Newbye quetion: Why a double can store more number than a long ?

[jose luis fernandez diaz]

Hi,

My OS is:

cronos:jdiaz:tm p>uname -a
HP-UX cronos B.11.11 U 9000/800 820960681 unlimited-user license
I compile in 64-bits mode the program below:

cronos:jdiaz:tm p>cat kk.C
#include <iostream.h>
#include <limits>

int main()
{
long l_min = numeric_limits< long>::min();
long l_max = numeric_limits< long>::max();
double d_min = numeric_limits< double>::min();
double d_max = numeric_limits< double>::max();
cout << sizeof(long) << " " << sizeof(double) << endl;
cout << l_min << " " << l_max << " " << d_min << " " << d_max <<
endl;
return 0;
}
cronos:jdiaz:tm p>a.out
8 8
-922337203685477 5808 922337203685477 5807 2.22507e-308 1.79769e+308

The double and long types have the same size, but the double limits
are bigger. Can anyone explein this to me ?

Thanks,
Jose Luis.

[Lew Pitcher]

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jose luis fernandez diaz wrote:

Hi,

My OS is:

cronos:jdiaz:tm p>uname -a
HP-UX cronos B.11.11 U 9000/800 820960681 unlimited-user license
I compile in 64-bits mode the program below:

cronos:jdiaz:tm p>cat kk.C
#include <iostream.h>
#include <limits>

int main()
{
long l_min = numeric_limits< long>::min();
long l_max = numeric_limits< long>::max();

[snip]

Well, since you are using a different language from C, we in the comp.lang.c
newsgroup cannot assist you. Your question does not relate to C, and should be
(presumably, from the crossposting, was) asked in a forum related to the
language in which you wrote your example. This appears to be C++, so that forum
would be comp.lang.c++.
- --
Lew Pitcher
IT Consultant, Enterprise Application Architecture,
Enterprise Technology Solutions, TD Bank Financial Group

(Opinions expressed are my own, not my employers')
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[Karl Heinz Buchegger]

jose luis fernandez diaz wrote:

Hi,

My OS is:

cronos:jdiaz:tm p>uname -a
HP-UX cronos B.11.11 U 9000/800 820960681 unlimited-user license

I compile in 64-bits mode the program below:

cronos:jdiaz:tm p>cat kk.C
#include <iostream.h>
#include <limits>

int main()
{
long l_min = numeric_limits< long>::min();
long l_max = numeric_limits< long>::max();
double d_min = numeric_limits< double>::min();
double d_max = numeric_limits< double>::max();
cout << sizeof(long) << " " << sizeof(double) << endl;
cout << l_min << " " << l_max << " " << d_min << " " << d_max <<
endl;

return 0;
}
cronos:jdiaz:tm p>a.out
8 8
-922337203685477 5808 922337203685477 5807 2.22507e-308 1.79769e+308

The double and long types have the same size, but the double limits
are bigger. Can anyone explein this to me ?

Well.

1.79769E308

does not mean that the number is accurate to the last digit. It means

179769000000000 0000000....0000 00.....000000
and the next smallest number is probably something like
179768000000000 0000000....0000 00.....000000

so there are large gaps between numbers. Those gaps get smaller when
the numbers get smaller.

--
Karl Heinz Buchegger
kb******@gascad .at

[Jeff Schwab]

jose luis fernandez diaz wrote:

Hi,

My OS is:

cronos:jdiaz:tm p>uname -a
HP-UX cronos B.11.11 U 9000/800 820960681 unlimited-user license
I compile in 64-bits mode the program below:

cronos:jdiaz:tm p>cat kk.C
#include <iostream.h>
#include <iostream> // No ".h" extension.
#include <limits>

int main()
{
using namespace std;
long l_min = numeric_limits< long>::min();
long l_max = numeric_limits< long>::max();
double d_min = numeric_limits< double>::min();
double d_max = numeric_limits< double>::max();
cout << sizeof(long) << " " << sizeof(double) << endl;
cout << l_min << " " << l_max << " " << d_min << " " << d_max <<
endl;
return 0;
}
cronos:jdiaz:tm p>a.out
8 8
-922337203685477 5808 922337203685477 5807 2.22507e-308 1.79769e+308

The double and long types have the same size, but the double limits
are bigger. Can anyone explein this to me ?

