为什么添加“ const”可以解决“无效的操作数到二进制表达式”

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?