C Exercises: Show the basic declaration of pointer

C Pointer : Exercise-1 with Solution

Write a program in C to show the basic declaration of pointer.
Expected Output :

```z stores the address of m  = 0x7ffe97a39854
*z stores the value of m = 10
&m is the address of m = 0x7ffe97a39854
&n stores the address of n = 0x7ffe97a39858
&o  stores the address of o = 0x7ffe97a3985c
&z stores the address of z = 0x7ffe97a39860
```

C Code:

```#include <stdio.h>
void main(void)
{
int m=10,n,o;
int *z=&m ;

printf("\n\n Pointer : Show the basic declaration of pointer :\n");
printf("-------------------------------------------------------\n");
printf(" Here is m=10, n and o are two integer variable and *z is an integer");
printf("\n\n z stores the address of m  = %p\n",  z); // z is a pointer so %p would print the address
printf("\n *z stores the value of m = %i\n",   *z);
printf("\n &m is the address of m = %p\n",   &m); // &m gives the address of the integer variable m
// so %p is the specifier for that address
printf("\n &n stores the address of n = %p\n",   &n);
printf("\n &o  stores the address of o = %p\n",   &o);
printf("\n &z stores the address of z = %p\n\n", &z); // &z gives the address, where the pointer z is
// stored -> still an address -> %p is the right
// specifier
}
```

Flowchart :

C Practice online:

```
#include <stdio.h>
void main(void)
{
int m=10,n,o;
int *z=&m ;

printf("\n\n Pointer : Show the basic declaration of pointer :\n");
printf("-------------------------------------------------------\n");
printf(" Here is m=10, n and o are two integer variable and *z is an integer");
printf("\n\n z stores the address of m  = %p\n",  z); // z is a pointer so %p would print the address
printf("\n *z stores the value of m = %i\n",   *z);
printf("\n &m is the address of m = %p\n",   &m); // &m gives the address of the integer variable m
// so %p is the specifier for that address
printf("\n &n stores the address of n = %p\n",   &n);
printf("\n &o  stores the address of o = %p\n",   &o);
printf("\n &z stores the address of z = %p\n\n", &z); // &z gives the address, where the pointer z is
// stored -> still an address -> %p is the right
// specifier
}
```

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