A signed int reserves one bit to signify whether a number is positive or
negative. In light of this, a colleague asked me whether there existed an
int in C++ that was -0, a zero with the negative bit set. I was intrigued
by this, so I tried the following code:
#include <stdio.h>
int main(int, char**) {
int a(-0);
printf("a=%d\n" , a);
if ( a==0 ) printf("a==0\n" );
if ( a==-0 ) printf("a==-0\n");
return 0;
}
The output is:
a=0
a==0
a==-0
So it seems that no such number can be used in C++, and speaking of
mathematical numbers, it makes perfect sense, but what is actually going on
"behing the curtains" in C++ since the compiled code can't discern between
+0 and -0 ?
--
Christian Stigen Larsen -- http://sublevel3.org/~csl/
Jul 19 '05
20  7562 
"Suzanne Vogel" <su************ *@hotmail.com> wrote... Thanks for the feedback.
(e.g., from -0 to 0?
Why do you think that would be a type conversion? Anyway, the compiler
may keep the bits as they are and just treat 0 and -0 as equal.
*Suppose* (this is hypothetical) there were different representations of
0 and -0 for longs but not for ints.
Consider the following (made-up example):
if (0 == -0)... // possibly convert one of these
To compare 0 and -0, they must have the same type. So in *this* made-up
case, -0 would be a long
No, it wouldn't. See the explanation below.
and 0 would be an int, so 0 would be converted
(converted, not cast) to a long.
Similarly to the following (real example):
if (float(0.0) == double(0.0))... // convert float(0.0) to a double
** Is this not right?
No. Not according to the C++ grammar. In it the code "-0" is
not a literal but an expression. "0" is a literal. It has the
type 'int' and base 8 (because it begins with 0). It's an octal
literal, and since it can be represented by 'int', it is an 'int'.
Now, -0 is a constant expression which involves negation operator
and the integer expression. The result is calculated by the
compiler on the fly and is (you guessed it!) 0.
const int NUM_BITS_PER_CH AR = 8;
Better use std::numeric_li mits<char>::dig its from the <limits> header or
CHAR_BIT from <climits>.
Good idea.
** Doesn't that just give the number of base-10 digits?
No. For those there is std::numeric_li mits<char>::dig its10.
If so, I guess I
could take the log base-2 of that to get the number of base-2 digits.
You don't need to.
Victor
"Suzanne Vogel" <su************ *@hotmail.com> wrote in message
news:3f******** **@news.unc.edu ... For readers who skim, I'll start with a question (instead of burying it
below):
*** Is there a C++ library function that prints binary representations
of numbers?
In a way, yes. You can ask C++ to convert the number to a string in binary
format and then print that string:
int a; // the variable you wish to represent in binary
format.
char* string = new char[50];
itoa(a, string, 2); // the 2 means binary
std::cout << string;
delete string;
"MiniDisc_2 k2" <Ma******@nospa m.com> wrote...
"Suzanne Vogel" <su************ *@hotmail.com> wrote in message
news:3f******** **@news.unc.edu ... For readers who skim, I'll start with a question (instead of burying it
below):
*** Is there a C++ library function that prints binary representations
of numbers?
In a way, yes. You can ask C++ to convert the number to a string in binary
format and then print that string:
int a; // the variable you wish to represent in binary
format.
char* string = new char[50];
itoa(a, string, 2); // the 2 means binary
Where did you dig out that function? Neither Standard C, nor
Standard C++ library has 'itoa'.
std::cout << string;
delete string;
On Thu, 3 Jul 2003 13:38:13 +0000 (UTC), cs******@newsca che.ntnu.no
(Christian Stigen Larsen) wrote in comp.lang.c++: A signed int reserves one bit to signify whether a number is positive or
negative. In light of this, a colleague asked me whether there existed an
int in C++ that was -0, a zero with the negative bit set. I was intrigued
by this, so I tried the following code:
#include <stdio.h>
int main(int, char**) {
int a(-0);
printf("a=%d\n" , a);
if ( a==0 ) printf("a==0\n" );
if ( a==-0 ) printf("a==-0\n");
return 0;
}
The output is:
a=0
a==0
a==-0
So it seems that no such number can be used in C++, and speaking of
mathematical numbers, it makes perfect sense, but what is actually going on
"behing the curtains" in C++ since the compiled code can't discern between
+0 and -0 ?
