Help: malloc and const variables??

Hi,
what happens if we allocate more than MAX_SIZE in below situation. Is the result unpredictable??

How is this related to corrupting "const" declarations??

Are they(malloc and const) both allocated from heap???


Thanks,
Tharakan.

const int p=100; <-- can a malloc like below corrupt a const declaration ???

{
int MAX_SIZE=10;
int *inptr[MAX_SIZE];

intptr[0] = alloc_and_set();
intptr[1] = alloc_and_set();
:
:
:
inptr[9] = alloc_and_set(); <-- reaches MAX_SIZE???
inptr[10] = alloc_and_set(); <-- Is this valid???
inptr[11] = alloc_and_set(); <-- Is this valid???
inptr[12] = alloc_and_set(); <-- Is this valid???
inptr[13] = alloc_and_set(); <-- Is this valid???
inptr[14] = alloc_and_set(); <-- Is this valid???

}

int * alloc_and_set(){
int *t;
t= (int*) malloc(sizeof(int));
*t = 10;
return t;
}

}

You would probably get a segmentation fault for trying to access an array outside its bounds. So I would suggest not to do this.

Just like any array, if you go out of bounds, you risk killing your program and other BadThingsTM.

Quzah.

Why is it corrupting the const declaration??? Isn't supposed to be protected?? Is the initialised variables also allocated in the Heap???

The protection means that code cannot change the value of variable p directly by assigning a value to it. However, if you write data beyond the bound of an array or at some other place which you haven't allocated, then it is possible that you write data at the place ware variable p is stored.

> Why is it corrupting the const declaration?
Once you have ANY undefined behaviour (like stepping off the end of the array), then ALL bets are off - your program is broken.
Attempting to rationalise things with 'but its const' is not going to help you.