allocate memory for pointer

[xuatla]

I encountered "segmentati on fault" and I checked my code, found the
following problem:

I want to reallocate memory for an array.
I defined the following function:

int reallocateMemor y( double *array, int newsize )
{
if (array) delete[] array;

array = new double[newsize];

return 1;
}

Now,
int main()
{
double *a = new double[2], *b = new double[10];

cout << a << endl; // address of a. (1)

for ( int i = 0; i < 10; i++ ) b[i] = i; // b = {0,1,....,9}

reallocateMemor y( a, 10 );

cout << a << endl; // the address of a is same as (1)! :(

for ( int i = 0; i < 10; i++ ) a[i] = 0.1*i;

for ( int i = 0; i < 10; i++ )
cout << a[i] << " ";
cout << endl; // fine, a is 0,0.1,...,0.9

for ( int i = 0; i < 10; i++ )
cout << b[i] << " ";
cout << endl;

// error: output = 0.3,0.4,...,0.9 ,7,8,9

return 0;
}

So the memory of "a" is not reallocated although I call the function to
change it.

What I think is that passed-by-pointer in function will change the
parameter directly. Therefore after I call the function
reallocateMemor y, "a" should be changed. But it seems that the result is
same as passed-by-value.

What's the problem in my function?

Thank you for your help!

X

[Sharad Kala]

"xuatla" <xu****@gmail.c om> wrote in message
news:40******** ******@gmail.co m...

I encountered "segmentati on fault" and I checked my code, found the
following problem:

I want to reallocate memory for an array.
I defined the following function:

int reallocateMemor y( double *array, int newsize )
{
if (array) delete[] array;

array = new double[newsize];
How do you return array back to main now?
return 1;
}

Now,
int main()
{
double *a = new double[2], *b = new double[10];

cout << a << endl; // address of a. (1)

for ( int i = 0; i < 10; i++ ) b[i] = i; // b = {0,1,....,9}

reallocateMemor y( a, 10 );

After this function a points to deleted memory. Henceforth trying anything
on that memory is UB.

So it resolve the problem change reallocate to -
int reallocateMemor y( double *&array, int newsize )
^^^^^^^^
You are now paasing a reference to a pointer. Any changes now made to array
will be reflected back in main.

-Sharad

[John Harrison]

On Thu, 15 Jul 2004 00:20:39 -0700, xuatla <xu****@gmail.c om> wrote:

I encountered "segmentati on fault" and I checked my code, found the
following problem:

I want to reallocate memory for an array.
I defined the following function:

int reallocateMemor y( double *array, int newsize )
{
if (array) delete[] array;

array = new double[newsize];

return 1;
}

Now,
int main()
{
double *a = new double[2], *b = new double[10];

cout << a << endl; // address of a. (1)

for ( int i = 0; i < 10; i++ ) b[i] = i; // b = {0,1,....,9}

reallocateMemor y( a, 10 );

cout << a << endl; // the address of a is same as (1)! :(

for ( int i = 0; i < 10; i++ ) a[i] = 0.1*i;

for ( int i = 0; i < 10; i++ )
cout << a[i] << " ";
cout << endl; // fine, a is 0,0.1,...,0.9

for ( int i = 0; i < 10; i++ )
cout << b[i] << " ";
cout << endl;

// error: output = 0.3,0.4,...,0.9 ,7,8,9

return 0;
}

So the memory of "a" is not reallocated although I call the function to
change it.

What I think is that passed-by-pointer in function will change the
parameter directly. Therefore after I call the function
reallocateMemor y, "a" should be changed. But it seems that the result is
same as passed-by-value.
Yes, pointers are passed by value just like everything else.

What's the problem in my function?

You are passing by value. If you want to use a function to change that
value of a variable you have three alternatives.

1) Use a return value

double* reallocateMemor y( double *array, int newsize )
{
delete[] array;
return new double[newsize];
}

a = reallocateMemor y( a, 10 );

2) Use a reference

void reallocateMemor y( double*& array, int newsize )
{
delete[] array;
array = new double[newsize];
}

reallocateMemor y( a, 10 );

3) Use a pointer (in your case this would be a pointer to a pointer)

void reallocateMemor y( double** array, int newsize )
{
delete[] *array;
*array = new double[newsize];
}

reallocateMemor y( &a, 10 );
BTW it is not necessary to test for NULL before deleteing a pointer

if (ptr) delete[] ptr;

works exactly the same as

delete[] ptr;

Deleting NULL is guaranteed to have no effect.

john

[Patrik Stellmann]

[...]

int reallocateMemor y( double *array, int newsize )
{
if (array) delete[] array;

array = new double[newsize];

return 1;
} [...]
What's the problem in my function?

your function changes only the local variable array which has no
influence to your variable a in main. Changing the functiondeclara tion to
int reallocateMemor y( double *&array, int newsize )
should do the trick. Now not the value of a is passed to the function
but the reference of it so you're able to modify its value...

