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copied!<p>There's no implicit conversion from <code>Nullable<bool></code> to <code>bool</code>. There <em>is</em> an implicit conversion from <code>bool</code> to <code>Nullable<bool></code> and that's what happens (in language terms) to each of the bool constants in the first version. The <code>bool operator==(Nullable<bool>, Nullable<bool></code> operator is then applied. (This isn't quite the same as other lifted operators - the result is just <code>bool</code>, not <code>Nullable<bool></code>.)</p> <p>In other words, the expression 'fred == false' is of type <code>bool</code>, whereas the expression 'fred' is of type <code>Nullable<bool></code> hence you can't use it as the "if" expression.</p> <p>EDIT: To answer the comments, there's never an implicit conversion from <code>Nullable<T></code> to <code>T</code> and for good reason - implicit conversions shouldn't throw exceptions, and unless you want <code>null</code> to be implicitly converted to <code>default(T)</code> there's not a lot else that could be done.</p> <p>Also, if there <em>were</em> implicit conversions both ways round, an expression like "nullable + nonNullable" would be very confusing (for types that support +, like <code>int</code>). Both +(T?, T?) and +(T, T) would be available, depending on which operand were converted - but the results could be very different!</p> <p>I'm 100% behind the decision to only have an explicit conversion from <code>Nullable<T></code> to <code>T</code>.</p>

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