how C++ telsl the number of element in delete []

[peter]

Hello,

for a C++ delete:

class Test{};
Test * A = new Test[50];
delete [] A;

We don't use delete [50] A;
How can C++ compiler can tell there are 50 destructor to deallocate?

The same if I have an array
char *b = new char[40];
delete [] b;

how C++ compiler can know the right number of 40 to deallocate?

Thanks,

Peter

 

[Jeff Schwab]

Hello,

for a C++ delete:

class Test{};
Test * A = new Test[50];
delete [] A;

We don't use delete [50] A;
How can C++ compiler can tell there are 50 destructor to deallocate?

The same if I have an array
char *b = new char[40];
delete [] b;

how C++ compiler can know the right number of 40 to deallocate?

Thanks,

Peter

It usually allocates a little extra space before the array, in which it
puts housekeeping data like the size of the array.

Way back when, you actually did have to cite the number of elements as
part of the delete statement.

 

[Clark Cox]

Hello,

for a C++ delete:

class Test{};
Test * A = new Test[50];
delete [] A;

We don't use delete [50] A;
How can C++ compiler can tell there are 50 destructor to deallocate?

The same if I have an array
char *b = new char[40];
delete [] b;

how C++ compiler can know the right number of 40 to deallocate?

Magic.

But seriously, the compiler is free to do whatever housekeeping it needs
to do behind your back, as long as it compiles the code in a manner
specified by the standard.

 

[Default User]

How can C++ compiler can tell there are 50 destructor to deallocate?

Magic. The runtime system uses some way to keep a count of the number of
bytes associated with a dynamic allocation. When the deallocation is
performed, it uses that number. A typical way is to record that number
preceeding the allocated block. IT IS NOT A NUMBER YOU CAN ACCESS.

So the bottom line is, don't worry about it. As far as you are concerned
it is magic. It just happens, in whatever way it happens.



Brian Rodenborn

 

[peter]

Hello,

for a C++ delete:

class Test{};
Test * A = new Test[50];
delete [] A;

We don't use delete [50] A;
How can C++ compiler can tell there are 50 destructor to deallocate?

The same if I have an array
char *b = new char[40];
delete [] b;

how C++ compiler can know the right number of 40 to deallocate?

Thanks,

Peter

It usually allocates a little extra space before the array, in which it
puts housekeeping data like the size of the array.

It might break all other code.

Way back when, you actually did have to cite the number of elements as
part of the delete statement.

If I don't cite the number of elements, it works fine to call
the deconstructor 50 times for delete [] A.

Thanks,

Peter

 

[peter]

peter wrote:





Magic. The runtime system uses some way to keep a count of the number of
bytes associated with a dynamic allocation. When the deallocation is

Not sure where to keep it internally for a pointer. As you said, just
treat it as a magic for C++ compiler to make it work.

Thanks,

Peter

 

[Jeff Schwab]

Hello,

for a C++ delete:

class Test{};
Test * A = new Test[50];
delete [] A;

We don't use delete [50] A;
How can C++ compiler can tell there are 50 destructor to deallocate?

The same if I have an array
char *b = new char[40];
delete [] b;

how C++ compiler can know the right number of 40 to deallocate?

Thanks,

Peter

It usually allocates a little extra space before the array, in which
it puts housekeeping data like the size of the array.

It might break all other code.

What, the compiler might? I'm not sure what you mean... It really does
work this way (for many implementations).

Way back when, you actually did have to cite the number of elements as
part of the delete statement.

If I don't cite the number of elements, it works fine to call
the deconstructor 50 times for delete [] A.

That doesn't free the memory, it just deconstructs the objects.

 

[one2001boy]

peter wrote:

Hello,

for a C++ delete:

class Test{};
Test * A = new Test[50];
delete [] A;

We don't use delete [50] A;
How can C++ compiler can tell there are 50 destructor to deallocate?

The same if I have an array
char *b = new char[40];
delete [] b;

how C++ compiler can know the right number of 40 to deallocate?

Thanks,

Peter



It usually allocates a little extra space before the array, in which
it puts housekeeping data like the size of the array.

It might break all other code.

What, the compiler might? I'm not sure what you mean... It really does
work this way (for many implementations).

I do not understand how compiler allocates extra space before the array
and then access it at the run time. Maybe that is the way it works.

Way back when, you actually did have to cite the number of elements
as part of the delete statement.

If I don't cite the number of elements, it works fine to call
the deconstructor 50 times for delete [] A.

That doesn't free the memory, it just deconstructs the objects.

I took it for granted that we don't need to specify the number of
elements for delete. Maybe I am wrong.

I just checked the previous thread discussing about delete [].

here is the quote from "Unforgiven <[email protected]>":

If you say:
One *B = new One;
Memory is allocated and the constructor is called.
If you say:
One *A = new One[100];
Memory is allocated and 100 constructors are called.

Similarly if you say:
delete[] A;
All 100 destructors are called and memory is deallocated