[Solved] C++ not modifiable lvalue

Question

getRadius() returns a copy of the object’s radius, not a reference to it, so you can’t use it for modification. So the following expression:

C.getRadius()=t;

attempts to modify a temporary copy, which isn’t allowed. (The reason that its not allowed is that otherwise that code would compile but do nothing useful, giving a subtle bug rather than an easily identified compile error).

There are various options:

return a reference, double & getRadius();
add a void setRadius(double); function, and use that for modification
make the radius public; there’s little point in using accessors unless they’re necessary to enforce invariants.

If you do want accessor functions, you should keep the existing double getRadius(), but declare it const. That will allow you to get, but not modify, the radius of a constant object.

Accepted Answer

getRadius() returns a copy of the object’s radius, not a reference to it, so you can’t use it for modification. So the following expression:

C.getRadius()=t;

attempts to modify a temporary copy, which isn’t allowed. (The reason that its not allowed is that otherwise that code would compile but do nothing useful, giving a subtle bug rather than an easily identified compile error).

There are various options:

return a reference, double & getRadius();
add a void setRadius(double); function, and use that for modification
make the radius public; there’s little point in using accessors unless they’re necessary to enforce invariants.

If you do want accessor functions, you should keep the existing double getRadius(), but declare it const. That will allow you to get, but not modify, the radius of a constant object.