My question is that if you were print out the resulting *(ary + i), which I know is another way to say ary[i] value, would the following output be a hexadecimal/garbage data or would that specific index be assigned a new value based on the result computed from *ary + i? This whole pointer stuff still throwing me off.
int ary[] = [7, 5, 3, 1, 2, 4, 6, 8];
for (int i = 0; i < 8; i++)
{
*(ary + i) = *ary + i;
}
ary decays to a pointer in the expressions you posted. And since it always produces the same address (the first element), *ary will evaluate to 7 at every iteration.
Since you understand that *(ary + i) is equivalent to ary[i], you should now gather that your loop body is akin to this:
ary[i] = 7 + i;
int ary[] = [7, 5, 3, 1, 2, 4, 6, 8]; is wrong assignment method,
one must use { } to assign values to array.
So, correct way to define is:
int ary[] = {7, 5, 3, 1, 2, 4, 6, 8};
Here *(ary + i) is equivalent to ary [i] , and *ary + i is equivalent to ary [0] + i.
Addressing your question, no it won't throw hex or garbage, you have dereferenced a pointer, so it will give the data stored at that location.
Code:
#include <iostream>
int main ()
{
int ary[] = {7, 5, 3, 1, 2, 4, 6, 8};
for (int i = 0; i < 8; i++)
{
*(ary + i) = *ary + i;
}
for (int j=0;j <8;j++)
{
std::cout <<ary [j];
}
}
Output:
7891011121314
Hope you draw the correlation.
if this helps great, if it doesnot, indicate so in comments, willing to edit
*ary + i is not equivalent to ary [i] + 1 it is equivalent to ary[0] + i(not the downvoter btw)
ary[i] = ary[0] + i.