Obviously there is other issues around not sanitising the data
yourself, but in the context of the reply - i.e. buffer overflows for
arbitrary code exec - java is fully protected.
any access to an array is checked by the vm.
-- Michael
On 3/29/06, Eliah Kagan <degeneracypressure@gmail.com> wrote:
On 3/29/06, Andrew van der Stock wrote:
This is not quite true.
Java does not prevent integer overflows (it will not throw an
exception). So you still have to be careful about array indexes.
On 3/28/06, michaelslists@gmail.com replied:
No you dont.
Arrays are all bounds checked; ..., that is, the following code will
throw an exception:
================================
class Foo {
static {
int[] m = new int[2];
System.out.println(m[34]);
}
}
================================
What do you mean by "overflow"? Do you mean this?
================================
class Foo {
static {
int m = Integer.MAX_VALUE;
int k = Integer.MAX_VALUE + Integer.MAX_VALUE;
System.out.println(m);
System.out.println(k);
System.exit(0);
}
}
================================
if so, I don't see how that is an issue.
-- Michael
That is an issue in a limited way--if you are trying to access a
record with a high enough number (say by adding a number to a previous
array index), you might end up accessing a record with a low number,
which could potentially compromise the security of an application if
certain assumptions are made. But this would only be within the same
array that is already being accessed. The risk is minimal compared to
the risks of accessing past the end of an array in, say, C.
Even with bounds checking, there is no general way for a programming
language to stop the programmer from writing a program that accesses
the wrong piece of data in within a data structure, causing a security
problem. Java was never designed to solve this sort of problem. Java
does abstract data access so that many common bugs like buffer
overflows are prevented, which is very useful.
-Eliah
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