Hi everyone, Thanks for the prev. read/write mutex replies. I wrote a simple container which support concurrent read support and exclusive write support. Is this code correct, I havent write test code for it yet. Tnx... #pragma once #ifndef ENTITYUPDATEBUFFER_H___ #define ENTITYUPDATEBUFFER_H___ // std includes #include <vector> // boost includes #include <boost/thread/shared_mutex.hpp> template <typename T> class EntityUpdateBuffer { public: typedef typename boost::shared_lock<boost::shared_mutex> read_lock; typedef typename boost::unique_lock<boost::shared_mutex> write_lock; EntityUpdateBuffer(void){} ~EntityUpdateBuffer(void){} void push_back(const T& value); const T& at(int index) const; T& at(int index); private: std::vector<T> m_buffer; boost::shared_mutex rw_mutex; }; template <typename T> void EntityUpdateBuffer<T>::push_back( const T& value ) { // acquire lock write_lock lock(rw_mutex); m_buffer.push_back(value); } template <typename T> const T& EntityUpdateBuffer<T>::at( int index ) const { // acquire lock read_lock lock(rw_mutex); return m_buffer.at(index); } template <typename T> T& EntityUpdateBuffer<T>::at( int index ) { // acquire lock read_lock lock(rw_mutex); return m_buffer.at(index); } #endif //ENTITYUPDATEBUFFER_H___ ---Sgo---
Selçuk Giray Özdamar wrote:
const T& at(int index) const;
This doesn't look good.
T& at(int index);
Or this.
private: std::vector<T> m_buffer; boost::shared_mutex rw_mutex; };
template <typename T> const T& EntityUpdateBuffer<T>::at( int index ) const { // acquire lock read_lock lock(rw_mutex); return m_buffer.at(index); }
The problem with this code is that you are returning a reference to data that might be immediately invalidated. A push_back via another thread can cause the referenced memory to no longer be valid.
template <typename T> T& EntityUpdateBuffer<T>::at( int index ) { // acquire lock read_lock lock(rw_mutex); return m_buffer.at(index); }
Same problem. KevinH -- Kevin Heifner heifner @ ociweb.com http://heifner.blogspot.com Object Computing, Inc. (OCI) www.ociweb.com
The problem with this code is that you are returning a reference > to data that might be immediately invalidated. A push_back via > another thread can cause the referenced memory to no longer be valid. Does push_back() really invalidate iterators to *previous* vector elements?!
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Igor R. <igor_rubinov@hotmail.com> writes:
The problem with this code is that you are returning a reference > to data that might be immediately invalidated. A push_back via > another thread can cause the referenced memory to no longer be valid. Does push_back() really invalidate iterators to *previous* vector elements?!
Yes: it might have to allocate a new chunk of memory for the vector, in which case it will have to move all the existing elements to the new memory, and free the old one. Anthony -- Anthony Williams | Just Software Solutions Ltd Custom Software Development | http://www.justsoftwaresolutions.co.uk Registered in England, Company Number 5478976. Registered Office: 15 Carrallack Mews, St Just, Cornwall, TR19 7UL
Hi, On Wed, May 14, 2008 at 7:44 PM, Anthony Williams <anthony_w.geo@yahoo.com> wrote:
Igor R. <igor_rubinov@hotmail.com> writes:
The problem with this code is that you are returning a reference > to data that might be immediately invalidated. A push_back via > another thread can cause the referenced memory to no longer be valid. Does push_back() really invalidate iterators to *previous* vector elements?!
Yes: it might have to allocate a new chunk of memory for the vector, in which case it will have to move all the existing elements to the new memory, and free the old one.
The write locks and read locks in your code are identical (shared mutex). On a call that does a push into the vector, you are modifying data or doing a write. You should upgrade the lock you are holding from a read to a write lock by a call to lock_upgrade (gets the thread ability to gain exclusive lock) followed by unlock_upgrade_and_lock() (gives up the ability and gains the exclusive lock atomically) (I may be wrong here, requesting others to chip in), do the push and call to unlock_upgrade(). That will be a true implementation of single write and multiple reader paradigm. -dhruva -- Contents reflect my personal views only!
