On 06.08.2013, at 17:40, Howard Hinnant wrote:

> This behavior mimics the defined/undefined behavior for pointers described in 
> [expr.eq] and [expr.rel].  j1 and j2 point to the same object, so they are 
> both equality comparable and less-than comparable.  i1, i2 both point to 
> different objects (none at all), and not to the same object as j1 and j2.  
> [expr.eq] says that all pointers are equality comparable, and this is 
> reflected in the above test.  However [expr.rel] defines operator<() only for 
> pointers pointing into the same object.

But comparison between pointers into different objects is unspecified, not 
undefined. I don't think trapping is a valid outcome.

Sebastian
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