Pointer of array in C++

Question

I saw the following line of code in C++. I have trouble understanding it. I hope that I can get some help here.

// Example program
#include <iostream>
#include <string>
using namespace std;

int main()
{
    int a[] = {1,2,3,4};
    cout<<*a+1<<;
    cout<<a[1];

}

Question: I don't understand how *a+1 works. It seems pretty unintuitive - are we adding 1 to the array here?

If is c++, why are you tagged [tag:c]????? Did you try to compile and launch it??? — LPs, Feb 02 '16 at 13:58
@LPs Yeah. Since I can run this both on C++ and C, I put both languages on the tag... — Mathemagician, Feb 02 '16 at 14:00
As far as you are able to compile it, launch it and see what is the output....BTW gcc does not compile namespaces... — LPs, Feb 02 '16 at 14:01
This code will definitely not compile as C. C has no namespaces, doesnt know about either of those headers and doesnt have overloaded operators. — Borgleader, Feb 02 '16 at 14:01
this example is as confusing as it could be. Change `{1,2,3,4}` to something like `{1,3,5,7}`, then you can run it and understand what is going on. — 463035818_is_not_an_ai, Feb 02 '16 at 14:03

score 0 · Answer 1 · answered Feb 02 '16 at 13:59

In the expression *a + 1, the array a is evaluated in pointer context, so it points to the first value in the array.

Next, *a dereferences that pointer (meaning it gets the value the pointer points to), and that gives you the first element in the array which is 1.

Finally, 1 is added to that value and printed, so it outputs 2.

score 0 · Answer 2 · answered Feb 02 '16 at 14:03

First of all this statement

cout<<*a+1<<;
          ^^^^

is syntactically wrong.

I think you mean something like

cout << *a+1 << endl;

Expression *a is equivalent to a[0]. So you could write instead

cout << a[0] + 1 << endl;

Take into account that semantically a[0] + 1 is not the same as a[1] though applied to your example the both expressions will yield the same result.:)

score 0 · Accepted Answer · answered Feb 02 '16 at 14:04

To understand the expression *a+1 you should consider the operator priority:

the dereference * has higher priority, so a (the name of an array is also the pointer to the first element) is dereferenced, giving the value 1

Then, 1 will be added to the value.

operator priorities are describe here http://en.cppreference.com/w/cpp/language/operator_precedence

Pointer of array in C++

3 Answers3