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Integer promotion, relational operators, and unequal integer ranks.

P: n/a
Hi all,

I had a question about the topics in the subject and posted to
comp.std.c, but feel it may also be appropriate here. Please excuse
this crosspost if it is in bad form.

I have a question about whether or not I am interpreting a nuance of
the standard correctly, and the implications of said nuance. The
sections in the C99 standard (and possibly older standards) that I will
reference are as follows (typed out hopefully to avoid people having to
flip through their own manuals; if you want to fast forward to the meat
of this post, skip down to "-------end reference-------" line):

-------begin reference-------
6.2.6.2p2 (specifically):

For signed integer types, the bits of the object representation shall
be divided into three groups: value bits, padding bits, and the sign
bit. There need not be any padding bits; there shall be exactly one
sign bit. Each bit that is a value bit shall have the same value as the
same bit in the object representation of the corresponding unsigned
type (if there are M value bits in the signed type and N in the
unsigned type, then M≤N)...

6.3.1.1p1:

Every integer type has an integer conversion rank defined as follows:

—No two signed integer types shall have the same rank, even if they
have the same representation.
—The rank of a signed integer type shall be greater than the rank of
any signed integer type with less precision.
—The rank of long long int shall be greater than the rank of long
int, which shall be greater than the rank of int, which shall be
greater than the rank of short int, which shall be greater than the
rank of signed char.
—The rank of any unsigned integer type shall equal the rank of the
corresponding signed integer type, if any.
—The rank of any standard integer type shall be greater than the rank
of any extended integer type with the same width.
—The rank of char shall equal the rank of signed char and unsigned
char.
—The rank of_Bool shall be less than the rank of all other standard
integer types.
—The rank of any enumerated type shall equal the rank of the
compatible integer type (see 6.7.2.2).
—The rank of any extended signed integer type relative to another
extended signed integer type with the same precision is
implementation-defined, but still subject to the other rules for
determining the integer conversion rank.
—For all integer types T1, T2, andT3, if T1 has greater rank than T2
and T2 has greater rank than T3, then T1 has greater rank than T3.

6.3.1.8p1 (specifically):

Otherwise, the integer promotions are performed on both operands. Then
the following rules are applied to the promoted operands:

If both operands have the same type, then no further conversion is
needed.

Otherwise, if both operands have signed integer types or both have
unsigned integer types, the operand with the type of lesser integer
conversion rank is converted to the type of the operand with greater
rank.

Otherwise, if the operand that has unsigned integer type has rank
greater or equal to the rank of the type of the other operand, then the
operand with signed integer type is converted to the type of the
operand with unsigned
integer type.

Otherwise, if the type of the operand with signed integer type can
represent all of the values of the type of the operand with unsigned
integer type, then the operand with unsigned integer type is converted
to the type of the
operand with signed integer type.

Otherwise, both operands are converted to the unsigned integer type
corresponding to the type of the operand with signed integer type.

6.5.8p3

If both of the operands have arithmetic type, the usual arithmetic
conversions are performed.
-------end reference-------

Alright, now for the background. While redoing some string libraries I
had created a while back, I ran across some interesting check code that
I had added and now feel may be subject to an oddity. We all know that
the return type of snprintf is int (I assume for historical reasons and
the fact that a negative is used to indicate failure) as opposed to a
size_t. In this library I had code that does the following check
(where n is a non negative return value from a prior snprintf call used
to determine the size of the required storage):

if ( n >= SIZE_MAX )
{
.... return error information indicating that the size required will not
appropriately fit into a size_t and thusly I can not call malloc...

}

The assumption was that there is nothing in the standard to my
knowledge that prohibits a size_t being typedef'd to an unsigned type
incapable of holding the full positive range of an int. I believe my
past assumption was that integer promotion would be performed in this
relational expression and that the expression would always work. I'm
now beginning to question the always work portion. If I'm reading the
above referenced sections correctly, it is completely possible for an
implementation to have the following setup:

size_t typedef'd to an unsigned type with lower rank than an int.
Following the above rules, if we assume that in this implementation
there is a corresponding signed type, then both the size_t and the
signed type would have equal rank. Let us call this rank R1. int and
unsigned int have equal and higher rank than R1, lets call this R2.
Let A be the precision of the size_t, B be the precision of the
corresponding size_t signed type, C be the precision of an unsigned
int, and D be the precision of an int.

According to the above mentioned rules then,

R2 R1
A >= B (Rank R1) [6.2.6.2p2]
C >= D (Rank R2) [6.2.6.2p2]
D >= B [6.3.1.1p1]

Therefor, there is no way to link A and C relationally. It is entirely
possible to have the size_t have greater precision than the unsigned
int, even though the unsigned int is higher ranked (seems counter
intuitive but if I'm correct, this is true). Therefor, in my original
comparison, if we're on an implementation where this is true, and where
an int can not sufficiently hold all the values of a size_t, than
according to 6.3.1.8p1, *both operands are converted to the unsigned
integer type corresponding to the type of the operand with signed
integer type*, in this case unsigned int. Therefor, the size_t
(SIZE_MAX) would get converted to a type incapable of holding that
value, and the value would wrap around to a smaller number and thus
possibly give incorrect results.

Can someone very familiar with the standards tell me if my assumption
is correct. I know I've written a book in this post and apologize, I
just find it difficult to explain without the elaboration. If it is
true, do their exist any ways to safely compare values of an unsigned
type in a lower rank to values of a signed type in a higher rank
without any possible "overflow" of the value in the converted type?

Any insight would be massively appreciated!

Thanks so much!

-Charlie

Jan 15 '07 #1
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P: n/a
ch**********@merck.com wrote:

You better recheck this, but
6.3.1.8p1 (specifically):

Otherwise, the integer promotions are performed on both operands. Then
the following rules are applied to the promoted operands:
which means that SIZE_MAX will be at least promoted to unsigned int.

Jan 15 '07 #2

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