#include <iostream>
using namespace std;
int main()
{
    int i;
    const char *arr[] = {"C", "C  ", "Java", "VBA"};
    const char *(*ptr)[4] = &arr;
    cout <<   (*ptr)[2];
    return 0;
}

Output is "ava"

I am not able to understand why code is giving this output. I tried to breakdown the code line by line to understand, but I am not that good in pointers so I am not able to understand what *(*ptr)[4] and (*ptr) is doing? So I want to know their functionality in this code

CodePudding user response：

Lets look at different statements in your program.

Case 1

Here we consider the statement:

 const char *arr[] = {"C", "C  ", "Java", "VBA"};

The above statement creates an array named arr that contains elements of type const char*. The size of this array is 4.

Case 2

Here we consider the statement:

const char *(*ptr)[4] = &arr;

The above statement creates a pointer named ptr that points to an array of size 4 that contains elements of type const char*. Moreover, this pointer ptr is initialized with &arr. This means, the pointer ptr is pointing to the array arr.

Since arr is of type const char* [4] therefore , &arr gives us a const char* (*)[4].

Case 3

Here we consider the statement:

cout <<   (*ptr)[2];

In the above statement, (*ptr) dereferences the pointer named ptr. This gives us the object to which ptr was pointing. But we know that(from case 2) ptr is pointing to the array arr. So the result of (*ptr) is nothing but the array named arr.

Now, the operator [] has higher precedence than operator . This means that the array arr that we got as a result of dereferencing, is subscripted for the value 2. That is we now get the 3rd element(which is at index 2) of the array arr. This means we now have a pointer that is pointing to the first element/character of the string literal "Java".

Finally, that pointer value is incremented. This is because of the . This mean now after incrementing, the pointer is pointing to the next character(which is 'a') in the string literal "Java".

So when you use cout on this pointer you get ava as the output because the pointer was pointing to the character 'a' and not 'J'.

CodePudding user response：

1. First line:

   const char *(*ptr)[4] = &arr;
   

   This declares ptr to be a pointer to an array of 4 pointers to const char, and initialize it to point to arr

2. second one (I willingly omitted the cout):

      (*ptr)[2];
   

   This increments the third element of the array pointed to by ptr, hence arr. After that, printing elements of the array would give C, C , ava and VBA.