*a is pointer a, so what is **a ?

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C++

*a is pointer a, so what is **a ?
Help me !

Posted 25-Aug-11 5:12am

nctit09

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Comments

Maximilien 25-Aug-11 11:24am

it's a pointer to a pointer. (too short for a real answer).

Simon Bang Terkildsen 25-Aug-11 11:28am

Short but true

Albert Holguin 25-Aug-11 16:30pm

don't think answers need to be a certain length... you can always follow up with a link for reference...

5 solutions

Solution 3

**a is pointer of *a.
Otherway its a representation of 2 dimention pointer.
concider this:

C++

a[] is array of a.
 AND
a[][] is array of a[].


suppose you have:
#define m=3
#define n=2
/////////////
...
int a[m][n];
int i,j;
for(i=0;i<m;i++)>
for(j=0;j<n;j++)>
{
///////
}

Like  this you can access:
int **a;
//use alloc or malloc to initialize the 2 dimentional pointer

for(i=0;i<m;i++)>
for(j=0;j<n;j++)>
{
///////
}

Posted 25-Aug-11 6:38am

Аslam Iqbal

Solution 2

It is called as double pointer,most of the times used to point to the dynamically created two dimensional arrays.
The following link has few more discussions:
http://www.daniweb.com/software-development/c/threads/69966[^]

Posted 25-Aug-11 5:57am

hakz.code

Updated 25-Aug-11 7:49am

v3

Comments

Smithers-Jones 25-Aug-11 13:50pm

I corrected your link.

hakz.code 26-Aug-11 0:14am

Thanks,I was wondering why the hyperlink was not working!

Stefan_Lang 26-Aug-11 4:56am

Actually it's a double dereference, not a double pointer. ;-)

Of course, everyone knows what we're talking about, but we should be more concise when explaining basic C syntax to a beginner. See C.Pallini's response (and mine - but C.Pallini beat me to it)

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Stefan_Lang · Accepted Answer · 2011-08-25T22:51:00

Actually, *a is only syntactically correct if a itself is a pointer, and in that case *a can not be a simply because the types don't match, no matter what type a is defined to point to.

What you probably mean is the syntax used to define a (pointer) variable:

C++

int f1() {
   int a; // this is a variable of type int
   a = 1;
   return a; // this will return the value 1
}
void f2() {
   int *a; // this is a pointer to a value of type int
   int b;
   a = &b; // now a points to the contents of b
   b = 2;
   return *a; // returns the value of b, which is 2
}
void f3() {
   int **a; // this is a pointer to a pointer to a value of type int
   int *b;
   int c = 3;
   b = &c; // ok, b points to the contents of c
   a = &b; // a now points to the contents of b which points to the contents of c
   return **a; // returns 3
}

CPallini · Accepted Answer · 2011-08-25T08:18:00

Solution 4

In the statement:

C++

int *p;

p is a pointer (not *p)
The variable declaration may be interpreted as follows:
*p is an int, hence p is a pointer to an int.

A similar reasoning gives

C++

int ** q;

q is a pointer to a pointer to an int.

Generally speaking, the expressions *a, **a cannot be interpreted correctly without context.
For instance, the statement

C++

printf("%d\n"), **q);

(if correct) tell us q is a double pointer.

Posted 25-Aug-11 8:18am

CPallini

Updated 25-Aug-11 23:00pm

v3

Comments

Sergey Alexandrovich Kryukov 25-Aug-11 19:48pm

Good stuff, a 5.
--SA

CPallini 26-Aug-11 2:48am

Thank you.

Stefan_Lang 26-Aug-11 4:58am

My 5 for being the first to not fall into the trap of mixing up dereferences with pointer definitions :-)

Or at least you beat me to the posting...

Marc A. Brown · Accepted Answer · 2011-08-25T05:51:00

Solution 1

Does this[^] help?

Posted 25-Aug-11 5:51am

Marc A. Brown

Comments

Albert Holguin 25-Aug-11 16:32pm

Good link, +5

Sergey Alexandrovich Kryukov 25-Aug-11 19:47pm

It is good, my 5.
--SA

*a is pointer a, so what is **a ?

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