Why does adding "const" solve "invalid operands to binary expression"

Question

I have overloaded the operator + by Integer operator+(Integer & a, Integer & b). But when I do a=b+c+d, it gives the error of invalid operands to binary expression. But by adding const to the parameters, no more errors. Why this happens?

Answer 1

b + c + d generates a temporary for the result of b + c. A reference to that temporary is then passed to the second call to operator+().

Only const references can be bound to temporaries.

For further discussion, see How come a non-const reference cannot bind to a temporary object?