pointer c++ explanation

Question

I am confused with this block of code:

ipPtr = ipPtr + 3; // 5 cout << *ipPtr << endl;

Why the cout is not 5 but some random large number? can anyone explain to me please. As my understanding I thought the cout << *ipPtr << endl; is pointed to the *ipPtr above it. Am I right ?

#include <iostream> void main(){ using namespace std; int iaArray[] = {1,2,3,4,5}; int* ipPtr = 0; ipPtr = &(iaArray[1]); cout << *ipPtr << endl;//2 ++ipPtr; cout << *ipPtr << endl;//3 ipPtr = ipPtr + 3; //not 5 but random number. cout << *ipPtr << endl; }

You are pointing beyond the end of the array. The array size is 5, and you are pointing at the 6th element that's why you are getting some random number. 2 + 1 + 3 = 6 not 5 :) — Master Yoda
– Master Yoda, Commented Jun 19, 2011 at 4:10

Ed Swangren · Accepted Answer · 2011-06-19 04:17:43Z

Because you have incremented the pointer past the end of the array. You seem to have forgotten that you wrote ++ipPtr before adding 3 to it.

 &(iaArray[1]) | iaArray = { 1, 2, 3, 4, 5 } ? | | ++ipPtr ipPtr + 3

Petar Ivanov · Accepted Answer · 2011-06-19 04:10:06Z

Because when you add 3 to the pointer it's already on the third position of the array, so it ends up after the last element.

zad · Accepted Answer · 2011-06-19 04:12:44Z

ipPtr = &(iaArray[1]);

//Pointing to the second position (first one is 0)

++ipPtr;

//Pointing to the third position

//3 + 3 = 6 ipPtr = ipPtr + 3;

The array only has 5 positions so it prints whatever is in that memory location not 5 which is in the fifth position.

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