What is the rationale for one past the last element of an array object?

Question

According to N1570 (C11 draft) 6.5.6/8 Additive operators:

Moreover, if the expression P points to the last element

Answer 1

The rationale is quite simple. The compiler is not allowed to place an array at the end of memory. To illustrate, assume that we have a 16-bit machine with 16-bit pointers. The low address is 0x0000. The high address is 0xffff. If you declare char array[256] and the compiler locates array at address 0xff00, then technically the array would fit into the memory, using addresses 0xff00 thru 0xffff inclusive. However, the expression

char *endptr = &array[256];   // endptr points one past the end of the array

would be equivalent to

char *endptr = NULL;          // &array[256] = 0xff00 + 0x0100 = 0x0000

Which means that the following loop would not work, since ptr will never be less than 0

for ( char *ptr = array; ptr < endptr; ptr++ )

So the sections you cited are simply lawyer-speak for, "Don't put arrays at the end of a memory region".

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Historical note: the earliest x86 processors used a segmented memory scheme wherein memory addresses where specified by a 16-bit pointer register and a 16-bit segment register. The final address was computed by shifting the segment register left by 4 bits and adding to the pointer, e.g.

pointer register    1234
segment register   AB00
                   -----
address in memory  AC234

The resulting address space was 1MByte, but there were end-of-memory boundaries every 64Kbytes. That's one reason for using lawyer-speak instead of stating, "Don't put arrays at the end of memory" in plain english.