C Programming - Pointers - Discussion

#include<stdio.h>

int main()
{
int arr[3] = {2, 3, 4};
char *p;
p = arr;
p = (char*)((int*)(p));
printf("%d, ", *p);
printf("\n address pointed by p%u",p);
p = (int*)(p+1);
printf("\n%d", *p);
printf("\n address pointed by p%u",p);
p++;
printf("\n%d", *p);
printf("\n address pointed by p%u",p);
return 0;
}

Output:

2.
Address pointed by p 2686704.

0.
Address pointed by p 2686705.

0.
Address pointed by p 2686706 // after int conversion why it only increment only one byte m/r location. And why not printing 3.

Don't get confused by type conversions they are just used to confuse you. You can also write it like this:

#include<stdio.h>
int main()
{
int arr[3] = {2, 3, 4};
char *p;
p = arr;
printf("%d, ", *p);
p = p+1;
printf("%d", *p);
return 0;
}

o/p = ?

char *p;p = arr
(char*)(int*)p
printf("%d,p");
p = (int*)p;
printf("%d,p");

(int*)p+1 comes to 2nd location in an array. Then how it prints zero?

I didn't understand this. If we are typecasting to int and giving some address then it will print the contents up to next int size (2 bytes or 4 bytes) (depending on compiler) so here 2 is fine but next thing it should print 3. Given below is little Indian representation.

First byte is 2. Incrementing it goes to next byte. Then typecasting to int and printing should print next two bytes (sizeof int) i.e. 3.

00000010 00000000 00000011 00000000 00000100 00000000.

The given problem should be like this, otherwise it may generate compile time error.

#include<stdio.h>

int main()
{
int arr[3] = {2, 3, 4};
char *p;
p = (char*)arr;
p = (char*)((int*)(p));
printf("%d, ", *p);
p = (p+1);
printf("%d", *p);
return 0;
}

Assume the Base Address of arr as 1024.

p=arr => p=1024.

here The arr in which the elements are arranging in the order as below,
1024 => 2.
1028 => 3.
1032 => 4.

Then *p=>2, means *(1024) =>2.

while perform p=(int*)(p+1); here p as a char pointer p+1 gives the 1025 instead of 1028,because the p incremented by 1 byte rather than 4 byte where there as an integer pointer.

so afterwards *p points no where as *(1025),that's why the o/p gives the Zero by default for second print statement.

Thus the output become 2,0.

Hope all understand.

What is this endian method?

(char*)((int*)(p))
Could someone explain this?

Why we type casted p again to char?

p=(char*)((int*)(p));