In the following I expected 13 to be printed.
I wanted to move arr (which is a pointer to the memory, where int values from array are stored, if i understand everything right) by the size of one array member, which is int.
Instead 45 is printed. So instead making one array-member-wide jump the 5th Array member is retrieved. Why?
int arr[] = {1,13,25,37,45,56};
int val = *( arr + 4 ); //moving the pointer by the sizeof(int)=4
std::cout << "Array Val: " << val << std::endl;
Your assumption is wrong. It moves the pointer 4 elements ahead, not 4 bytes ahead.
*(arr + 4) is like saying in that logic *(arr + 4 * sizeof (arr [0])).
The statement *(arr + 4) is equivalent to arr [4]. It does make for some neat syntax, though, as *(4 + arr) is equally valid, meaning so is 4 [arr].
Your behaviour could be achieved through the following example:
#include <iostream>
int main()
{
int a[3] = {65,66,67};
char *b = reinterpret_cast<char *>(a);
std::cout << *(b + sizeof (int)); //prints 'B'
}
I wouldn't recommend using reinterpret_cast for this purpose though.
(arr+x) accesses the array? How can I change the memory address then?
arr + x part is part of pointer arithmetic rules. arr + 1 advances to the next element of the type that arr points to.void* ?
B? There's no justification for it, and, on a big endian machine, it won't.B for me and doesn't give any warnings, but you're correct in saying that. I did recommend not to do it.arr + 4 will give you 4 items on from the start address of the array, not 4 bytes. That's why you get 45 which is the zeroth item plus 4.
It is performing pointer arithmetic. See this: http://www.eskimo.com/~scs/cclass/notes/sx10b.html
arr is your array and this decomposes to the first element arr[0] which will be 1, then you + 4 to it which moves the pointer by 4 elements arr[4] so it is now pointing to 45.
int val = (*arr) + 4 ; this would produce 5 as you dereference the first element which is 1 and then add 4 what you are doing is incrementing the pointer by 4 positions