Why is the following code not valid? I mean, I see the code and it
doesn't allow it, but I am curious about the rationale?
boost::shared_p tr<intpi = new int;
pi = new int;
Thanks
Tim
9  4732 
On 1 Jun, 02:21, Tim H <thoc...@gmail. comwrote:
Why is the following code not valid? I mean, I see the code and it
doesn't allow it, but I am curious about the rationale?
boost::shared_p tr<intpi = new int;
pi = new int;
(you are aware that there is no operator=(raw ptr) in shared_ptr)
IMHO the clue is in the explicit constructor of shared_ptr
so something like
void foo(shared_ptr< intconst & p)
can not be used with
int * p = new int;
void foo(p);
delete p; // UB !
so
boost::shared_p tr<intpi = new int;
can not unwind into
boost::shared_p tr<intpi = boost::shared_p tr<int>(new int);
DS
On 1 Jun, 11:35, dasjotre <dasjo...@googl email.comwrote:
On 1 Jun, 02:21, Tim H <thoc...@gmail. comwrote:
Why is the following code not valid? I mean, I see the code and it
doesn't allow it, but I am curious about the rationale?
boost::shared_p tr<intpi = new int;
pi = new int;
(you are aware that there is no operator=(raw ptr) in shared_ptr)
IMHO the clue is in the explicit constructor of shared_ptr
so something like
void foo(shared_ptr< intconst & p)
can not be used with
int * p = new int;
void foo(p);
delete p; // UB !
so
boost::shared_p tr<intpi = new int;
can not unwind into
boost::shared_p tr<intpi = boost::shared_p tr<int>(new int);
DS
also, the first
boost::shared_p tr<intpi = new int;
doesn't call the operator= but the constructor.
On 1 Jun, 11:36, dasjotre <dasjo...@googl email.comwrote:
On 1 Jun, 11:35, dasjotre <dasjo...@googl email.comwrote:
On 1 Jun, 02:21, Tim H <thoc...@gmail. comwrote:
Why is the following code not valid? I mean, I see the code and it
doesn't allow it, but I am curious about the rationale?
boost::shared_p tr<intpi = new int;
pi = new int;
(you are aware that there is no operator=(raw ptr) in shared_ptr)
IMHO the clue is in the explicit constructor of shared_ptr
so something like
void foo(shared_ptr< intconst & p)
can not be used with
int * p = new int;
void foo(p);
delete p; // UB !
so
boost::shared_p tr<intpi = new int;
can not unwind into
boost::shared_p tr<intpi = boost::shared_p tr<int>(new int);
DS
also, the first
boost::shared_p tr<intpi = new int;
doesn't call the operator= but the constructor.
(sorry I haven't had my coffee yet)
I meant
pi = new int;
can not unwind into
pi = boost::shared_p tr<int>(new int);
On Jun 1, 3:35 am, dasjotre <dasjo...@googl email.comwrote:
On 1 Jun, 02:21, Tim H <thoc...@gmail. comwrote:
Why is the following code not valid? I mean, I see the code and it
doesn't allow it, but I am curious about the rationale?
boost::shared_p tr<intpi = new int;
pi = new int;
(you are aware that there is no operator=(raw ptr) in shared_ptr)
IMHO the clue is in the explicit constructor of shared_ptr
so something like
void foo(shared_ptr< intconst & p)
can not be used with
int * p = new int;
void foo(p);
delete p; // UB !
Ahh, of course. I get that, then.
boost::shared_p tr<intpi = new int;
can not unwind into
boost::shared_p tr<intpi = boost::shared_p tr<int>(new int);
This is supposed to call the constructor, isn't it? Why can it not
figure out (or not be allowed) to call the (int *) ctor?
Thanks
Tim
On 1 Jun, 16:32, Tim H <thoc...@gmail. comwrote:
On Jun 1, 3:35 am, dasjotre <dasjo...@googl email.comwrote:
On 1 Jun, 02:21, Tim H <thoc...@gmail. comwrote:
Why is the following code not valid? I mean, I see the code and it
doesn't allow it, but I am curious about the rationale?
boost::shared_p tr<intpi = new int;
pi = new int;
(you are aware that there is no operator=(raw ptr) in shared_ptr)
IMHO the clue is in the explicit constructor of shared_ptr
so something like
void foo(shared_ptr< intconst & p)
can not be used with
int * p = new int;
void foo(p);
delete p; // UB !
Ahh, of course. I get that, then.
boost::shared_p tr<intpi = new int;
can not unwind into
boost::shared_p tr<intpi = boost::shared_p tr<int>(new int);
This is supposed to call the constructor, isn't it? Why can it not
figure out (or not be allowed) to call the (int *) ctor?
Thanks
Tim
(sorry for my clumsy reply)
boost::shared_p tr<intpi = new int;
actually calls the constructor. the consequent
pi = new int;
can not. since pi is already constructed object
the assignment operator is looked up and there
are two assignment operators in shared_ptr
one that takes shared_ptr and one that takes
std::auto_ptr. neither of these two are implicitly
constructible from raw pointer, so neither of them
can be used.
that is, compiler can not convert
pi = new int;
into either
pi = shared_ptr<int> (new int);
or
pi = auto_ptr<int>(n ew int);
DS
On Jun 4, 2:33 am, dasjotre <dasjo...@googl email.comwrote:
On 1 Jun, 16:32, Tim H <thoc...@gmail. comwrote:
On Jun 1, 3:35 am, dasjotre <dasjo...@googl email.comwrote:
On 1 Jun, 02:21, Tim H <thoc...@gmail. comwrote:
Why is the following code not valid? I mean, I see the code and it
doesn't allow it, but I am curious about the rationale?
boost::shared_p tr<intpi = new int;
pi = new int;
(you are aware that there is no operator=(raw ptr) in shared_ptr)
IMHO the clue is in the explicit constructor of shared_ptr
so something like
void foo(shared_ptr< intconst & p)
can not be used with
int * p = new int;
void foo(p);
delete p; // UB !
