In the absence of IDE features, you can't differentiate between value arguments and reference arguments unless you view the callee's prototype. Pointers do not have this problem.
In the absence of IDE features, you also can't differentiate between a function taking a pointer and a const pointer. You must be aware in all cases that the value that you pass to the function may be modified. Yes, the function would not be modifying the pointer itself, but the outcome is exactly the same. A reference is pretty much equal to a non-nullable pointer in this context.
> In the absence of IDE features, you also can't differentiate between a function taking a pointer and a const pointer.
So? It still helps me with decoding `foo(bar, baz)`, as without the `&` I KNOW that bar and baz would not change after the call. And since both bar and baz are in the local scope, I already know whether they are pointers or not.
With pointers, all the information necessary to determine if bar and baz would change after a call to foo is in the local scope. With references that information is elsewhere.
TonyTrapp|2 years ago
lelanthran|2 years ago
So? It still helps me with decoding `foo(bar, baz)`, as without the `&` I KNOW that bar and baz would not change after the call. And since both bar and baz are in the local scope, I already know whether they are pointers or not.
With pointers, all the information necessary to determine if bar and baz would change after a call to foo is in the local scope. With references that information is elsewhere.