Hello i would like to find out how its doing. I have code like that:
int tab[] = {1,2,3};
int* p;
p = tab;
cout <<p<<endl; // cout adress here
cout <<p[0]<<endl; // cout first element of the array
How its doing that p is storing address of first element but p[0] is already storing first element?
its working like p[0] = *p ? and p[1] is working like p[1] = *(p+1)?
I know how to use it but I'm little confused because i don't really understand it
How its doing that p is storing address of first element
p is a pointer. A pointer stores an address.
but p[0] is already storing first element?
p[0] is an an element of the array. It stores an integer.
It is possible for a program to store both of these at the same time.
its working like p[0] = *p ? and p[1] is working like p[1] = *(p+1)?
Correct. This is how the subscript operator works on pointer and integer.
How its doing that
pis storing address of first element
The data type int tab[] is the same as int*, so does the type of p.
So both these variables point to the same memory location, and that will be the start of the array.
but
p[0]is already storing first element?
Variable tab contains a pointer to the first element of tab[0], so assigning the same value to another variable will have an equivalent effect.
its working like
p[0] = *p ?andp[1]is working likep[1] = *(p+1)?
Yes, *(p+i) is same as p[i].
If you need a detailed explanation please check this link
Pointer p points to the first variable of an array and as the array has a contiguous allocation of memory, so on *(p+1) it starts pointing to the second variable of the array. You can understand this by further example.
#include <iostream>
using namespace std;
int main()
{
// Pointer to an integer
int *p;
// Pointer to an array of 5 integers
int (*ptr)[5];
int arr[5];
// Points to 0th element of the arr.
p = arr;
// Points to the whole array arr.
ptr = &arr;
cout << "p =" << p <<", ptr = "<< ptr<< endl;
p++;
ptr++;
cout << "p =" << p <<", ptr = "<< ptr<< endl;
return 0;
}
