Accessing an array out of bounds gives no error, why?

250

I am assigning values in a C++ program out of the bounds like this:

#include <iostream>
using namespace std;
int main()
{
    int array[2];
    array[0] = 1;
    array[1] = 2;
    array[3] = 3;
    array[4] = 4;
    cout << array[3] << endl;
    cout << array[4] << endl;
    return 0;
}

The program prints 3 and 4. It should not be possible. I am using g++ 4.3.3

Here is compile and run command

$ g++ -W -Wall errorRange.cpp -o errorRange
$ ./errorRange
3
4

Only when assigning array[3000]=3000 does it give me a segmentation fault.

If gcc doesn't check for array bounds, how can I be sure if my program is correct, as it can lead to some serious issues later?

I replaced the above code with

vector<int> vint(2);
vint[0] = 0;
vint[1] = 1;
vint[2] = 2;
vint[5] = 5;
cout << vint[2] << endl;
cout << vint[5] << endl;

and this one also produces no error.

c++ arrays

Edit Report

1 Answer

C or C++ will not check the bounds of an array access.

You are allocating the array on the stack. Indexing the array via array[3] is equivalent to *(array + 3), where array is a pointer to &array[0]. This will result in undefined behavior.

One way to catch this sometimes in C is to use a static checker, such as splint. If you run:

splint +bounds array.c

on,

int main(void)
{
    int array[1];

    array[1] = 1;

    return 0;
}

then you will get the warning:

array.c: (in function main) array.c:5:9: Likely out-of-bounds store: array[1] Unable to resolve constraint: requires 0 >= 1 needed to satisfy precondition: requires maxSet(array @ array.c:5:9) >= 1 A memory write may write to an address beyond the allocated buffer.

answered 2009-08-06T16:18:44.420

Accessing an array out of bounds gives no error, why?

1 Answer

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