I see... Then the whole problem could be solved just by using deque instead of vector - since deque does't guarantee contiguous memory block for its elements, and thus it shouldn't reallocate the storage.> >> > The problem with this code is that you are returning a reference > to data that might be immediately invalidated. A push_back via > another thread can cause the referenced memory to no longer be valid.> > Does push_back() really invalidate iterators to *previous* vector elements?!> > Yes: it might have to allocate a new chunk of memory for the vector, in which> case it will have to move all the existing elements to the new memory, and> free the old one.> > Anthony _________________________________________________________________ News, entertainment and everything you care about at Live.com. Get it now! http://www.live.com/getstarted.aspx
What you need is something that guarantees not to move its elements rather than something that doesn't guarantee contiguous memory. I don't know if deque does this but you could end up with code that only works if it's built with the right compiler... ________________________________ From: boost-users-bounces@lists.boost.org [mailto:boost-users-bounces@lists.boost.org] On Behalf Of Igor R. Sent: 16 May 2008 12:04 To: boost-users@lists.boost.org Subject: Re: [Boost-users] is read/write mutex example correct I see... Then the whole problem could be solved just by using deque instead of vector - since deque does't guarantee contiguous memory block for its elements, and thus it shouldn't reallocate the storage.
The problem with this code is that you are returning a reference to data that might be immediately invalidated. A push_back via > another thread can cause the referenced memory to no longer be valid. Does push_back() really invalidate iterators to *previous* vector elements?!
Yes: it might have to allocate a new chunk of memory for the vector, in which case it will have to move all the existing elements to the new memory, and free the old one.
Anthony
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Patrick Loney wrote:
What you need is something that guarantees not to move its elements rather than something that doesn’t guarantee contiguous memory. I don’t know if deque does this but you could end up with code that only works if it’s built with the right compiler…
The spec says for a deque, "An insert at either end of the deque invalidates all the iterators to the deque, but has no effect on the validity of references to elements of the deque." However, is the requirement to only append and never erase? That seems unlikely. KevinH -- Kevin Heifner heifner @ ociweb.com http://heifner.blogspot.com Object Computing, Inc. (OCI) www.ociweb.com
-----BEGIN PGP SIGNED MESSAGE----- Hash: SHA1 On Wednesday 14 May 2008 10:03 am, Igor R. wrote:
The problem with this code is that you are returning a reference to data that might be immediately invalidated. A push_back via another thread can cause the referenced memory to no longer be valid.
Does push_back() really invalidate iterators to *previous* vector elements?!
It can, if it causes the vector to reallocate memory. - -- Frank -----BEGIN PGP SIGNATURE----- Version: GnuPG v1.4.6 (GNU/Linux) iD8DBQFIKvQv5vihyNWuA4URAiFnAKDVpSdlLmqGSCuxOPH4WitCpgiPNQCcCfC3 TkZFyiDH5n10xeZrF6soptc= =hxAV -----END PGP SIGNATURE-----
Igor R. wrote:
The problem with this code is that you are returning a reference to data that might be immediately invalidated. A push_back via another thread can cause the referenced memory to no longer be valid.
Does push_back() really invalidate iterators to *previous* vector elements?!
Yes, and I quote the spec for insert which push_back() is based on: "Remarks: Causes reallocation if the new size is greater than the old capacity. If no reallocation happens, all the iterators and references before the insertion point remain valid." So unless you set capacity of the vector so that it never grows then push_back() can invalidate a held reference. Also it probably should be noted that even if you setup capacity so that the vector never grows and only allowed push_back() you still need mutex guards around any access to the shared object. KevinH -- Kevin Heifner heifner @ ociweb.com http://heifner.blogspot.com Object Computing, Inc. (OCI) www.ociweb.com
participants (7)
-
Anthony Williams -
dhruva -
Frank Mori Hess -
Igor R. -
Kevin Heifner -
Patrick Loney -
Selçuk Giray Özdamar