[xuatla]

Thanks all of you for the kind help.

I made a mistake. As I mentioned I thought passed-by-pointer is
different with passed-by-value and the change will be kept after calling
the function. Now I know I made a mistake. If "a" is a double value, not
an array, then

testfun( double *a)
{ *a = newvalue; }
and
{
...
double a;
testfun(&a);
}
will change the value of a. But what I used is the pointer to an array.
So it's differnt.

Thanks for pointing out my mistake and giving me the good answers.

I have one more related (and naive) question
(1) double* a --- a is a double array
(2) double *a --- a is a double pointer

what's the difference of (1) & (2)? both a are address and one points to
an array while another points to a value. is there any other characters
in expression tell c++ that there're different? Thanks!

X

John Harrison wrote:

On Thu, 15 Jul 2004 00:20:39 -0700, xuatla <xu****@gmail.c om> wrote:

I encountered "segmentati on fault" and I checked my code, found the
following problem:

I want to reallocate memory for an array.
I defined the following function:

int reallocateMemor y( double *array, int newsize )
{
if (array) delete[] array;

array = new double[newsize];

return 1;
}

Now,
int main()
{
double *a = new double[2], *b = new double[10];

cout << a << endl; // address of a. (1)

for ( int i = 0; i < 10; i++ ) b[i] = i; // b = {0,1,....,9}

reallocateMemor y( a, 10 );

cout << a << endl; // the address of a is same as (1)! :(

for ( int i = 0; i < 10; i++ ) a[i] = 0.1*i;

for ( int i = 0; i < 10; i++ )
cout << a[i] << " ";
cout << endl; // fine, a is 0,0.1,...,0.9

for ( int i = 0; i < 10; i++ )
cout << b[i] << " ";
cout << endl;

// error: output = 0.3,0.4,...,0.9 ,7,8,9

return 0;
}

So the memory of "a" is not reallocated although I call the function
to change it.

What I think is that passed-by-pointer in function will change the
parameter directly. Therefore after I call the function
reallocateMemor y, "a" should be changed. But it seems that the result
is same as passed-by-value.

Yes, pointers are passed by value just like everything else.

What's the problem in my function?

You are passing by value. If you want to use a function to change that
value of a variable you have three alternatives.

1) Use a return value

double* reallocateMemor y( double *array, int newsize )
{
delete[] array;
return new double[newsize];
}

a = reallocateMemor y( a, 10 );

2) Use a reference

void reallocateMemor y( double*& array, int newsize )
{
delete[] array;
array = new double[newsize];
}

reallocateMemor y( a, 10 );

3) Use a pointer (in your case this would be a pointer to a pointer)

void reallocateMemor y( double** array, int newsize )
{
delete[] *array;
*array = new double[newsize];
}

reallocateMemor y( &a, 10 );
BTW it is not necessary to test for NULL before deleteing a pointer

if (ptr) delete[] ptr;

works exactly the same as

delete[] ptr;

Deleting NULL is guaranteed to have no effect.

john

[Sharad Kala]

"xuatla" <xu****@gmail.c om> wrote in message
news:40******** ******@gmail.co m...

I have one more related (and naive) question
(1) double* a --- a is a double array
(2) double *a --- a is a double pointer

No both are pointers.
double* a, b;
a is a pointer to double while b is a double.
Arrays are indicated by the subscript operator [] like arr[10] . Arrays
decay down to pointers though.

-Sharad

[John Harrison]

"xuatla" <xu****@gmail.c om> wrote in message
news:40******** ******@gmail.co m...

Thanks all of you for the kind help.

I made a mistake. As I mentioned I thought passed-by-pointer is
different with passed-by-value and the change will be kept after calling
the function. Now I know I made a mistake. If "a" is a double value, not
an array, then

testfun( double *a)
{ *a = newvalue; }
and
{
...
double a;
testfun(&a);
}
will change the value of a. But what I used is the pointer to an array.
So it's differnt.

Thanks for pointing out my mistake and giving me the good answers.

I have one more related (and naive) question
(1) double* a --- a is a double array
(2) double *a --- a is a double pointer

what's the difference of (1) & (2)?
There is no difference at all. Whitespace is not significant in C++. Some
people say that stylistically 2 is better than 1 and a few say the opposite,
but that's all it is, a style issue.
both a are address and one points to
an array while another points to a value. is there any other characters
in expression tell c++ that there're different? Thanks!

Both are pointers. Strictly speaking if you want a pointer to an array its a
different syntax

int a[5] = { 1, 2, 3, 4, 5 };
int (*p)[5] = &a; // a pointer to an array
cout << (*p)[0]; // prints 1

but true pointers to arrays are not used very much, so people often say
pointer to an array when really they mean a pointer which happens to be
pointing to the first element of an array.

int a[5] = { 1, 2, 3, 4, 5 };
int *p = a; // a pointer which is pointing to the first element of a
cout << *p; // prints 1

john

[JKop]

John Harrison posted:

"xuatla" <xu****@gmail.c om> wrote in message
news:40******** ******@gmail.co m...