Typically, some of the bits representing a double are interpreted as an
exponent, and the rest as a multiplier. Floating-point math results in
approximate results, whereas any operation closed over the field of
integers will give an exact result when applied to long int's (if we
discount {over,under}flo w).

http://en.wikipedia.org/wiki/IEEE_fl...point_standard

[James McIninch]

<posted & mailed>

The code below isn't C language code, but the question still applies.

A double can represent larger numbers because it lacks the resolution of the
long type. For a long integer, you have one number with absolute precision
for each integer within the range. For a double, you have exactly the same
number of values as the long, but on average, the difference between each
individual value and the next is much greater. Instead of counting by 1's,
you are counting by tens of thousands.

It may be easier to understand if you think about it this way, a integer
simply stores a number. A double uses the same number of bits to store two
numbers: one an integer, and one to tell it where to stick the decimal
point. The range of integers (where there is precision) is much smaller
because only a portion of the double is used to store the integral part
(mantissa) while another portion of the bits is used to store the position
of the decimal point (exponent). That's a simplification, but you get the
idea.

Integers are perfectly precise with a narrow range, and doubles are, on
average, very imprecise but cover a much wider range. Both doubles and
integers (in this case, since they are the same number of bits on your
system) have the same number of values.

jose luis fernandez diaz wrote:

Hi,

My OS is:

cronos:jdiaz:tm p>uname -a
HP-UX cronos B.11.11 U 9000/800 820960681 unlimited-user license
I compile in 64-bits mode the program below:

cronos:jdiaz:tm p>cat kk.C
#include <iostream.h>
#include <limits>

int main()
{
long l_min = numeric_limits< long>::min();
long l_max = numeric_limits< long>::max();
double d_min = numeric_limits< double>::min();
double d_max = numeric_limits< double>::max();
cout << sizeof(long) << " " << sizeof(double) << endl;
cout << l_min << " " << l_max << " " << d_min << " " << d_max <<
endl;
return 0;
}
cronos:jdiaz:tm p>a.out
8 8
-922337203685477 5808 922337203685477 5807 2.22507e-308 1.79769e+308

The double and long types have the same size, but the double limits
are bigger. Can anyone explein this to me ?

Thanks,
Jose Luis.

--
remove .spam from address to reply by e-mail.

[Terry]

"Lew Pitcher" <Le*********@td .com> wrote in message
news:MH******** ************@ne ws20.bellglobal .com...

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jose luis fernandez diaz wrote:

Hi,

My OS is:

cronos:jdiaz:tm p>uname -a
HP-UX cronos B.11.11 U 9000/800 820960681 unlimited-user license
I compile in 64-bits mode the program below:

cronos:jdiaz:tm p>cat kk.C
#include <iostream.h>
#include <limits>

int main()
{
long l_min = numeric_limits< long>::min();
long l_max = numeric_limits< long>::max(); [snip]

Well, since you are using a different language from C, we in the

comp.lang.c newsgroup cannot assist you. Your question does not relate to C, and should be (presumably, from the crossposting, was) asked in a forum related to the
language in which you wrote your example. This appears to be C++, so that forum would be comp.lang.c++.
- --
Lew Pitcher
IT Consultant, Enterprise Application Architecture,
Enterprise Technology Solutions, TD Bank Financial Group

Not very helpful.

Regards

[CBFalconer]

jose luis fernandez diaz wrote:

I compile in 64-bits mode the program below:

cronos:jdiaz:tm p>cat kk.C
#include <iostream.h>
#include <limits>

int main()
{
long l_min = numeric_limits< long>::min();
long l_max = numeric_limits< long>::max();
double d_min = numeric_limits< double>::min();
double d_max = numeric_limits< double>::max();
cout << sizeof(long) << " " << sizeof(double) << endl;
cout << l_min << " " << l_max << " " << d_min << " " << d_max << endl;

return 0;
}

C++ is off topic on c.l.c. Please refrain from any such
cross-postings.

