Newbye quetion：为什么双人可以存储更多的数字而不是长？ [英] Newbye quetion: Why a double can store more number than a long ?
问题描述
我的操作系统是：
cronos：jdiaz：tmp> uname -a
HP-UX cronos B.11.11 U 9000/800 820960681无限用户许可证
我在64位模式下编译以下程序：
cronos：jdiaz ：tmp> 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（长）<< " " << sizeof（double）<< endl;
cout<< l_min<< " " << l_max<< " " << d_min<< " " << d_max<<
endl;
返回0;
}
cronos：jdiaz：tmp> a .out
8 8
-9223372036854775808 9223372036854775807 2.22507e-308 1.79769e + 308
双和长类型有相同的大小，但双重限制
更大。任何人都能告诉我这个吗？
谢谢，
何塞路易斯。
Hi,
My OS is:
cronos:jdiaz:tmp>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:tmp>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:tmp>a.out
8 8
-9223372036854775808 9223372036854775807 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.
推荐答案
-----开始PGP签名消息-----
哈希：SHA1
jose luis fernandez diaz写道：
-----BEGIN PGP SIGNED MESSAGE-----
Hash: SHA1
jose luis fernandez diaz wrote:
我的操作系统是：
cronos：jdiaz：tmp> uname -a
HP-UX cronos B.11.11 U 9000/800 820960681无限用户许可证
我在64位模式下编译下面的程序：
cronos：jdiaz：tmp> cat kk.C
#include< iostream.h>
#include< limits>
{/ l}长l_min = numeric_limits< long> :: min（）;
long l_max = numeric_limits< long> :: max（）;
Hi,
My OS is:
cronos:jdiaz:tmp>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:tmp>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]
嗯，因为你使用的是另一种来自C的语言，我们在comp.lang.c
新闻组中无法为您提供帮助。你的问题与C无关，应该是
（大概来自交叉发布，是）在与你编写你的例子的
语言相关的论坛中被问到的。这似乎是C ++，所以论坛
将是comp.lang.c ++。
- -
Lew Pitcher
IT顾问，企业应用架构，
企业技术解决方案，道明银行金融集团
（表达的意见是我自己的，不是我的雇主' '）
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版本：GnuPG v1.2.4（MingW32）
iD8DBQFAljf6agVFX4UWr64RAnujAJwJ4mOsZZ9THxxul4vwzr vTo / acYwCcDLiJ
+ EqfYBtEqbJjlf / u1b7p2DQ =
= 7HZL
----- END PGP SIGNATURE -----
[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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Version: GnuPG v1.2.4 (MingW32)
iD8DBQFAljf6agVFX4UWr64RAnujAJwJ4mOsZZ9THxxul4vwzr vTo/acYwCcDLiJ
+EqfYBtEqbJjlf/u1b7p2DQ=
=7HZL
-----END PGP SIGNATURE-----
jose luis fernandez diaz写道：
jose luis fernandez diaz wrote:
我的操作系统是：
cronos：jdiaz：tmp> uname -a
HP-UX cronos B.11.11 U 9000/800 820960681无限用户许可证
我在64位模式下编译程序如下：
cronos：jdiaz：tmp> cat kk.C
#include< iostream.h>
#include< limits>
int m ain（）
{l / l长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（长）<< " " << sizeof（double）<< endl;
cout<< l_min<< " " << l_max<< " " << d_min<< " " << d_max<<
endl;
返回0;
}
cronos：jdiaz：tmp> a.out
8 8
-9223372036854775808 9223372036854775807 2.22507e-308 1.79769e + 308
双和长型具有相同的尺寸，但双重限制
更大。有人可以把这个发给我吗？
Hi,
My OS is:
cronos:jdiaz:tmp>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:tmp>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:tmp>a.out
8 8
-9223372036854775808 9223372036854775807 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 ?
好吧。
1.79769E308
并不意味着该数字精确到最后一位数。这意味着
1797690000000000000000 .... 000000 ..... 000000
和下一个最小的数字可能类似
1797680000000000000000 .... 000000 ..... 000000
因此数字之间存在较大差距。当数字变小时，这些差距会变小。
-
Karl Heinz Buchegger
kb ****** @ gascad.at
jose luis fernandez diaz写道：
jose luis fernandez diaz wrote:
我的操作系统是：
cronos：jdiaz：tmp> uname -a
HP-UX cronos B.11.11 U 9000/800 820960681无限用户许可证
我在64位模式下编译以下程序：
cronos：jdiaz：tmp> cat kk.C
#include< iostream.h>
#include< iostream> //没有.h扩展。
#include< limits>
{
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（长）<< " " << sizeof（double）<< endl;
cout<< l_min<< " " << l_max<< " " << d_min<< " " << d_max<<
endl;
返回0;
}
cronos：jdiaz：tmp> a.out
8 8
-9223372036854775808 9223372036854775807 2.22507e-308 1.79769e + 308
双和长型具有相同的尺寸，但双重限制
更大。任何人都可以向我发布这个吗？
Hi,
My OS is:
cronos:jdiaz:tmp>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:tmp>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:tmp>a.out
8 8
-9223372036854775808 9223372036854775807 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 ?
通常，代表double的一些位被解释为
指数，其余的为乘数。浮点数学结果会产生近似结果，而任何关于
整数字段的操作在应用于long int时会产生精确的结果（如果我们
折扣{over，under} flow）。
http://en.wikipedia.org/wiki/IEEE_fl...point_standard
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