Thanks all of you for the kind help.

I made a mistake. As I mentioned I thought passed-by- pointer is different with passed-by-value and the change will be kept after calling the function. Now I know I made a mistake. If "a" is a double value, not an array, then

testfun( double *a)
{ *a = newvalue; } and { ...
double a;
testfun(&a);
}
will change the value of a. But what I used is the pointer to an array. So it's differnt.

Thanks for pointing out my mistake and giving me the good answers.
I have one more related (and naive) question
(1) double* a --- a is a double array
(2) double *a --- a is a double pointer

what's the difference of (1) & (2)?
There is no difference at all. Whitespace is not

significant in C++. Some people say that stylistically 2 is better than 1 and a few say the opposite, but that's all it is, a style issue.

both a are address and one points to
an array while another points to a value. is there any other characters in expression tell c++ that there're different? Thanks!
Both are pointers. Strictly speaking if you want a

pointer to an array its a different syntax

int a[5] = { 1, 2, 3, 4, 5 };
int (*p)[5] = &a; // a pointer to an array
cout << (*p)[0]; // prints 1

but true pointers to arrays are not used very much, so people often say pointer to an array when really they mean a pointer which happens to be pointing to the first element of an array.

int a[5] = { 1, 2, 3, 4, 5 };
int *p = a; // a pointer which is pointing to the first element of a cout << *p; // prints 1

john

The following is a mess. What I tried to get was a
multidemensiona l array with new.

It compiles, but don't ask me what it does. Note how at one
point I've write int[5][5][5][5], as opposed to just three
5's, it won't compile otherwise and I don't know why.

Anyway,
#ifndef INCLUDE_ACCESSO R_CAST
#define INCLUDE_ACCESSO R_CAST

template<typena me TO, class FROM>
inline TO& accessor_cast(F ROM& from)
{
return *(reinterpret_c ast<TO*>(&from) );
}

template<typena me TO, class FROM>
inline const TO& accessor_cast(c onst FROM& from)
{
return *(reinterpret_c ast<const TO*>(&from));
}

#endif
int main()
{
int* blah = new int[125];

int (*krt)[5][5][5] = accessor_cast<i nt([5][5][5][5])>
(blah);

*krt[1][3][2] = 45;

delete [] blah;
}
-JKop

[John Harrison]

>

The following is a mess. What I tried to get was a
multidemensiona l array with new.

It compiles, but don't ask me what it does. Note how at one
point I've write int[5][5][5][5], as opposed to just three
5's, it won't compile otherwise and I don't know why.

Anyway,
#ifndef INCLUDE_ACCESSO R_CAST
#define INCLUDE_ACCESSO R_CAST

template<typena me TO, class FROM>
inline TO& accessor_cast(F ROM& from)
{
return *(reinterpret_c ast<TO*>(&from) );
}

template<typena me TO, class FROM>
inline const TO& accessor_cast(c onst FROM& from)
{
return *(reinterpret_c ast<const TO*>(&from));
}

#endif
int main()
{
int* blah = new int[125];

int (*krt)[5][5][5] = accessor_cast<i nt([5][5][5][5])>
(blah);

*krt[1][3][2] = 45;

delete [] blah;
}

int[5][5][5][5] has 625 elements and your original array has 125. I think
you meant this

int main()
{
int* blah = new int[125];
int (*krt)[5][5] = accessor_cast<i nt[5][5][5]> (blah);

krt[1][3][2] = 45;
delete [] blah;
}

It just an example of an array to pointer conversion. int[5][5][5] is an
array of size 5, where each element happens to be an int[5][5]. Therefore
int[5][5][5] can convert to a pointer to int[5][5]. The syntax for pointer
to int[5][5] is int (*)[5][5].

Another way of writing your cast would be

int (*krt)[5][5] = accessor_cast<i nt(*)[5][5]> (blah);

Nothing was compiled so apologies in advance for any stupid mistakes.

john

[JKop]

Success! I think...

I got three seperate codes to compile, but only one would
run without run-time errors, ie. bad memory
access/allocation.

Here it is:

#ifndef INCLUDE_ACCESSO R_CAST
#define INCLUDE_ACCESSO R_CAST

template<typena me TO, class FROM>
inline TO& accessor_cast(F ROM& from)
{
return *(reinterpret_c ast<TO*>(&from) );
}

template<typena me TO, class FROM>
inline const TO& accessor_cast(c onst FROM& from)
{
return *(reinterpret_c ast<const TO*>(&from));
}

#endif
int main()
{
int* blah = new int[125];

int (*krt)[5][5] = accessor_cast<i nt(*)[5][5]> (blah);

krt[1][3][2] = 45;

delete [] blah;
}

Are we the first people to get a multidimensiona l array out
of the heap?!
Next step is making a template!
(But I still don't understand why we've turned [5][5][5]
into [5][5] in the casts.)
-JKop