Re: Const constructor
On 19 Jun., 00:37, Edward Rosten <Edward.Ros...@gmail.com> wrote:
On 18 June, 14:26, Dragan Milenkovic <dra...@plusplus.rs> wrote:
typedef int * B;
const B <=> int * const
They behave the same. The only difference being that you can
add a "left-side const" to a pointer where you see fit.
That's my point! You can get different behavior from built-in types as
compared to user defined ones.
But there is no difference, really. Please read this example
carefully:
typedef int* ptr_t;
typedef std::list<int>::iterator listiter_t;
const ptr_t === int*const #1
const listiter_t === std::list<int>::iterator const #2
#1 is a const pointer to a non-const int. #2 is a const iterator to a
non-const int. It's exactly the same behaviour.
But it's unfortunate that the STL uses names like "const iterator". In
STL-land "const iterator" means that it's an iterator for const value
types. The iterator itself is not const.
You can use a pointer to represent a
slice and left side const does what you expect.
Well, the const in "const int*" applies to the int and not to the
pointer. It's a DIFFERENT pointer type much like you can provide
DIFFERENT user-defined iterators.
These may be the same:
typedef int * B
const B <=> int * const
These are not:
void foo(const double* d);
void foo(const ArraySilce& s);
Of course not. This comparison is nonsense. See below.
It almost allows you do write the class once, but instead you have to
duplicate loads of code. Code duplication is IMO bad.
You can parameterize the type of elements your ArraySlice points to to
avoid code duplication:
void foo(const double* d);
void foo(ArraySilce<const couble> s);
would be the equivalent. (!)
Anyway, I have already tried to explain the mentioned distinction
in my previous post. There is even a fictional (ugly? stupid?)
language feature that would allow adding a "left-side const".
I fully understand the distinction.
I don't think you do.
Cheers!
SG
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