and for a user defined type, the final results will
also depend on how the user defined operator++, as well as the
compiler dependent order of evaluation.
James Kanze
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Yes, this is undefined behaviour. The compiler will not make any promises on when the increment will happen if you reuse the same variable in the statement.
Sergey L.
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It's only undefined behavior if `C` has a built-in type. Otherwise, it's only unspecified (and with a limited set of possible values---supposing that the user defined `operator++` is deterministic). – James Kanze Feb 13 '14 at 15:53
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yes this is UB. From C99, Section 6.5
An expression is a sequence of operators and operands that specifies computation of a value
Except as specified later (for the function-call (), &&, ||, ?:, and comma operators), the order of evaluation of subexpressions and the order in which side effects take place are both unspecified
Therefore the is no guarantee in the express C++ - C when the post increment is executed.
Jimbo
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