Ahh, of course. I get that, then.
boost::shared_p tr<intpi = new int;
can not unwind into
boost::shared_p tr<intpi = boost::shared_p tr<int>(new int);
This is supposed to call the constructor, isn't it? Why can it not
figure out (or not be allowed) to call the (int *) ctor?
Thanks
Tim
(sorry for my clumsy reply)
boost::shared_p tr<intpi = new int;
actually calls the constructor. the consequent
pi = new int;
can not.
g++ is warning about both, which is what confuses me
On 4 Jun, 16:14, Tim H <thoc...@gmail. comwrote:
On Jun 4, 2:33 am, dasjotre <dasjo...@googl email.comwrote:
On 1 Jun, 16:32, Tim H <thoc...@gmail. comwrote:
On Jun 1, 3:35 am, dasjotre <dasjo...@googl email.comwrote:
On 1 Jun, 02:21, Tim H <thoc...@gmail. comwrote:
Why is the following code not valid? I mean, I see the code and it
doesn't allow it, but I am curious about the rationale?
boost::shared_p tr<intpi = new int;
pi = new int;
(you are aware that there is no operator=(raw ptr) in shared_ptr)
IMHO the clue is in the explicit constructor of shared_ptr
so something like
void foo(shared_ptr< intconst & p)
can not be used with
int * p = new int;
void foo(p);
delete p; // UB !
Ahh, of course. I get that, then.
boost::shared_p tr<intpi = new int;
can not unwind into
boost::shared_p tr<intpi = boost::shared_p tr<int>(new int);
This is supposed to call the constructor, isn't it? Why can it not
figure out (or not be allowed) to call the (int *) ctor?
Thanks
Tim
(sorry for my clumsy reply)
boost::shared_p tr<intpi = new int;
actually calls the constructor. the consequent
pi = new int;
can not.
g++ is warning about both, which is what confuses me
(I don't use g++).
in the code:
boost::shared_p tr<intpi = new int; // 1
pi = new int; // 2
line 1 should not give you any warning at
all since it is equivalent to
boost::shared_p tr<intpi (new int);
line 2 should give you a compile error.
DS
dasjotre wrote:
>g++ is warning about both, which is what confuses me
(I don't use g++).
in the code:
boost::shared_p tr<intpi = new int; // 1
pi = new int; // 2
line 1 should not give you any warning at
all since it is equivalent to
boost::shared_p tr<intpi (new int);
line 2 should give you a compile error.
shared.cpp:1: error: conversion from 'int*' to non-scalar type
'boost::shared_ ptr<int>' requested
shared.cpp:2: error: no match for 'operator=' in 'pi = (int*)operator new(4u)'
line 1 is not equivalent to boost::shared_p tr<intpi (new int);
also,
boost::shared_p tr<intpi = boost::shared_p tr<int>(new int);
is not equivalent to
boost::shared_p tr<intpi (new int);
since you need a copy constructor for the first (the compiler complains if
you haven't, but it may optimize it away).
So in both lines you need to _explicitly_ construct a temporary shared_ptr:
boost::shared_p tr<intpi = boost::shared_p tr<int>(new int);
pi = boost::shared_p tr<int>(new int);
This was made so that the compiler does not accidentaly construct a
shared_ptr from a raw pointer and free it afterwards. This could cause bugs
like double frees or calling delete on automatic (stack)-objects.
--
Thomas http://www.netmeister.org/news/learn2quote.html
On 4 Jun, 16:45, "Thomas J. Gritzan" <Phygon_ANTIS.. .@gmx.dewrote:
dasjotre wrote:
g++ is warning about both, which is what confuses me
(I don't use g++).
in the code:
boost::shared_p tr<intpi = new int; // 1
pi = new int; // 2
line 1 should not give you any warning at
all since it is equivalent to
boost::shared_p tr<intpi (new int);
line 2 should give you a compile error.
shared.cpp:1: error: conversion from 'int*' to non-scalar type
'boost::shared_ ptr<int>' requested
shared.cpp:2: error: no match for 'operator=' in 'pi = (int*)operator new(4u)'
line 1 is not equivalent to boost::shared_p tr<intpi (new int);
also,
boost::shared_p tr<intpi = boost::shared_p tr<int>(new int);
is not equivalent to
boost::shared_p tr<intpi (new int);
since you need a copy constructor for the first (the compiler complains if
you haven't, but it may optimize it away).
So in both lines you need to _explicitly_ construct a temporary shared_ptr:
boost::shared_p tr<intpi = boost::shared_p tr<int>(new int);
pi = boost::shared_p tr<int>(new int);
This was made so that the compiler does not accidentaly construct a
shared_ptr from a raw pointer and free it afterwards. This could cause bugs
like double frees or calling delete on automatic (stack)-objects.
I stand corrected.
1. will call
boost::shared_p tr<intpi = new int;
will try to call boost::shared_p tr<int>::operat or=(raw ptr)
and will fail for the same reason as 2.
pi = new int;
most compilers will call constructor on 1 (not standard
behaviour), and I got lost thinking about the rationale
for explicit shared_ptr/auto_ptr constructors.
DS This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics | |
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