Exercise 6.15

Explain the rationale for the type of each of find_char's parameters. In particular, why is s a reference to const but occurs is a plain reference? Why are these parameters references, but the char parameter c is not? What would happen if we made s a plain reference? What if we make occurs a reference to const?

Solution

s is a reference to const because the function does not edit s, while the function does edit occurs. c could be a reference to const, such that passing a char literal would still work, but c cannot be a plain reference, otherwise find_char(s, 'o', ctr) would not work. The other parameters are references to avoid unneccesary copying of the arguments. If s was a plain reference, then the function would have to copy the entire string every time called. If occurs was a reference to const, then its value could not be edited by the function and ++occurs would not work.