Why is Out-of-Bounds Pointer Arithmetic in C Considered Undefined Behavior?

Why is Out-of-Bounds Pointer Arithmetic Undefined Behavior?

Out-of-bounds pointer arithmetic, as exemplified by the code snippet below, is considered undefined behavior in C .

int arr[4] = {0, 1, 2, 3};
int* p = arr + 5; // Undefined behavior

Contrary to expectations that pointers should behave like integers, they possess unique characteristics. The crux of the issue lies in the C standard itself, explicitly classifying out-of-bounds pointer manipulation as undefined behavior.

While it's true that on most platforms, pointer arithmetic beyond array bounds won't result in a crash or anomalous behavior as long as the pointer remains undereferenced, one must question the purpose of such arithmetic if it's not intended for use.

However, the C 11 specification explicitly acknowledges that an expression exceeding an array's end by exactly one is technically "correct" and will not cause a crash. However, its result is undefined, while expressions going more than one over the array bounds are strictly undefined behavior.

It's important to emphasize that despite allowing access to within one position past the array end, this does not imply safety. Reading or writing data within this extended range will likely manipulate data outside the array's bounds, leading to memory corruption and state inconsistencies.

The rationale behind this strict stance on out-of-bounds pointer arithmetic is the complexity of potential scenarios where pointer arithmetic could lead to dangerous situations. Therefore, to avoid inconsistencies and maintain consistency, it's deemed simpler to prohibit such behavior altogether.

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