Write a c plus plus program of array to pointers?

Answer 1

#include
void

printarr(int

a[]);
void

printdetail(int

a[]);
main()
{
int

a[5];
for

(int

i = 0;i<5;i++)
{
a[i]=i;
}
printdetail(a);
}
void

printarr(int

a[])
{
for

(int

i = 0;i<5;i++)
{
printf("value in array %d\n"

,a[i]);
}
}
void

printdetail(int

a[])
{
for

(int

i = 0;i<5;i++)
{
printf("value in array %d and address is %8u\n"

,a[i],&a[i]);
}
}
void

print_usingptr(int

a[])
{
int

*b;

Answer 2

The following program demonstrates how to create both static and dynamic arrays of pointers. In both cases, the pointers in the arrays are allocated new memory (which means the individual elements must be released), however the pointers could just as easily point to existing allocations (which means the individual elements must not be released, since the memory would not actually belong to these pointers).

Note that the static array, a, has the same address and value. This is because a is a reference to the start address of array. References are not variables so they have no address of their own, they are simply aliases for the memory they refer to. Compare with b which is not a reference, it is a pointer, and therefore has its own separate address from the array it points to.

The two dereferenced values show the value of the first element in the array, since the first element always resides at the start of the array. The first dereference is the value of the pointer in that element, the second is the indirect value of that pointer.

Note that after we destroy the elements in the dynamic array, b, we must also destroy the array itself.

#include<iostream>

int main()

{

// declare static array of pointer to int

int* a[5];

// initialise array

for(int i=0; i<5; ++i)

a[i] = new int(i+1);

// print values

std::cout<<"Name\tAddress\t\tValue\t\tDereference\tDereference"<<std::endl;

std::cout<<"a\t"<<&a<<"\t"<<a<<'\t'<<*a<<'\t'<<**a<<'\n'<<std::endl;

std::cout<<"Name\tAddress\t\tValue\t\tDereference"<<std::endl;

for(int i=0; i<5; ++i)

std::cout<<"a["<<i<<"]\t"<<&a[i]<<"\t"<<a[i]<<"\t"<<*a[i]<<std::endl;

std::cout<<std::endl;

// release pointers

for(int i=0; i<5; ++i)

delete( a[i] );

// declare dynamic array of pointer to int

int** b = new int*[5];

// initialise array

for(int i=0; i<5; ++i)

b[i] = new int(i+1);

// print values

std::cout<<"Name\tAddress\t\tValue\t\tDereference\tDereference"<<std::endl;

std::cout<<"b\t"<<&b<<"\t"<<b<<'\t'<<*b<<'\t'<<**b<<'\n'<<std::endl;

std::cout<<"Name\tAddress\t\tValue\t\tDereference"<<std::endl;

for(int i=0; i<5; ++i)

std::cout<<"b["<<i<<"]\t"<<&b[i]<<"\t"<<b[i]<<"\t"<<*b[i]<<std::endl;

std::cout<<std::endl;

// release pointers

for(int i=0; i<5; ++i)

delete( b[i] );

delete[]b;

}

Output:

Name Address Value Dereference Dereference

a 0021FF10 0021FF10 00230990 1

Name Address Value Dereference

a[0] 0021FF10 00230990 1

a[1] 0021FF14 002309C0 2

a[2] 0021FF18 002309F0 3

a[3] 0021FF1C 00230A20 4

a[4] 0021FF20 00230A50 5

Name Address Value Dereference Dereference

b 0021FEE0 00333870 00230A50 1

Name Address Value Dereference

b[0] 00333870 00230A50 1

b[1] 00333874 00230A20 2

b[2] 00333878 002309F0 3

b[3] 0033387C 002309C0 4

b[4] 00333880 00230990 5

Answer 3

char x[32], *px;

px= x;

...

from now on simply use 'px' instead of 'x', and that will work perfectly well

Answer 4

It is not exactly a question, but here is a little hint:

printf ("%d: %p\n", i, &array[i]);