Question

I was trying some basic pointer manipulation and have a issue i would like clarified. Here is the code snippet I am referring to

    int arr[3] = {0};
*(arr+0) = 12;
*(arr+1) = 24;
*(arr+2) = 74;
*(arr+3) = 55;
cout<<*(arr+3)<<"\t"<<(long)(arr+3)<<endl;
//cout<<"Address of array arr : "<<arr<<endl;
cout<<(long)(arr+0)<<"\t"<<(long)(arr+1)<<"\t"<<(long)(arr+2)<<endl;;
for(int i=0;i<4;i++)
    cout<<*(arr+i)<<"\t"<<i<<"\t"<<(long)(arr+i)<<endl;
//*(arr+3) = 55;
cout<<*(arr+3)<<endl<<endl;

My problem is:
When I try to acces arr+3 outside the for-loop , I get the desired value 55 printed. But when I try to access it through the for loop, I get some different value(3 in this case). After the for loop, it is printing the value as 4. Could someone explain to me what is happening? Thanks in advance..

Answer 1

You have created an array of size 3 and you are trying to access the 4th element. The outcome is therefore undefined.

Since you allocate the array in the stack, the first time you try to write the 4th element, you are actually writing beyond the space that was allocated for the stack. In Debug mode this will work, but in Release your program will probably crash.

The second time you are reading the value at the 4th place you are reading the value 4. This makes sense, as the compiler has allocated the stack space after the array for variable i, which after the loop has finished executing will have the value 4.