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POUndefined behavior and sequence points reloaded
 

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    <p>Consider this topic a sequel of the following topic:</p> <blockquote> <p><strong>Previous installment</strong> <br/> <em><a href="https://stackoverflow.com/questions/4176328">Undefined behavior and sequence points</a></em></p> </blockquote> <p>Let's revisit this <em>funny</em> and <em>convoluted</em> expression (the italicized phrases are taken from the above topic *smile* ):</p> <pre><code>i += ++i; </code></pre> <p>We say this invokes undefined-behavior. I presume that when say this, we implicitly assume that <em>type</em> of <code>i</code> is one of the built-in types.</p> <p>What if the <em>type</em> of <code>i</code> is a user-defined type? Say its type is <code>Index</code> which is defined later in this post (see below). Would it still invoke undefined-behavior?</p> <p>If yes, why? Is it not equivalent to writing <code>i.operator+=(i.operator++());</code> or even syntactically simpler <code>i.add(i.inc());</code>? Or, do they too invoke undefined-behavior?</p> <p>If no, why not? After all, the object <code>i</code> gets modified <strong>twice</strong> between consecutive sequence points. Please recall the rule of thumb: <a href="http://msdn.microsoft.com/en-us/library/d45c7a5d%28v=vs.90%29.aspx" rel="nofollow noreferrer">an expression can modify an object's value only once between consecutive "sequence points</a>. And if <code>i += ++i</code> is an expression, then it must invoke undefined-behavior. If so, then its equivalents <code>i.operator+=(i.operator++());</code> and <code>i.add(i.inc());</code> must also invoke undefined-behavior which seems to be untrue! (as far as I understand)</p> <p>Or, <code>i += ++i</code> is not an <em>expression</em> to begin with? If so, then what is it and what is the definition of <em>expression</em>?</p> <p>If it's an expression, and at the same time, its behavior is <strong>also</strong> well-defined, then it implies that the number of sequence points associated with an expression somehow depends on the <em>type</em> of operands involved in the expression. Am I correct (even partly)?</p> <hr> <p>By the way, how about this expression?</p> <pre><code>//Consider two cases: //1. If a is an array of a built-in type //2. If a is user-defined type which overloads the subscript operator! a[++i] = i; //Taken from the previous topic. But here type of `i` is Index. </code></pre> <p>You must consider this too in your response (if you know its behavior for sure). :-)</p> <hr> <p>Is</p> <pre><code>++++++i; </code></pre> <p>well-defined in C++03? After all, this is this,</p> <pre><code>((i.operator++()).operator++()).operator++(); </code></pre> <hr> <pre><code>class Index { int state; public: Index(int s) : state(s) {} Index&amp; operator++() { state++; return *this; } Index&amp; operator+=(const Index &amp; index) { state+= index.state; return *this; } operator int() { return state; } Index &amp; add(const Index &amp; index) { state += index.state; return *this; } Index &amp; inc() { state++; return *this; } }; </code></pre>
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    1. CO+1 great question, which inspired great answers. I feel I ought to say that it's still horrible code which should be refactored to be more readable, but you probably know that anyway :)
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    2. COsince when s++ is same as ++s?
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    3. CO@What is the Question: who said it's same? or who said it's not same? Does it not depend on how you implement them? (Note: I'm assuming type of `s` is user-defined type!)
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