Understanding * operator in pointers.

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Hello,
I am not used to C programming so i would like to understand the useage of * operator in C pointers.I tried to google it but nothing is quite convincing.(An example will be very good)
I would like to know What the following indicate or function as:-
1. int *i;
2. Struct My_DataBase *My_Name; (My_DataBase is defined earlier as a struct)
3. int* i;
4. j = *(int *)k; (j and k are defined as integers).
5. Struct My_DataBase *My_Name1 = (My_DataBase *) (j + (k * l));

Kindly help and let me know if you have any

Posted 4-Nov-10 0:33am

asjadazeez

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CPallini · Answer 1 · 2010-11-04T00:42:00

(1) declares i as a pointer to an int.
(2) declares My_Name as a pointer to a My_Database struct.
(3) has the same meaning of (1).
(4) interpreting k as pointer to int (i.e. assuming that k is an address) set j to the int value pointed.
(i.e. get the int value having address k).
(5) interpreting the expression (j+(k*l)) as a pointer to a struct My_DataBase, set My_Name1 pointing to the same location.

I guess (2),(5) should be written:

2. struct My_DataBase *My_Name;
5. struct My_DataBase *My_Name1 = (struct My_DataBase *) (j + (k * l));

:)

Sauro Viti · Answer 2 · 2010-11-04T01:07:00

Solution 2

Be careful about the little differences between C and C++:

~~in C++ is admitted to write int* p; and int *p; to declare p as a pointer to integer. In C you should write int *p;~~
in C to declare a variable of a struct type you should always use the following syntax:

C
```
struct MyStruct
{
   ...
};

struct MyStruct a;
struct MyStruct *b;
```
In C++ you can omit the struct keyword when declaring variables:

C++
```
struct MyStruct
{
   ...
};

MyStruct a;
MyStruct *b;
```

Posted 4-Nov-10 1:07am

Sauro Viti

Updated 4-Nov-10 3:15am

v2

Comments

CPallini 4-Nov-10 7:42am

Are you sure about C pointer declaration? I guess most C compilers would accept 'int* p' (at least Microsoft CL and GCC do).

Sauro Viti 4-Nov-10 9:13am

You make me doubt! It 's been so long since I studied books on C!