Does Accessing a 2D Array Element Using Pointer Arithmetic Beyond Its Bounds Result in Undefined Behavior?

Can a 2D Array Be Treated as a Concatenated 1D Array?

Consider the following code snippet:

int a[25][80];
a[0][1234] = 56;
int* p = &a[0][0];
p[1234] = 56;

Question: Does the code access the array beyond its bounds, resulting in undefined behavior?

Answer: Yes, both lines 2 and 4 exhibit undefined behavior.

In C , array indexing is essentially pointer arithmetic. When accessing a[i][j], the compiler effectively translates it to *(a[i] j). Similarly, p[i] refers to *(p i).

In this case, the array a has dimensions 25 x 80. The first row of the array, a[0], contains 80 elements (ranging from a[0][0] to a[0][79]) and is allocated at a contiguous memory location.

By accessing a[0][1234], the code attempts to access an element at index 1234 in the first row, despite a[0] only containing elements within the range [0, 79]. This is out of bounds and triggers undefined behavior.

The same logic applies to line 4. While p points to the first element of a[0], the array is still dimensioned as 25 x 80. Accessing p[1234] is essentially an out-of-bounds pointer arithmetic operation, once again leading to undefined behavior.

Furthermore, as Language Lawyer pointed out in the comments, a similar code snippet using a constexpr would fail to compile due to the compiler detecting such undefined behaviors in a constant expression:

constexpr int f(const int (&a)[2][3])
{
    auto p = &a[0][0];
    return p[3];
}

int main()
{
    constexpr int a[2][3] = { 1, 2, 3, 4, 5, 6, };
    constexpr int i = f(a);
}

This compiles correctly because accessing p[3] using a constexpr triggers the compiler to detect the undefined behavior and throw an error.

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