Why do we get different results for the two follow statements:
int x = 5;
x = x * x++;
the output is 26; whereas the next example returns 30, though they are same?!
int x = 5;
x *= x++;
Thank you
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Abdulkarim Kanaan
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9Isn't that undefined? (both of them) – user2345215 Jan 01 '14 at 22:26
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4Both are undefined behavior. The stored value of a variable shall not be modified more than once between sequence points. – IInspectable Jan 01 '14 at 22:27
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Do you mean that the behaviour is unexpected?! – Abdulkarim Kanaan Jan 01 '14 at 22:28
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The behaviour is implementation-dependent. You can get different results depending on the compiler used. – kuroi neko Jan 01 '14 at 22:30
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2*Undefined behaviour* means, you are wrong no matter what you expect. *Undefined behaviour* means, you do not have any justified reason to expect any particular result. – Oswald Jan 01 '14 at 22:30
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1@kuroi No, the behavior is genuinely undefined. *Implementation-defined* is another class, for example, how the `NULL` macro is defined is implementation-defined. – IInspectable Jan 01 '14 at 22:31
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1@llnspectable : you're right. I was too quick to answer this one. My bad. – kuroi neko Jan 01 '14 at 22:42
2 Answers
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These both exhibit undefined behaviour in both C++03 and C++11. In C++11 terminology, you can't have two unsequenced modifications of the same scalar or a modification and a value computation using the same scalar, otherwise you have undefined behaviour.
x = x * x++;
In this case, incrementing x (a modification) is unsequenced with the other two value computation of x.
x *= x++;
In this case, incrementing x is unsequenced with the value computation of x on the left.
For the meaning of undefined behaviour, see C++11 §1.3.24:
undefined behavior
behavior for which this International Standard imposes no requirements
Joseph Mansfield
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1*"incrementing `x` is unsequenced"* that is, the modification of `x` is unsequenced to the other two value computations. (It is sequenced after the value computation of the `x++`.) – dyp Jan 01 '14 at 22:33
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I'm just wondering, what's the reasoning for undefined and not just unspecified evaluation order? – user2345215 Jan 01 '14 at 22:33
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@user2345215 I'm fairly sure it's to allow the compilers to perform optimizations under the assumption that the programmer doesn't do this. Somebody else might have a better answer though. – Joseph Mansfield Jan 01 '14 at 22:38
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2The compiler is supposed to let some operations run simultaneously (on different CPU cores). Hardware might do bizarre things (e.g. generate junk or throw an exception) when 2 cores read and write the same variable simultaneously. Since UB allows doing these things, compiler has much less to worry about when deciding how to do parallelization. – anatolyg Jan 01 '14 at 22:56
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Assigning a ++'d value to the original value is undefined behavior. So it makes sense that assigning a ++'d then multiplied value is also undefined.
BWG
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