Re: arrays decaying to pointers

From:
=?ISO-8859-1?Q?=D6=F6_Tiib?= <ootiib@hot.ee>
Newsgroups:
comp.lang.c++.moderated
Date:
Mon, 30 May 2011 16:15:46 CST
Message-ID:
<3d8f68c9-0af2-4220-b58a-37d74837ec82@h9g2000yqk.googlegroups.com>
On May 30, 3:56 pm, Marc <marc.gli...@gmail.com> wrote:

Hello,

arrays decay to pointers, which allows for:

template<class Iter>
void f(Iter first, Iter theend);

int tab[]={1,2,3};
f(tab,tab+3);

and Iter is deduced as int*.

However, if I go through a forwarding function:
template<class A,class B>
void g(A const&a, B const&b){ f(a,b); }
g(tab,tab+3);

f now receives a reference to an array of _const_ int and a reference
to a const pointer to int. But just passing f a pointer to const
int and a pointer to int is already an error as the compiler can't
deduce Iter anymore (yes, one could dream about a compiler clever
enough to deduce Iter as const int* in this case).


Alf P. Steinbach proposed a generic solution to your problem in his:

http://alfps.wordpress.com/2010/05/10/how-to-avoid-disastrous-integer-wrap-around/

Do not pay attention to that int/uint "unholy" thing but the
startOf(T), endOf(T) and templates for iterator type deduction.

In short no matter what the type of tab is ... a container or raw
array your call of f() can be made to look like:

f( startOf( tab ), endOf( tab ) );

Also your magic number 3 did walk away from picture. ;)

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