On 01/10/13 00:54, Kevin O'Connor wrote:
On Wed, Jan 09, 2013 at 08:34:18AM -0600, Dave Frodin wrote:
Here's a patch that's been lingering awhile. thanks,
Thanks. I don't receive a warning for this - what is the exact warning you receive? I don't see why gcc would convert (datalow_end - datalow_start) to a long.
In the expression "datalow_end - datalow_start", both operands - (have incomplete array type (size unknown), ISO C99 6.2.5p22), - are converted ("decay") to type "pointer-to-u8" (ISO C99 188.8.131.52p3).
The expression "datalow_end - datalow_start" invokes undefined behavior, because the (decayed) operands are not pointers into the same array (or to the element one past the last element in the array).
Anyway, the result type of "datalow_end - datalow_start" would be ptrdiff_t, whose size is implementation-defined.
From ISO C99, 6.5.6 Additive operators (normative):
9 When two pointers are subtracted, both shall point to elements of the same array object, or one past the last element of the array object; the result is the difference of the subscripts of the two array elements. The size of the result is implementation-defined, and its type (a signed integer type) is ptrdiff_t defined in the <stddef.h> header. If the result is not representable in an object of that type, the behavior is undefined. In other words, if the expressions P and Q point to, respectively, the i-th and j-th elements of an array object, the expression (P)-(Q) has the value i-j provided the value fits in an object of type ptrdiff_t. Moreover, if the expression P points either to an element of an array object or one past the last element of an array object, and the expression Q points to the last element of the same array object, the expression ((Q)+1)-(P) has the same value as ((Q)-(P))+1 and as -((P)-((Q)+1)), and has the value zero if the expression P points one past the last element of the array object, even though the expression (Q)+1 does not point to an element of the array object.88)
Footnote 88 (informative)
Another way to approach pointer arithmetic is first to convert the pointer(s) to character pointer(s): In this scheme the integer expression added to or subtracted from the converted pointer is first multiplied by the size of the object originally pointed to, and the resulting pointer is converted back to the original type. For pointer subtraction, the result of the difference between the character pointers is similarly divided by the size of the object originally pointed to.
When viewed in this way, an implementation need only provide one extra byte (which may overlap another object in the program) just after the end of the object in order to satisfy the "one past the last element" requirements.
I can see two ways to solve this "problem" (many are possible probably):
(1) print the difference (of type ptrdiff_t) with the "%td" printf() conversion specification. It was first defined in SUSv3 http://pubs.opengroup.org/onlinepubs/000095399/functions/fprintf.html, ie. not standard C. However this leaves the undefined behavior (the subtraction) in place.
(2) Convert the operands first to pointer-to-void (safe), then to uintptr_t ((a) an optional type that is required on XSI conformant systems, (b) the conversion is safe from void*), then take their difference, convert it to uintmax_t, and print it with "%"PRIuMAX:
dprintf(1, "Relocating low data from %p to %p (size %"PRIuMAX")\n" , (void *)datalow_start, (void *)final_datalow_start, , (uintmax_t)( (uintptr_t)(void *)datalow_end - (uintptr_t)(void *)datalow_start));
(I'm of course aware that you won't do this, bit I think it explains the "problem" and you could simplify from here, perhaps exploiting characteristics that are guaranteed for any platform that runs SeaBIOS.)