I'm struggling to understand some code from "Ray Tracing in One Weekend" chapter 3.1.

We have a 3d vector class defined, and are overloading some operators. The following two lines confuse me:

        double operator[](int i) const { return e[i]; }
        double& operator[](int i) { return e[i]; }

From what I understand the first line runs when the object is a const and returns e[i] by value. The second line runs when the object is not const and returns e[i] by reference. What is the point in doing this? I think there's some potential for speedup by using references to mutable objects instead of returning them by value. Are there any other reasons?

CodePudding user response：

When the operator returns by non-const reference, the value can be modified by the caller. Normally, the const version would return const reference, but in case of double, returning by value is just as fast.

CodePudding user response：

A non-‘’’const’’’ function can only be called on a mutable object. A const function can be called on either a ‘’’const’’’ or non-‘’’const’’’ object, but will default to the mutable function. The mutable function allows the object to access a reference the double value (therefore it can be modified). The ‘’’const’’’ function lets a const object access the double value; it returns a value because it cannot modify the object. It is unlikely to have any performance impact; both a reference and a double are likely to have 64 bits (on a modern desktop), although this is an implementation detail and will likely be optimized as needed by the compiler anyways.