C Programming - Pointers - Discussion

Please tell me how this program will execute.

Good explanation viraj.

It's give error in turbo c3
p = arr; \\ can not convert "int *" to "char *"
p = (int*)(p+1); \\ can not convert "int *" to "char *"

First it will perform outer typecast then it will perform inner typecast.

And in second printf since pointer type is char so it point to 1byte data. Hence it will fetch the first byte of value 2 and p+1 point to second byte of value 2.

So it is zero (2=0000000000000010).

@viraj accepted your idea. its the same

p = (char*)((int*)(p));
// till now the pointer p is type casted to store the variable of type character.

printf("%d, ", *p); // %d means integer value so value at first address i.e. 2 will be printed.

p = (int*)(p+1); // here p is still of type character as type casted in step 1 so p(i.e address) and plus 1 will increase only by one byte so

as per viraj suggested

Assuming that integer requires 2 bytes of storage
the integer array will be stored in memory as


value 2 3 4
address 00000010 00000000 00000011 00000000 00000100 00000000
pointer p+1


so p+1 points to that location which is unfilled as during intialization 2,3,4 were stored in variable of type integer(2 bytes).

so p+1 will point to 00000000.

(int*)p+1 // p+1 is type casted again to integer

printf("%d", *p); // this will print 0 as output as by default integer contains 0 as value.

It gives error "cannot convert 'int*' to 'char*'" in both lines

p=arr;
p=(int*)(p+1);

p = (char*)((int*)(p));
even without this line program will run fine

char p stored in int type arr. which is not possible without conversion.

I didn't get this one can anyone explain me?