Thread: help with pointers

i have a question.
why do we need to declare pointers as int *p when int p also does the same work?
eg:

Code:

//eg1
int a=10, *p;
p=&a;
printf("%p", p);
//eg2
int a=10, p;
p=&a;
printf("%p", p);

both give the same result

Although you can get the memory address just fine with this method, you wouldn't be able to dereference to get the value without declaring it as a pointer. With eg2 try changing the printf statement to:

Code:

printf("%d", *p);

You will get a compiler error on this one, thus the value isn't retrievable and the pointer is pointless.

Originally Posted by parthpk92

i have a question.
why do we need to declare pointers as int *p when int p also does the same work?
eg:

Code:

//eg1
int a=10, *p;
p=&a;
printf("%p", p);
//eg2
int a=10, p;
p=&a;
printf("%p", p);

both give the same result

For most things you don't need pointers... in fact there is some wisdom in avoiding them as much as possible.

However, they do have their uses... For example I routinely deal with very large strings, text files loaded as one big piece, into memory that is allocated with malloc(). Then I will take with strtok() and build an array of pointers to the beginning of each line... None of it is the actual string but it does tell me where the strings are.

What a pointer does is allow you to store the address of something... not it's value, it's location... For example: "The house at 37 anystreet has a red door"... the pointer stores the address of the red door (37 anystreet) but not the red door itself. To see the red door you need to go to the address, or in computer terms you need to "dereference" the address to get at the data.


@megafiddle... The second example will compile and it will get the address of the data in a ... The integer variable p becomes just a number that happens to be the same as the address of a that will be utterly useless as a pointer.

thanks everyone!
especially Crptix for a lucid explanation. i'm just starting out so your simple was extremely helpful.