How to access pointer inside a structure pointer

[mohammadthalif]

Hello friends,

struct datas{
int *a,*b;
} *name;

can you explain me how can i get the data at location pointed by *a and *b.

i tried,

name->*a = 10; // but this produces an error...
name->*b = 5; // but this produces an error...

Thanks...
H.Mohamed Thalib

[JosAH]

I'll move your question to the C/C++ forum section where it belongs.

kind regards,

Jos

[ilikepython]

Hello friends,

struct datas{
int *a,*b;
} *name;

can you explain me how can i get the data at location pointed by *a and *b.

i tried,

name->*a = 10; // but this produces an error...
name->*b = 5; // but this produces an error...

Thanks...
H.Mohamed Thalib

You need to dereference it in front:
[code=cpp]
*name->a = 10;
[/code]
Use parenthesis to make it clearer:
[code=cpp]
*(name->a) = 10;
[/code]

[JosAH]

Even more important: a pointer must point to something; in the OPs example
the name pointer doesn't point to a struct with two int pointers so nothing can
be dereferenced at all.

kind regards,

Jos

[mohammadthalif]

You need to dereference it in front:
[code=cpp]
*name->a = 10;
[/code]
Use parenthesis to make it clearer:
[code=cpp]
*(name->a) = 10;
[/code]

Thank you guys ... i t works,....

[sajet]

try this

name = (struct datas *)malloc(sizeof *name);

now u can access

name->a = 10;
name->b = 5;

[TamusJRoyce]

Close Sajet... But your allocation only allocates space for an unsigned integer/pointer, since name is a pointer. You need the sizeof(datas).

name = (struct datas *)malloc(sizeof (datas))

Now that you have a valid datas allocated to name, you need *a and *b to be something.

// Always check to see if name truely has been allocated.
// The system could be out of memory.
if (name != NULL)
{
name->a = (int)malloc(siz eof(int))
name->b = (int)malloc(siz eof(int))

if (name->a != NULL)
{
*name->a = 5

// Done using it, so lets free it
free(name->a)
} // End if

if (name->b != NULL)
{
*name->b = 10

// Done using it, so lets free it
free(name->b)
} // End if

// Done using it, so lets free it
free(name)
} // End if

Search how pointers work with the below if you're interested...
1. Dynamic Arrays
2. Polymorphism

1. But dynamic memory allocation is exciting/useful when you get into linked lists and dynamic arrays.

2. It's also useful in C++ when dynamically allocating a class which inherits a base class (using new/delete instead of malloc/free, of course), and that class can be set to a base class pointer with that class abstractly performing like the base class, but doing much more stuff in the extended part.

Hopefully helpful,

TamusJRoyce

[Banfa]

Close Sajet... But your allocation only allocates space for an unsigned integer/pointer, since name is a pointer. You need the sizeof(datas).

Not true notice Sajet wrote

name = (struct datas *)malloc(sizeof *name);

not

name = (struct datas *)malloc(sizeof name);

since *name is a struct of type datas sizeof *name returns the size of the struct name points to. In fact for me

name = malloc(sizeof *name);

is preferable to

name = malloc(sizeof (datas));

because if the type of name is changed (which I admit is notterribly likely) the first form of the malloc still works without alteration where as the second requires a change to the sizeof to work correctly. For any type T using

T *pointer;
pointer = malloc(sizeof *pointer);

is about the most robust way you can write the malloc. NOTE I have removed the casts because they are not necessary (in fact they are dangerous) in C and if you are using C++ you should be using new anyway to ensure and constructors get called.