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    <p>Well, the key moment here is that the result of <code>++i</code> is an <em>lvalue</em>. And in order to participate in binary <code>+</code> that lvalue has to be converted to an rvalue by lvalue-to-rvalue conversion. Lvalue-to-rvalue conversion is basically an act of reading the variable <code>i</code> from memory.</p> <p>That means that the value of <code>++i</code> is supposed to be obtained <em>as if</em> it was read directly from <code>i</code>. That means that conceptually the new (incremented) value of <code>i</code> must be ready (must be physically stored in <code>i</code>) before the evaluation of binary <code>+</code> begins. </p> <p>This extra sequencing is what inadvertently made the behavior defined in this case.</p> <p>In C++03 there was no strict requirement to obtain the value of <code>++i</code> directly from <code>i</code>. The new value could have been predicted/precalulated as <code>i + 1</code> and used as an operand of binary <code>+</code> even before it was physically stored in the the actual <code>i</code>. Although it can be reasonably claimed that the requirement was implicitly there even in C++03 (and the fact that C++03 did not recognize its existence was a defect of C++03)</p> <hr> <p>In case of <code>i++</code>, the result is an <em>rvalue</em>. Since it is already an rvalue, there's no lvalue-to-rvalue conversion involved and there's absolutely no requirement to store it in <code>i</code> before we start evaluating the binary <code>+</code>.</p>
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    1. POSo why is i = ++i + 1 well-defined in C++11?
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    1. COBut that still doesn't make it sequenced with respect to the value computation of `i`, does it?
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      1. USJoseph Mansfield
    2. CO@sftrabbit: I'm not sure what you mean. You see, the result of `++i` has two properties: 1) it is an lvalue, 2) numerically, it is the new (incremented) value of `i`. The only way to make these two requirements to coexist is to require that the modification of `i` by `++i` is sequenced *before* any further use of the result of `++i` in a surround expression. In fact, this logic applied to C++03 as well, but the intent of C++03 was to keep this behavior undefined, so C++03 refused to accept this logic.
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    3. COIt was expected that further changes to the standard will keep it as UB, i.e. something else in the standard was expected to be redesigned in order to preserve the "UB freedom" in `i = ++i + 1` and at the same time remove the above issue from the `++i` specification. However, in the end things played out differently and the changes made in C++11 actually supported the above logic instead of preventing it. So, we ended up with defined behavior in `i = ++i + 1`, even though no one was trying to legalize this code deliberately.
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