dynamic allocation

[venkatapr]

int(*p)[5];

what this statement does..?????


does it allocate some memory or does it just declare p......what it actually does ..???i am new to c++ please help

[JLBorges]

> int(*p)[5];

p is an (uninitialized) pointer to array of 5 int.

You could now do:

int array[5] = {0} ;
p = &array ;

[jain amit 14]

yes it allocate some memory. try this

int (*p)[5];
cout<<"Address "<<&(*p)[0]<<endl;
cout<<"Address "<<&(*p)[1]<<endl;
cout<<"Address "<<&(*p)[2]<<endl;
cout<<"Address "<<&(*p)[3]<<endl;
cout<<"Address "<<&(*p)[4]<<endl;

[andywestken]

int (*p)[5]; does not allocate memory; it is, as JLBorge said, a pointer (to an array of int of size 5.)

#include <iostream>
#include <typeinfo>  // for 'typeid' to work 
using namespace std;

int main()
{
    int (*p)[5];
    int* q[5];
    int  r[5];

    cout << "type of p = " << typeid(p).name() << endl;
    cout << "type of q = " << typeid(q).name() << endl;
    cout << "type of r = " << typeid(r).name() << endl;
    cout << endl;

    cout << "sizeof p = " << sizeof(p) << endl;
    cout << "sizeof q = " << sizeof(q) << endl;
    cout << "sizeof r = " << sizeof(r) << endl;
    cout << endl;

    return 0;
}

Output:

type of p = int (*)[5]
type of q = int * [5]
type of r = int [5]

sizeof p = 4
sizeof q = 20
sizeof r = 20

So you can see that p has a size of just 4 bytes (the size of a pointer when compiling 32-bit, as I did.)

Andy

PS @jain amit 14

Please see:
How to use code tags
http://www.cplusplus.com/articles/jEywvCM9/