--
Some useful references:
<http://www.ungerhu.com/jxh/clc.welcome.txt >
<http://www.eskimo.com/~scs/C-faq/top.html>
<http://benpfaff.org/writings/clc/off-topic.html>
<http://anubis.dkuug.dk/jtc1/sc22/wg14/www/docs/n869/> (C99)

[joe durusau]

CBFalconer wrote:

jose luis fernandez diaz wrote:

I compile in 64-bits mode the program below:

cronos:jdiaz:tm p>cat kk.C
#include <iostream.h>
#include <limits>

int main()
{
long l_min = numeric_limits< long>::min();
long l_max = numeric_limits< long>::max();
double d_min = numeric_limits< double>::min();
double d_max = numeric_limits< double>::max();
cout << sizeof(long) << " " << sizeof(double) << endl;
cout << l_min << " " << l_max << " " << d_min << " " << d_max << endl;

return 0;
}

C++ is off topic on c.l.c. Please refrain from any such
cross-postings.

--
Some useful references:
<http://www.ungerhu.com/jxh/clc.welcome.txt >
<http://www.eskimo.com/~scs/C-faq/top.html>
<http://benpfaff.org/writings/clc/off-topic.html>
<http://anubis.dkuug.dk/jtc1/sc22/wg14/www/docs/n869/> (C99)

Well, the question doesn't have anything to do with unix, either, but
you might see whether your sompiler has sizeof() in it. That could tell
you the size of long and of double, and might provide a clue.

The best place to persue this sort of thing is a textbook. There are
very few newsgroups that devote themselves to trying to run down
questions relating to problems that probably pertain to a specific
compiler/platform issue such as this one. Too bad, but that's the
way it is.

Speaking only for myself,

Joe Durusau

[Jack Klein]

On Mon, 3 May 2004 17:09:27 +0100, "Terry"
<te***@tbean.fr eeserve.co.uk> wrote in comp.lang.c:

[snip]

Not very helpful.

Regards

Nor was your follow up.

--
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++
http://www.contrib.andrew.cmu.edu/~a...FAQ-acllc.html

[jose luis fernandez diaz]

Hi,

I have made the program below to try to understand how a floating
point number is made by my computer (HP-UX cronos B.11.11 U 9000/800):

#include <stdio.h>
#include <math.h>

union
{
float f;
unsigned char uc[sizeof(float)];
} f_union;
int main()
{
f_union.f=1.0F;

unsigned char u1=1, u2;
u1 <<= sizeof(unsigned char)*8-1;

for (int i=0; i < sizeof(float); ++i)
{
u2=f_union.uc[i];
for(int j=0; j<sizeof(unsign ed char)*8; j++)
{
printf("%d", ((u1 & u2)!=0));
u2 <<= 1;
}
}
printf("\n");

int exp;
double man = frexp(f_union.f , &exp);
printf("%lf %d\n%f\n", man, exp,f_union.f);
return 0;
}
but I didn't. Perhaps the program is wrong. Here are some outputs:

001111111000000 000000000000000 00
0.500000 1
1.000000

010000000000000 000000000000000 00
0.500000 2
2.000000

010000000100000 000000000000000 00
0.750000 2
3.000000
010000001000000 000000000000000 00
0.500000 3
4.000000

010000001010000 000000000000000 00
0.625000 3
5.000000

010000010000000 000000000000000 00
0.500000 4
8.000000


Any hint are welcome.

Thanks,
Jose Luis.

jo************* *********@yahoo .es (jose luis fernandez diaz) wrote in message news:<c2******* *************** ****@posting.go ogle.com>...

Hi,

My OS is:

cronos:jdiaz:tm p>uname -a
HP-UX cronos B.11.11 U 9000/800 820960681 unlimited-user license
I compile in 64-bits mode the program below:

cronos:jdiaz:tm p>cat kk.C
#include <iostream.h>
#include <limits>

int main()
{
long l_min = numeric_limits< long>::min();
long l_max = numeric_limits< long>::max();
double d_min = numeric_limits< double>::min();
double d_max = numeric_limits< double>::max();
cout << sizeof(long) << " " << sizeof(double) << endl;
cout << l_min << " " << l_max << " " << d_min << " " << d_max <<
endl;
return 0;
}
cronos:jdiaz:tm p>a.out
8 8
-922337203685477 5808 922337203685477 5807 2.22507e-308 1.79769e+308

The double and long types have the same size, but the double limits
are bigger. Can anyone explein this to me ?

Thanks,
Jose Luis.