Hello,

In the code below, why the first Stdout throws a Exception when the second doesn't ?

/*  -------- CODE --------   */
import tango.io.Stdout;


struct VoidPtr(T)
{
  void* vPtr;

  void value(T val)
  {
    vPtr = &val;
  }

  T value()
  {
    return *(cast(T*)vPtr);
  }
}


void main()
{
  VoidPtr!(char[][]) vPtr ;

  char[][]  arr = [ "hello" , "you" ];

  vPtr.value =  arr ;

  Stdout( vPtr.value ).newline; // <-- [tango.core.Exception.OutOfMemoryException: Memory allocation failed

  Stdout( *(cast(char[][]*) vPtr.vPtr ) ); // <-- This works good
}

/*  ------ END CODE ------  */


Thanks in advance for your help,
TSalm

TSalm wrote:
> Hello,
> 
> In the code below, why the first Stdout throws a Exception when the second doesn't ?
> 
> /*  -------- CODE --------   */
> import tango.io.Stdout;
> 
> 
> struct VoidPtr(T)
> {
>   void* vPtr;
> 
>   void value(T val)
>   {
>     vPtr = &val;

Here you're storing a pointer to a non-ref parameter. This is the bug; the parameter itself is implicitly deallocated on returning from this function.
Change to 'ref T val' to fix it.

Essentially, an array is a struct { size_t length; T ptr; }. This means 'val' refers to a copy in the stack frame of this function, not to 'arr' in main().

>   }
> 
>   T value()
>   {
>     return *(cast(T*)vPtr);
>   }
> }
> 
> 
> void main()
> {
>   VoidPtr!(char[][]) vPtr ;
> 
>   char[][]  arr = [ "hello" , "you" ];
> 
>   vPtr.value =  arr ;
> 
>   Stdout( vPtr.value ).newline; // <-- [tango.core.Exception.OutOfMemoryException: Memory allocation failed

Here you're calling a new function, overwriting *vPtr with something else (probably vptr itself), resulting in a huge array when you try to read it later.

> 
>   Stdout( *(cast(char[][]*) vPtr.vPtr ) ); // <-- This works good

This reads the (implicitly deallocated) 'val' parameter before it gets overwritten, hiding the bug. It's still there: just because the code doesn't crash doesn't mean it's correct.

> }
> 
> /*  ------ END CODE ------  */
> 
> 
> Thanks in advance for your help,
> TSalm

Excellent explication !
Thank you Frits

Le Sat, 14 Feb 2009 17:58:35 +0100, Frits van Bommel <fvbommel@remwovexcapss.nl> a écrit:

> TSalm wrote:
>> Hello,
>>  In the code below, why the first Stdout throws a Exception when the second doesn't ?
>>  /*  -------- CODE --------   */
>> import tango.io.Stdout;
>>   struct VoidPtr(T)
>> {
>>   void* vPtr;
>>    void value(T val)
>>   {
>>     vPtr = &val;
>
> Here you're storing a pointer to a non-ref parameter. This is the bug; the parameter itself is implicitly deallocated on returning from this function.
> Change to 'ref T val' to fix it.
>
> Essentially, an array is a struct { size_t length; T ptr; }. This means 'val' refers to a copy in the stack frame of this function, not to 'arr' in main().
>
>>   }
>>    T value()
>>   {
>>     return *(cast(T*)vPtr);
>>   }
>> }
>>   void main()
>> {
>>   VoidPtr!(char[][]) vPtr ;
>>    char[][]  arr = [ "hello" , "you" ];
>>    vPtr.value =  arr ;
>>    Stdout( vPtr.value ).newline; // <-- [tango.core.Exception.OutOfMemoryException: Memory allocation failed
>
> Here you're calling a new function, overwriting *vPtr with something else (probably vptr itself), resulting in a huge array when you try to read it later.
>
>>    Stdout( *(cast(char[][]*) vPtr.vPtr ) ); // <-- This works good
>
> This reads the (implicitly deallocated) 'val' parameter before it gets overwritten, hiding the bug. It's still there: just because the code doesn't crash doesn't mean it's correct.
>
>> }
>>  /*  ------ END CODE ------  */
>>   Thanks in advance for your help,
>> TSalm