undefined behaviour

Suppose

int i = 2;

i = ++i + 1;

Please explain whether the evaluation of this statement, has undefined
behaviour ?

subramanian100in@yahoo.com, India wrote:

Quote:

Suppose
>
int i = 2;
>
i = ++i + 1;
>
Please explain whether the evaluation of this statement, has undefined
behaviour ?

Yes, you're modifying an object twice without an intervening sequence
point. In such cases, the order of evaluation of the expression is
undefined by the standard, though a particular implementation might
happen to produce results that you might consider appropriate.

subramanian100in@yahoo.com, India wrote:

Quote:

Suppose
>
int i = 2;
>
i = ++i + 1;
>
Please explain whether the evaluation of this statement, has undefined
behaviour ?

In case you don't know yet, this group has a very useful FAQ at
http://www.c-faq.com/

It'll help you avoid asking questions, like this one, that've been
beaten to death over the years here.

What is a sequence point

subramanian100in@yahoo.com, India wrote:

Quote:

What is a sequence point

See comp.lang.c's FAQ at http://www.c-faq.com/

Your question is explained at:

<http://c-faq.com/expr/seqpoints.html>

On Mar 5, 9:47 am, "subramanian10...@yahoo.com, India"
<subramanian10...@yahoo.comwrote:

Quote:

Suppose
>
int i = 2;
>
i = ++i + 1;
>
Please explain whether the evaluation of this statement, has undefined
behaviour ?

Yes, it does have undefined behavior. Because the value of 'i' is
modified more than once between sequence points.

On Sun, 04 Mar 2007 20:47:20 -0800, subramanian100in@yahoo.com, India
wrote:

Quote:

Suppose
>
int i = 2;
>
i = ++i + 1;
>
Please explain whether the evaluation of this statement, has undefined
behaviour ?

You go first.

undefined behaviour

re: undefined behaviour

re: undefined behaviour

re: undefined behaviour

re: undefined behaviour

re: undefined behaviour

re: undefined behaviour

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