Thread: Typecasting pointers

Hi!

I was studying typecasting with pointers when I got stuck on the following exercise:

Code:

#include <stdio.h>

int main() {

int i;

char char_array[5] =  {'a','b','c','d','e'};
int int_array[5] = {1,2,3,4,5};

char_pointer = (char*) int_array;
int_pointer = (int*) char_array;

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

printf("[integer pointer] points to %p, which contains the char '%c'\n", int_pointer, *int_pointer);

int_pointer = (int*)((char*) int_pointer+1;
}
}

My question is the following and concerns about the instruction in bold:

• Why do I have to typecast back the pointer? I know what happens if I don't typecast the pointer in order to see good results, but if I run the program without putting that (int*) before the ((char*) int_pointer+1) it works fine, with just a warning from the compiler.
  Actually I don't get what happens in memory in order to be necessary to typecast it back.

Thanks in advance!

I see, so in memory nothing changes. Thank you, I just wanted to know this.

Actually this exercise is just a part of different versions of the same task: this was just an example on how the same thing can be done using pointer that points do different types, but I couldn't get the reason of that typecasting back.

Thanks a lot!