If a1, a2, and a3 are of a user defined type, T, and you overload the * operator for T, then it may be possible to write:
(a1*a2) = a3;
This looks non-sensical. Yet, it compiles and runs, but I think the statement behaves like a noop.
Could you show me from the C++ standard how this can be? I thought the LHS of assignment operator is an lvalue. But a1*a2 does not look like an lvalue to me. I don't see what possible usefulness the above statement could have.
(a1*a2) = a3;
This looks non-sensical. Yet, it compiles and runs, but I think the statement behaves like a noop.
Could you show me from the C++ standard how this can be? I thought the LHS of assignment operator is an lvalue. But a1*a2 does not look like an lvalue to me. I don't see what possible usefulness the above statement could have.
Learn from the best
Network and collaborate with thousands of CTOs, CISOs, and IT Pros rooting for you and your success.
Andrew Hancock - VMware vExpert
See if this solution works for you by signing up for a 7 day free trial.
Unlock 2 Answers and 7 Comments.
Try for 7 days”The time we save is the biggest benefit of E-E to our team. What could take multiple guys 2 hours or more each to find is accessed in around 15 minutes on Experts Exchange.
Our community of experts have been thoroughly vetted for their expertise and industry experience.
The Distinguished Expert awards are presented to the top veteran and rookie experts to earn the most points in the top 50 topics.