Constructing a shared pointer from a const pointer
Hi All, I have been in trouble for a day solving this seemingly simple problem - constructing shared_ptr<X> from const X& without copying the X object. The programming logic does not allow me to construct the shared_ptr like this: SX x = SX( new X( x ) ); Consider the following piece of code: void foo( const X& x ) { typedef shared_ptr<X> SX; SX px = SX( &x ); bar( px ); } The compiler gives the following error: In shared_ptr.hpp at line 184: 'initializing' cannot convert from 'const X *' to 'X*'. Please suggest any procedure to solve this problem. -- View this message in context: http://old.nabble.com/Constructing-a-shared-pointer-from-a-const-pointer-tp2... Sent from the Boost - Users mailing list archive at Nabble.com.
To add more to the above query,
assume that 'bar' is defined as follows:
map
veerus wrote:
Hi All,
I have been in trouble for a day solving this seemingly simple problem - constructing shared_ptr<X> from const X& without copying the X object.
The programming logic does not allow me to construct the shared_ptr like this: SX x = SX( new X( x ) );
Consider the following piece of code:
void foo( const X& x ) { typedef shared_ptr<X> SX; SX px = SX( &x );
bar( px ); }
The compiler gives the following error: In shared_ptr.hpp at line 184: 'initializing' cannot convert from 'const X *' to 'X*'.
To make it compile, you could simply change the typedef to "typedef shared_ptr<X const> SX;". However, the code is suspect - depending on how the caller of foo declares and manages the variable used as "x", you're likely to either encounter a delete of a (invalid reference to a) stack object sooner or later, or perhaps a double delete. If I saw the above during a code review I would say foo needs to be redesigned. If you're relying on identity here, or if X is non-copyable, and want to transfer ownership of "x" into "bar" you could define foo as: void foo(std::auto_ptr<X> px) { bar(shared_ptr<X>(px)); } [It's hard to tell without knowing the full context] / Johan
I got out of the problem. The signature of bar is bar( int xpointer ); and foo calls bar( (int) &x). All I have to pass to bar is an identity of x. Johan Nilsson-4 wrote:
veerus wrote:
Hi All,
I have been in trouble for a day solving this seemingly simple problem - constructing shared_ptr<X> from const X& without copying the X object.
The programming logic does not allow me to construct the shared_ptr like this: SX x = SX( new X( x ) );
Consider the following piece of code:
void foo( const X& x ) { typedef shared_ptr<X> SX; SX px = SX( &x );
bar( px ); }
The compiler gives the following error: In shared_ptr.hpp at line 184: 'initializing' cannot convert from 'const X *' to 'X*'.
To make it compile, you could simply change the typedef to "typedef shared_ptr<X const> SX;".
However, the code is suspect - depending on how the caller of foo declares and manages the variable used as "x", you're likely to either encounter a delete of a (invalid reference to a) stack object sooner or later, or perhaps a double delete. If I saw the above during a code review I would say foo needs to be redesigned.
If you're relying on identity here, or if X is non-copyable, and want to transfer ownership of "x" into "bar" you could define foo as:
void foo(std::auto_ptr<X> px) { bar(shared_ptr<X>(px)); }
[It's hard to tell without knowing the full context]
/ Johan
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-- View this message in context: http://old.nabble.com/Constructing-a-shared-pointer-from-a-const-pointer-tp2... Sent from the Boost - Users mailing list archive at Nabble.com.
participants (2)
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Johan Nilsson -
veerus