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Rule 14–6–1 and injected-class-names
#3
I agree. The example should probably be changed to:


Code:
typename A::B::TYPE t2 = 0; // Compliant - explicit use of B and of TYPE.

and for simplicity even:

Code:
typename A::B::TYPE t2 = 0; // Compliant - explicit use of B and of TYPE.

'A' and 'A' match the current specialization.

The rule text (or at least rationale) should be updated to highlight that this is required.


Regards,

Richard
<r>-- <br/>
Richard Corden<br/>
Programming Research Ltd.<br/>
<EMAIL email="[email protected]">[email protected]</EMAIL><br/>
+ 44 845 0048478</r>
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