troubleshooting Question

lvalue c++ question

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phoffric asked on
C++Programming Theory
7 Comments2 Solutions494 ViewsLast Modified:
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.
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