Re: have question on 'passing array to reference to pointer'
On 3=BF=F926=C0=CF, =BF=C0=C0=FC5=BD=C331=BA=D0, blargg....@gishpuppy.com (=
blargg) wrote:
Juha Nieminen wrote:
ljh131 wrote:
but i want to know why exactly c++ prohibits this casting. someone sa=
y
array should be considered as 'const pointer'.
I assume you know the difference between "const pointer" and "pointer
to const".
The latter is a pointer variable, which has been declared to point to
a value which cannot be modified through that pointer (ie. the pointer
points to a const).
The former means that the pointer variable itself is const (rather
than the value it's pointing to). In other words, you cannot modify the
pointer itself to point somewhere else.
You can get a long way by, indeed, thinking that array names act like
const pointers (ie. pointer variables which cannot be changed).
[...]
You can get all the way by simply noting that an array can be implicitly
converted into a pointer to its first element, similar to how a double ca=
n
be implicitly converted into an int. In both cases, you can't take a
non-const reference to this implicit conversion:
// using int and double
typedef double T;
typedef int U;
void f( U& );
void fc( U const& );
T t;
void example()
{
f( t ); // error
fc( t ); // OK, creates temporary
U const& temp = t; // equivalent to previous statement
fc( temp );
}
// using array
typedef int T [2];
typedef int* U;
void f( U& );
void fc( U const& );
T t;
void example()
{
f( t ); // error
fc( t ); // OK, creates temporary
U const& temp = t; // equivalent to previous statement
fc( temp );
}
The code after the typedefs is identical in both cases.
thanks for all of your answers.
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