book1 chapter3, vec3, what is the difference of "double operator[](int i) const {return e[i];}" and "double& operator[](int i) {return e[i];}" ?

[jia-heng]

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

Do these two sentences have the same effect?

[hollasch]

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

The const following the parameter list (before the function body) indicates that this method is available for constant vec3 objects. It will return a double-precision floating-point value.

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

This one returns a reference (pointer) to the actual double-precision floating-point value inside the vec3. This reference can be used to mutate (change) the vec3, so it is not available to use on constant vec3 objects.

[hollasch]

I'm afraid this is some C++ low-level detail that's unfortunately necessary here. We really try to keep C++ specifics to a minimum. This could look quite different in your own code, depending on your language of choice.

To be a bit more concrete, the first form lets you grab elements out of any vec3 object, like

vec3 v = foo();
double x = v[0];  // Deconstruct the vector v
double y = v[1];
double z = v[2];

The second form lets you change/set vector values, like so:

vec3 v;
v[0] = 0.0;
v[1] = 10.0;
v[2] = 100.0;

[jia-heng]

I got it, thx so much