Hi, as i understand c doesnot perform bound checking , and any acess of previous index will just get the prev indexed mem contents and same for beyond index.


main()
{
int y = 98;
int a[5]={11,12,13,14,15};
int z = 541;

printf("a[0]=%d a[4]=%d\n",a[0], a[4]);

printf("a[-1]=%d\n",a[-1]);
printf("a[5]=%d\n",a[5]);
return 0;
}

[santosh@localhost test]$ ./a.out
a[0]=11 a[4]=15
a[-1]=134518428 .. why it is not 98? as it refrers to prev mem location
a[5]=98 ... and i 'm confused how its 98 but should be 541.


If the same set of local variables is made global s it works , so is theat difference because of stack ?

Plz provide clarification, thanks in advance.

Yes, the difference is because of the stack.

Modify your program to print out the addresses of the stack variables, and you will see how your compiler uses the stack.

Disassemble your program, and you will become even more enlightened.

What architecture (x86, ARM, etc.) and what compiler (GCC, MSVC++, etc.) are you using?

As far as I am aware the language doesn't dictate to the compiler in what order it should allocate the memory for your program variables. So one way that a compiler could work would be to collect details of each variable as it parses your code. In the example that you have given the compiler would identify three variables y, a and z. When it is parsing the code it will notice when these variables are being used. "Heck," says the compiler, "You don't even use the variables y and z" and so it will throw a warning and possibly not even allocate memory for those two. But given that it will allocate memory how might it do that? Well if the identified variables have been placed on a stack data structure then it will pop them off one at a time and generate the code to allocate the memory for them. Given that it will have them in reverse order.

As I said this is just one way that a compiler could approach the problem; and I've not mentioned any approaches to optimisation.