N
nifsmith
Is
*(pointer1->pointer2)
The same as
(*(*pointer1).pointer2)
TYIA
nifsmith
*(pointer1->pointer2)
The same as
(*(*pointer1).pointer2)
TYIA
nifsmith
J
John Harrison
nifsmith said:
Yes.
john
S
Sharad Kala
nifsmith said:
Yes.
Section 5.2.5 paragraph 3 - "If E1 has the type ``pointer to class X,'' then
the expression E1->E2 is converted to the equivalent form (*(E1)).E2;"
Sharad
J
JKop
nifsmith posted:
Yes, it is for *pointers*.
Note however, that for acutal objects, it doesn't always hold true.
For instance:
struct Monkey
{
int* tail;
};
class Blah
{
public:
Monkey a;
Monkey b;
Monkey& operator*()
{
return a;
}
Monkey& operator->()
{
return b;
}
};
int main()
{
Blah poo;
int k;
*( poo->tai l) = &k;
*( (*poo).tail ) = &k;
}
In that above, the statements do two totally different things, ie. there's
two different "tail"s!
(Actually now as I'm writing this, I'm not 100% sure that you can overload
->)
-JKop
Yes, it is for *pointers*.
Note however, that for acutal objects, it doesn't always hold true.
For instance:
struct Monkey
{
int* tail;
};
class Blah
{
public:
Monkey a;
Monkey b;
Monkey& operator*()
{
return a;
}
Monkey& operator->()
{
return b;
}
};
int main()
{
Blah poo;
int k;
*( poo->tai l) = &k;
*( (*poo).tail ) = &k;
}
In that above, the statements do two totally different things, ie. there's
two different "tail"s!
(Actually now as I'm writing this, I'm not 100% sure that you can overload
->)
-JKop
C
Catalin Pitis
[...]
Yes you can
Catalin
Yes you can
Catalin
J
JKop
Catalin Pitis posted:
This pleases me!
( C++, 1 point, Bullshit, 0 points )
-JKop
This pleases me!
( C++, 1 point, Bullshit, 0 points )
-JKop
B
Brian Riis
JKop said:
It should!
That would be how smart pointers work...(Well.. part of it.)