Why is the maximum size of an array “too large”?

Question

I\'m under the same impression as this answer, that size_t is always guaranteed by the standard to be large enough to hold the largest possible type of a given

Answer 1

The of range size_t is guaranteed to be sufficient to store the size of the largest object supported by the implementation. The reverse is not true: you are not guaranteed to be able to create an object whose size fills the entire range of size_t.

Under such circumstances the question is: what does SIZE_MAX stand for? The largest supported object size? Or the largest value representable in size_t? The answer is: it is the latter, i.e. SIZE_MAX is (size_t) -1. You are not guaranteed to be able to create objects SIZE_MAX bytes large.

The reason behind that is that in addition to size_t, implementations must also provide ptrdiff_t, which is intended (but not guaranteed) to store the difference between two pointers pointing into the same array object. Since type ptrdiff_t is signed, the implementations are faced with the following choices:

1. Allow array objects of size SIZE_MAX and make ptrdiff_t wider than size_t. It has to be wider by at least one bit. Such ptrdiff_t can accommodate any difference between two pointers pointing into an array of size SIZE_MAX or smaller.

2. Allow array objects of size SIZE_MAX and use ptrdiff_t of the same width as size_t. Accept the fact that pointer subtraction can overflow and cause undefined behavior, if the pointers are farther than SIZE_MAX / 2 elements apart. The language specification does not prohibit this approach.

3. Use ptrdiff_t of the same width as size_t and restrict the maximum array object size by SIZE_MAX / 2. Such ptrdiff_t can accommodate any difference between two pointers pointing into an array of size SIZE_MAX / 2 or smaller.

You are simply dealing with an implementation that decided to follow the third approach.