CODEWITHSUNDEEP
c++c++11language-lawyerundefined-behavior
Khurshid
Khurshid

Reputation: 2724

here are more undefined behaviour in c++, sorry for this que[s]tion, but again UB

There is very simple UB example:

int i = 1;
i = i++; // classic example of UB.

I recently saw, how to use Pascal style inc operation. Eric Niebler github

// this structure little difference than original.
struct inc_t 
{   
  template< typename T> 
  T operator()(T& t ) const { return t++; } 
};

constexpr inc_t inc{};

//usage
int i = 1;
inc(i);
//or
int j = inc(i);

So, combine :

int i = 1;
i = inc(i); // Is there still UB ?

Thanks.

Upvotes: 2

Views: 128

Answers (1)

Brian Bi
Brian Bi

Reputation: 119099

No.

i = i++ is UB because it's not specified when i is incremented; it could be at any point after the value computation of i++ (i.e., lvalue-to-rvalue conversion of i) and before the end of the full-expression. Therefore this write is unsequenced with respect to the assignment statement. Two unsequenced writes to the same int is undefined behaviour.

But in i = inc(i) the incrementation of i occurs before the function returns, because the full-expression in which t++ occurs lies inside the function. In turn, the function has to return before the value computation of the right-hand side, and the assignment is sequenced after the value computation of both sides. Therefore the incrementation of i is sequenced before the assignment, and there is no UB.

Relevant quotes from the C++11 standard:

§1.9/14

Every value computation and side effect associated with a full-expression is sequenced before every value computation and side effect associated with the next full-expression to be evaluated.

§5.17/1

In all cases, the assignment is sequenced after the value computation of the right and left operands, and before the value computation of the assignment expression.

Upvotes: 9

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