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    <p>I think it's allowed to print out true, or false, or for somewhat unrelated reasons, nothing at all.</p> <p>The true or false part is (as you've said), that the destruction of the temporary <code>A</code> object is not ordered with respect to the dynamic initialization of <code>b</code>.</p> <p>The nothing at all possibility is because the initialization of <code>b</code> is not ordered with respect to the creation/initialization of <code>std::cout</code>; when you try to destroy the temporary, <code>cout</code> may not have been created/initialized yet, so attempting to print something may not work at that point at all. [Edit: this is specific to C++98/03, and does not apply to C++11.]</p> <p>Edit: here is how I, at least, see the sequence:</p> <p><img src="https://i.stack.imgur.com/INTq5.png" alt="enter image description here"></p> <p>Edit2: After rereading §12.2/4 (yet again), I've changed the diagram again. §12.2/4 says:</p> <blockquote> <p>There are two contexts in which temporaries are destroyed at a different point than the end of the full expression. The first context is when an expression appears as an initializer for a declarator defining an object. In that context, the temporary that holds the result of the expression shall persist until the object’s initialization is complete. The object is initialized from a copy of the temporary; during this copying, an implementation can call the copy constructor many times; the temporary is destroyed after it has been copied, before or when the initialization completes.</p> </blockquote> <p>I believe this expression is an initializer for a declarator defining an object, so it's required to initialize the object from a copy of the value of the expression (<code>true</code>, in this case), not directly from the return value. In the case of <code>true</code>, this is probably a distinction without a difference, but I think the diagram is technically more accurate as it stands right now.</p> <p>This also makes fairly clear (I think) that the temporary holding <code>true</code> does <em>not</em> have to be destroyed at the end of the full expression, so I've re-drawn the diagram to reflect that as well.</p> <p>This section is gone in C++0x/C++11, so I've re-drawn the diagram (yet again) to show the difference between the two (and how much simpler this piece has gotten in C++11).</p>
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    1. POWhen exactly is an initializer temporary destroyed?
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    1. USJerry Coffin
    1. USJerry Coffin
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    1. CO+1 (when I get more votes in 16 minutes) for pointing out that `cout` may not even exist yet.
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      1. USLightness Races in Orbit
    2. CO@Jerry if I do `bool b = true; b = false;` does that mean i'm doing undefined behavior? Because I don't see the writes to `b` being ordered. In C++03 it seems they could happen in between the same sequence points, and in C++0x they seem to be two unordered side effects.
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      1. USJohannes Schaub - litb
    3. CO@litb: Two statements with a sequence point between them, no?
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