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Re: What is truncation for (singed) integer in C ?

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On 09/19/2015 03:25 PM, fl wrote:
> Hi,
> 
> I see the following on wikipedia and doubt about negative integer when
>  truncating.
> 
> https://en.wikipedia.org/wiki/Type_conversion
> ///////
>  Data can be lost when floating-point representations are converted to integer
>  representations as the fractional components of the floating-point values
>  will be truncated (rounded towards zero)
> .......
> 
> I agree for positive integer, truncation in C is towards 0. But for negative
> integer, it looks like away from '0'. For example, for a unsigned char 83H,
> one bit (LSB) truncation result will be 41H, whose non-truncation value 
> (8-bit) is equal to 82H. The real value of 82H plus '1' is 83H. That is,

You have a LOT of confusion here.

The relevant clause in the C standard says: "When a finite value of real
floating type is converted to an integer type other than _Bool,
the fractional part is discarded (i.e., the value is truncated toward
zero)." (6.3.1.4p1)

The truncation it's talking about here involves only floating point
values, and your example doesn't involve any floating point values. What
it's saying is that if, for instance, a floating point value of 3.21 is
converted to an integer type, the result is 3, obtained by truncating
the "21" at the end. Similarly, value of -5.67 is converted to an
integer type by truncating (removing) the "67", giving a value of -5. In
both cases, the truncated value is closer to 0, which is why it's
described as being truncated towards zero.

> 82H < 83H. For these negative numbers, it is not rounded towards zero.

What are "these negative numbers"? I don't see any negative numbers in
your message. The only data type you mentioned is unsigned char, which
is incapable of representing a negative number.

The form 83H has no special meaning in C. It would get parsed as a
pp-number during translation phase 4 (6.4.8p1), and then rejected during
translation phase 7 due to the fact that it doesn't match the form of
any of the five permitted types of tokens. I would guess that what you
mean by 83H is equivalent to the C hexadecimal constant 0x83. A key
point to remember about C is that decimal constants, octal constants,
and hexadecimal constants never have a negative value. -3 is NOT a
decimal constant. It is the unary negation operator applied to the
decimal constant 3. 0x83 therefore has the decimal value 131.

My best guess is that you're talking about an 8-bit signed char which is
represented by the same bit pattern that would represent a 0x83 as an
unsigned char; such a char object would represent a negative value: -125
if using 2's complement (by far the most common choice), -3 if using
sign-magnitude, and -124 if using 1's complement (assuming I did the
conversions correctly, which is not guaranteed). C allows for all three
possibilities. C also allows CHAR_BIT > 8; CHAR_BIT==16 is not uncommon
in C compilers targeting DSPs.

> Is the above link not talking the same thing with my understanding?

I'm not sure what you're understanding is, but your words make it pretty
clear that your understanding is incorrect, and that the above link is
referring to something quite different.

I'm not sure what your comments "one bit LSB truncation" are intended to
mean. There appears to be some resemblance between what you're talking
about and division by 2, so I'll use such a division in my example below.

There is a "toward zero" rule that applies to integers. It applies
specifically to integer division. "When integers are divided, the result
of the / operator is the algebraic quotient with any
fractional part discarded." (6.5.5p6). Thus, 8/3 is algebraically 2 2/3.
Discarding  the 2/3 leaves the C value of 8/3 == 2. Algebraically,
-125/2 is  -62 1/2; discarding the 1/2 leaves the C value of (-125)/2 ==
-62. In both cases, the result of integer division is closer to 0 than
the algebraic quotient would be, which is why footnote 105 says that
this rule 'is often called "truncation toward zero"'.

Note that the C rule is entirely in terms of the meaning of the value
being represented, it is not in terms of the bit pattern being used to
represent it. A signed char variable containing a value of -125 will,
when divided by 2, give a result of -62, regardless of whether it's
represented in 2's complement, 1's complement, or sign-magnitude, and
regardless of whether char has 8 bits, 9 bits, 16 bits, or any other
permitted size.

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What is truncation for (singed) integer in C ? fl <rxjwg98@gmail.com> - 2015-09-19 12:25 -0700
  Re: What is truncation for (singed) integer in C ? fir <profesor.fir@gmail.com> - 2015-09-19 12:55 -0700
    Re: What is truncation for (singed) integer in C ? fir <profesor.fir@gmail.com> - 2015-09-19 13:20 -0700
  Re: What is truncation for (singed) integer in C ? Geoff <geoff@invalid.invalid> - 2015-09-19 13:05 -0700
  Re: What is truncation for (singed) integer in C ? Ben Bacarisse <ben.usenet@bsb.me.uk> - 2015-09-19 21:28 +0100
    Re: What is truncation for (singed) integer in C ? Tim Prince <tprince@computer.org> - 2015-09-19 16:38 -0400
      Re: What is truncation for (singed) integer in C ? James Kuyper <jameskuyper@verizon.net> - 2015-09-19 16:52 -0400
  Re: What is truncation for (singed) integer in C ? Keith Thompson <kst-u@mib.org> - 2015-09-19 13:45 -0700
  Re: What is truncation for (singed) integer in C ? James Kuyper <jameskuyper@verizon.net> - 2015-09-19 16:46 -0400
  Re: What is truncation for (singed) integer in C ? Eric Sosman <esosman@comcast-dot-net.invalid> - 2015-09-19 16:47 -0400
  Re: What is truncation for (singed) integer in C ? Joe Pfeiffer <pfeiffer@cs.nmsu.edu> - 2015-09-19 22:55 -0600
  Re: What is truncation for (singed) integer in C ? James Kuyper <jameskuyper@verizon.net> - 2015-09-20 10:54 -0400
    Re: What is truncation for (singed) integer in C ? Joe Pfeiffer <pfeiffer@cs.nmsu.edu> - 2015-09-20 10:11 -0600

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