Question

In the following code, why is not possible to do this dereferencing: *arr = 'e'. Shouldn’t the output be the character 'e'?

int main ()
{
    char *arr = "abt";
    arr++;
    * arr='e'; // This is where I guess the problem is occurring.
    cout << arr[0];

    system("pause");

}

I am getting the following error:

Unhandled exception at 0x00a91da1 in Arrays.exe: 0xC0000005: Access violation writing location 0x00a97839.

Answer 1

int main ()
{
    char *arr= "abt"; // This could be OK on some compilers ... and give an access violation on others
    arr++;
    *arr='e'; // This is where I guess the problem is occurring.
    cout<<arr[0];

    system("pause");

}

In contrast:

int main ()
{
    char arr[80]= "abt"; // This will work
    char *p = arr;
    p++;    // Increments to 2nd element of "arr"
    *(++p)='c'; // now spells "abc"
    cout << "arr=" << arr << ",p=" << p << "\n"; // OUTPUT: arr=abc,p=c
return 0;    
}

This link and diagram explains “why”:

http://www.geeksforgeeks.org/archives/14268

Memory Layout of C Programs

A typical memory representation of C program consists of following
sections.

1. Text segment
2. Initialized data segment
3. Uninitialized data segment
4. Stack
5. Heap

And a C statement like const char* string = “hello world” makes the
string literal “hello world” to be stored in initialized read-only
area and the character pointer variable string in initialized
read-write area.