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    <p>First, just to clear up the paragraph that was previously here, using <code>b</code> in its own (dynamic) initialisation here is not UB. Before the expression is evaluated, <code>b</code> is not uninitialised but zero-initialised.</p> <hr> <p>The temporary <code>A</code> must live for precisely as long as the full expression:</p> <blockquote> <p>Temporary objects are destroyed as the last step in evaluating the full-expression (1.9) that (lexically) contains the point where they were created.</p> <p>[ISO/IEC 14882:2003(E) 12.2/3]</p> </blockquote> <p>The line <code>bool b = A(b).yield();</code> is a declaration, which is a statement, which is not an expression. The expression at hand is found only to the RHS of the <code>=</code>. [ISO/IEC 14882:2003(E) A.6]</p> <p>This would mean that the temporary should be destroyed <em>before</em> the dynamic initialisation takes place, no? Sure, the value <code>true</code> is held in the temporary that contains the result of the expression<sup>1</sup> until the initialisation completes, but the original <code>A</code> temporary should be destroyed before <code>b</code> is actually modified.</p> <p>Therefore I'd expect the output <code>false</code>, every time.</p> <hr> <p><sup>1</sup></p> <blockquote> <p>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"</p> <p>[ISO/IEC 14882:2003(E) 12.2/4]</p> </blockquote>
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    1. POWhen exactly is an initializer temporary destroyed?
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    1. USLightness Races in Orbit
    1. USLightness Races in Orbit
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    1. CO"*I realise that b is zero-initialised to false before the specified initialisation takes place. Regardless, isn't using it in its own initialiser still UB?*" : No, because `b` has static storage duration, referring to itself during *dynamic* initialization is fine, since it is guaranteed to have been zero-initialized already. If `b` were defined/initialized so in local scope (e.g. inside `main`), or if it were initialized with a constant expression referring to itself, it would be UB.
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    2. COahh that makes a lot of sense. But I think that it doesn't say anywhere that the initialization of `b` has to happen before or after the full expression sequence point. What if the implementation initializes `b` directly after `yield()` returns, and before destroying the temporary `A`? The temporary would still be destroyed " as the last step in evaluating the full-expression".
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      1. USJohannes Schaub - litb
    3. CO@ildjarn: I know that it'll already have a value from zero-initialization, but I can't seem to find any clause regarding this UB that says it makes a distinction for dynamic initialization. I'm probably just missing it though. Still looking.
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