Sign preserving Vs value preserving

[sophia.agnes]

Dear all,

I was going through the book "C a software engineering approach by
darnell & Margolis" there was a section named sign preserving vs value
preserving

it is as follows

sign preserving rule:-
when signed and un signed objects meet in an expression, the result is
always unsigned

ex:- if a is an unsigned short whose value is 2,then the expression
a - 3 evaluates to a very large un signed value rather than
the signed value of -1

value preserving rule:-

This method converts unsigned chars and unsigned shorts to
int,assuming that the int type is larger than unsigned char and
unsigned short respectively. if int is not larger the object is
converted to unsigned int

assuming 16-bit shorts and 32-bit ints in the previous example,
a would be converted to int rather than unsigned int,so the result of
the expression would be -1

note that the difference between sign-preserving and value-preserving
rules only becomes manifest when an unsigned type is shorter than
int.

if both operands are unsigned ints, the result is unsigned, so that
the expression 2u-3u always evaluates to a large unsigned value

1) how valid is the above explanation?

2)is there any implementation where the int type is smaller in size
when compared to unsigned short and unsigned char?

3)Does the condition

sizeof(short) <= sizeof(int) <= sizeof(long)

always holds in ANSI-C ? or is it true for unsigned values ?

[vippstar]

On Dec 5, 1:44 pm, [e-mail] wrote:

Dear all,

I was going through the book "C a software engineering approach by
darnell & Margolis" there was a section named sign preserving vs value
preserving

it is as follows

sign preserving rule:-
when signed and un signed objects meet in an expression, the result is
always unsigned

true

if both operands are unsigned ints, the result is unsigned, so that
the expression 2u-3u always evaluates to a large unsigned value

To be precise, its value is equal to UINT_MAX

1) how valid is the above explanation?

valid.

2)is there any implementation where the int type is smaller in size
when compared to unsigned short and unsigned char?

no, because your 3) is true

3)Does the condition

sizeof(short) <= sizeof(int) <= sizeof(long)

always holds in ANSI-C ? or is it true for unsigned values ?

yes.

[Peter Nilsson]

[e-mail] wrote:

[e-mail] wrote:>

I was going through the book "C a software
engineering approach by darnell & Margolis" there was
a section named sign preserving vs value preserving

it is as follows

sign preserving rule:-
when signed and un signed objects meet in an expression,
the result is always unsigned

true

No, it's false. Let's look at the OP's (snipped) example.

"ex:- if a is an unsigned short whose value is 2, then
the expression a - 3 evaluates to a very large un signed
value rather than the signed value of -1"

It depends on whether an unsigned short promotes to int
or to unsigned int, and that depends on whether unsigned
short values can be represented by an int.

If USHRT_MAX <= INT_MAX, then the value is -1, otherwise
unsigned short promotes to unsigned int and the value is
UINT_MAX.

if both operands are unsigned ints, the result is
unsigned,

Unless by ints you mean integers. Arithmetic on two
unsigned short values may be signed or unsigned for
reasons stated above.

so that the expression 2u-3u always evaluates to a
large unsigned value

To be precise, its value is equal to UINT_MAX

1) how valid is the above explanation?

valid.

No, it's invalid. If you've been fooled it's probably
because unsigned short has the same width as unsigned
int on the implementations you've seen.

2)is there any implementation where the int type is
smaller in size when compared to unsigned short and
unsigned char?

No addressable type is smaller than a character type.
The standard has rules on the values that are representable
by types of the same signedness but different rank. It
has rules that the unsigned integer must be able to
represent all the non-negative values of it's signed
counterpart (same rank.)

It does not talk about the byte size of the representations
beyond imposing implicit minimums based on how many value
(and sign) bits are required to support the minimum range.

no, because your 3) is true

No, 3) is not necessarily true.

3)Does the condition

sizeof(short) <= sizeof(int) <= sizeof(long)

[Assuming a mathematical relation, rather than C's <=...]

always holds in ANSI-C ? or is it true for unsigned values ?

yes.

No. The standard allows sizeof(short) sizeof(int). All it
requires is that SHRT_MAX <= INT_MAX and INT_MIN <= SHRT_MIN.

You're not likely to find an implementation where shorts are
larger than ints, but the standard doesn't exclude them.

--
Peter

[sophia.agnes]

what about this result

when an integer value x is converted to a smaller unsigned integer
type, the result is the non negative remainder of

x/(U_MAX+1)

where U_MAX is the largest number that can be represented in the
shorter unsigned type.

i checked it using the following program

#include<stdio.h>
#include<stdlib.h>
#include<limits.h>
int main(int argc, char* argv[])
{
unsigned short j;

j = 71124;
printf("\n USHRT_MAX = %d",USHRT_MAX);
printf("\n j = %d",j);
printf("\n ......... = %d",(71124) % (USHRT_MAX+1));

puts("");
return(EXIT_SUCCESS);
}

I got the o/p as

USHRT_MAX = 65535
j = 5588
......... = 5588