There are three defined formats for the representation of negative
signed integer values. Two of them have the physical possibility of
representing -0, one does not. Of the two that can physically
represent -0, that might actually be a valid value, or it might be a
trap value that causes undefined behavior.
On implementations that support -0, if you could actually find one, it
is required to compare equal to ordinary 0.
--
Jack Klein
Home: http://JK-Technology.Com
FAQs for
comp.lang.c http://www.eskimo.com/~scs/C-faq/top.html
comp.lang.c++ http://www.parashift.com/c++-faq-lite/
alt.comp.lang.l earn.c-c++ ftp://snurse-l.org/pub/acllc-c++/faq
"Jack Klein" <ja*******@spam cop.net> wrote in message news:2c******** *************** *********@4ax.c om... There are three defined formats for the representation of negative
signed integer values.
Only in C99.
"Ron Natalie" <ro*@sensor.com > wrote...
"Jack Klein" <ja*******@spam cop.net> wrote in message
news:2c******** *************** *********@4ax.c om... There are three defined formats for the representation of negative
signed integer values.
Only in C99.
What is that supposed to mean? The representation of negative
integers is not language-specific, it's hardware-specific.
"Victor Bazarov" <v.********@att Abi.com> wrote in message news:vg******** ****@corp.super news.com... "Ron Natalie" <ro*@sensor.com > wrote...
"Jack Klein" <ja*******@spam cop.net> wrote in message
news:2c******** *************** *********@4ax.c om... There are three defined formats for the representation of negative
signed integer values.
Only in C99.
What is that supposed to mean? The representation of negative
integers is not language-specific, it's hardware-specific.
C99 defines (well limits the allowable) integer formats to three (signed mag, 1's comp.,
and 2's comp.). There is NO such restriction in C++. Any encoding that meets the
requirement that the positive numbers share representation with the corresponding unsigned
type meets the requirement.
"Ron Natalie" <ro*@sensor.com > wrote...
"Victor Bazarov" <v.********@att Abi.com> wrote in message
news:vg******** ****@corp.super news.com... "Ron Natalie" <ro*@sensor.com > wrote...
"Jack Klein" <ja*******@spam cop.net> wrote in message
news:2c******** *************** *********@4ax.c om...
> There are three defined formats for the representation of negative
> signed integer values.
Only in C99.
What is that supposed to mean? The representation of negative
integers is not language-specific, it's hardware-specific.
C99 defines (well limits the allowable) integer formats to three (signed
mag, 1's comp., and 2's comp.). There is NO such restriction in C++. Any encoding
that meets the requirement that the positive numbers share representation with the
corresponding unsigned type meets the requirement.
Would you say, then, that C99 is smarter about it or that C++
is more flexible about it? AFAIK, there are no other schemes
than the three mentioned in this thread...
Victor
"Victor Bazarov" <v.********@att Abi.com> wrote in message news:vg******** ****@corp.super news.com...
..
Would you say, then, that C99 is smarter about it or that C++
is more flexible about it? AFAIK, there are no other schemes
than the three mentioned in this thread...
Got me. I've never understood why they were suddenly enumerated
in C99. To my knowledge I've never worked on a signed-mag machine
and I haven't been near a 1's complement machine in over a decade.
Ron Natalie wrote:
[...] Got me. I've never understood why they were suddenly enumerated
in C99. To my knowledge I've never worked on a signed-mag machine
and I haven't been near a 1's complement machine in over a decade.
How about offset binary (aka "Excess-K"), various BCD schemes, uhmm,
and "gray code"? ;-)
regards,
alexander. This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics | |
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