Dear all, I'm a newbie in boost! I want to know what the following means: int a = 5; const boost::shared_ptr<int> sp(&a); Does it mean that the following line is an error? *sp = 7; If so, what are the guarantees? Is there any documented guarantee? (I couldn't find anything in the documentation. Excuseme if I'm careless!) Thanks, --Hossein __________________________________ Do you Yahoo!? The New Yahoo! Search - Faster. Easier. Bingo. http://search.yahoo.com
On Wednesday, April 30, 2003, at 08:28 AM, Hossein Haeri wrote:
I want to know what the following means:
int a = 5; const boost::shared_ptr<int> sp(&a);
Does it mean that the following line is an error?
*sp = 7;
No, if you want that then you need boost::shared_ptr<int const> sp2(&a); The way const works with shared_ptr is similar to the way it works with plain pointers. You can have a constant pointer, which means the pointer can't be modified, or a pointer to constant, which means the object that's pointed to can't be modified.
If so, what are the guarantees? Is there any documented guarantee?
The fact that sp is constant means you can't do this: int b = 6; sp = &b; The fact that sp2 is constant means you can't do this: *sp2 = 7; Hope that helps. -- Darin
On Wednesday, April 30, 2003, at 09:29 AM, Darin Adler wrote:
The fact that sp2 is constant means you can't do this:
*sp2 = 7;
Oops, make that: The fact that sp2 points to a constant reference means you can't do this: *sp2 = 7; -- Darin
Dear Darin,
boost::shared_ptr<int const> sp2(&a);
The way const works with shared_ptr is similar to the way it works with plain pointers. You can have a constant pointer, which means the pointer can't be modified, or a pointer to constant, which means the object that's pointed to can't be modified.
OK! Now what are the "cast" guarantees? I mean something such as this: boost::shared_ptr<int> sp3(&a); boost::shared_ptr<const int> sp4(sp3); Is the above code legal? What is the result then?
Hope that helps. Yes, it was.
Thanks a lot, --Hossein __________________________________ Do you Yahoo!? The New Yahoo! Search - Faster. Easier. Bingo. http://search.yahoo.com
On Wednesday, April 30, 2003, at 09:40 AM, Hossein Haeri wrote:
OK! Now what are the "cast" guarantees? I mean something such as this:
boost::shared_ptr<int> sp3(&a); boost::shared_ptr<const int> sp4(sp3);
Is the above code legal? What is the result then?
Yes, it's legal. Both sp3 and sp4 point to a, and sp3 can be used to modify the value of a, but sp4 can't. -- Darin
Thank you very much Darin. That was a great help. Wishes, --Hossein __________________________________ Do you Yahoo!? The New Yahoo! Search - Faster. Easier. Bingo. http://search.yahoo.com
--- In Boost-Users@yahoogroups.com, Hossein Haeri
Dear all,
I'm a newbie in boost!
I want to know what the following means:
int a = 5; const boost::shared_ptr<int> sp(&a);
Does it mean that the following line is an error?
*sp = 7;
If so, what are the guarantees? Is there any documented guarantee? (I couldn't find anything in the documentation. Excuseme if I'm careless!)
I'd say you could call any of the 'const' member functions (e.g. operator*(), operator->(), get(),...,etc), but the returned type is a non-const pointer. Since operator*() is a const member function, it is legal on a const shared_ptr<T>. The '=', then, applies to the non-const shared ptr returned, not to the shared_ptr<T> object, itself. Note also, that you're supposed to initialized shared_ptr<T> with a *dynamically* allocated T (use new), because it will call 'delete' on the ptr when the last reference disappears. So you are wrong to initialize it with the address of a stack variable. jh
Dear John,
I'd say you could call any of the 'const' member functions (e.g. operator*(), operator->(), get(),...,etc), but the returned type is a non-const pointer.
Since operator*() is a const member function, it is legal on a const shared_ptr<T>. The '=', then, applies to the non-const shared ptr returned, not to the shared_ptr<T> object, itself.
I can't understand! Can you please describe your idea more? Do you mean that calling a non-const member function for shared_ptr<T> will result in acting the desired operatio on a temporal mutable copy of the original const version?
Note also, that you're supposed to initialized shared_ptr<T> with a *dynamically* allocated T (use new), because it will call 'delete' on the ptr when the last reference disappears. So you are wrong to initialize it with the address of a stack variable.
Thank you very much because of recalling that to me. But a subtle point here is that the asserion you made is not true for all types. If T is not a scaler type, it has some destructor, so I can call that destructor explicitely, before the compiler does it, and before the end of shared_ptr<T>. This way, there would be no problem had I initiialised the share_ptr<T> with an object of type T. Wishes, --Hossein __________________________________ Do you Yahoo!? The New Yahoo! Search - Faster. Easier. Bingo. http://search.yahoo.com
Dear John, I have done a mistake in the answer to your following assertion:
Note also, that you're supposed to initialized shared_ptr<T> with a *dynamically* allocated T (use new), because it will call 'delete' on the ptr when the last reference disappears. So you are wrong to initialize it with the address of a stack variable.
By the destructor of the object, I meant some mechanism for "realeasing" the contained object whithin boost::shared_ptr<>. Silly! Not? So the true answer is not to forget to call the reset() member function for every shared_ptr<>, before the end of it's scope. Thanks, --Hossein __________________________________ Do you Yahoo!? The New Yahoo! Search - Faster. Easier. Bingo. http://search.yahoo.com
On Thursday, May 1, 2003, at 04:57 AM, Hossein Haeri wrote:
So the true answer is not to forget to call the reset() member function for every shared_ptr<>, before the end of it's scope.
No, that won't do any good. When you do a reset(), the shared_ptr relinquishes ownership of the thing it points to. If it's the sole owner, it will destroy the object. It's simply an error to put a pointer to a stack object inside a shared_ptr; you can't correct that error with a reset() call. -- Darin
On Thursday, May 1, 2003, at 04:57 AM, Hossein Haeri wrote:
So the true answer is not to forget to call the reset() member function for every shared_ptr<>, before the end of it's scope.
No, that won't do any good. When you do a reset(), the shared_ptr relinquishes ownership of the thing it points to. If it's the sole owner, it will destroy the object. It's simply an error to put a pointer to a stack object inside a shared_ptr; you can't correct
--- In Boost-Users@yahoogroups.com, Darin Adler
error with a reset() call.
-- Darin
I agree with Darin. Shared_ptr is used for automatically deleting a heap object when the last reference to it (that is, the last copy of the shared_ptr managing it) goes out of scope. It keeps count as the shared_ptr is copied around through parameters and return values and is constructed and destroyed. The underlying count goes up on construction and down on destruction. When the count gets to zero, 'delete' is called on your object. It is used to prevent copies of an object from being made. Hossein, if you'll post a description of what you really want to do, then we might be able to suggest a better idiom for you to follow. jh
Dear John,
Hossein, if you'll post a description of what you really want to do, then we might be able to suggest a better idiom for you to follow.
You are too nice to me because of keeping track of my issues so much! But I don't have any seroius work to do on this latter topic, and your previouse answers were enough for my application. The only point is that I was trying to purge my mistake. That's all! But if you want to know about the reason behind my first question, I'm going to define a new "holder" object (named so by Bjarne Stroustrup), that is described in an artocle, which I'm writting these days. I would be pleased if I can post a proposal for that object to Boost, but I don't know how. You will help me if you can tell me about the way. Thanks, --Hossein __________________________________ Do you Yahoo!? The New Yahoo! Search - Faster. Easier. Bingo. http://search.yahoo.com
participants (3)
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Darin Adler -
Hossein Haeri -
John Harris