C ++中的sign_command（例如not_negative）缺少（？）（/ C） [英] sign_command (e.g not_negative) missing(?) in C++ (/C)

问题描述

我不是C ++（/ C）的专家，但也不是一个完整的初学者。


我正在写一个更快的cos（和sin）近似基于

预先计算的数据（取得了一定的成功）当我发现从双倍到整数时，
非常昂贵，所以可能是cos / sin近似值

没有预先计算会更好（更快）。


尽管如此我仍然需要找出装置上的位置 -

角度

（从0到2PI的双倍（v））是。我找了一个像

这样的命令
int not_negative（double d）//如果d ==（+）0.0则返回1，d == 1.2 aso else 0

（和int not_positive（double d）//返回1如果d == - 0.0 d == - 1.6（否则

0））


我的猜测是这样一个命令可以快速实现（有些

比特错误 - 我在这里错了吗？）我不知道dobbles看起来怎么样？
比特

如果它在所有机器上都是一样的，但我认为可以快速制作

....


如果我有一个像这样的快速命令（并且gcc / g ++与-O3不会使



比较x> = 0快 - 也许答案就是那个是真正的

问题

，但如果可以在不进行比较的情况下制作快速功能，那么

语言就会发出信号比较，因此

更快）


可以编写如下内容：

取决于函数的速度是的可能他比一个

二进制文件更快

搜索v。


有没有人对此有任何意见？


双cos（双v）

{

：重新运行

v = fabs（v） - （2.0 * PI）;

int part = 0;


part = part + not_negative（v）; // v大于2PI

v = v + PI / 4.0;

part = part + not_negative（v）;

v = v + PI / 4.0;

part = part + not_negative（v）;

v = v + PI / 4.0;

part = part + not_negative（ v）;

v = v + PI / 4.0;

part = part + not_negative（v）;

v = v + PI / 4.0;

part = part + not_negative（v）;

v = v + PI / 4.0;

part = part + not_negative（v）;

v = v + PI / 4.0;

part = part + not_negative（v）;


开关（部分）

{

案例0：为case0返回一些内容;

案例1：为case1返回一些内容;

案例2：返回适用于case2;

案例3：为case3返回一些内容;

案例4：为case4返回一些内容;

案例5：返回一些内容case5;

案例6：为case6返回一些内容;

案例7：为case7返回一些内容;

案例8：x不在[ 0; 2PI]。

//这不是部分这笔交易但是

//修复x并转到goto重新运行

}

}

I am not an expert on C++(/C), but also not a complete beginner.

I was writing a faster cos (and sin) approximation based on
precalculated data (with some success) when I discovered that going
from double to int is quite expensive, so maybe a cos/sin approximation
without precalculation would be better (faster).

Nevertheless I (still) needed to find out where on the unit-cirkle the
angle
(a double (v) from 0 to 2PI) was. And I looked for a command like

int not_negative(double d) // return 1 if d==(+)0.0, d==1.2 aso else 0
(and int not_positive(double d) // return 1 if d==-0.0 d==-1.6 (else
0))

My guess is that such a command could be made real fast (with some
bit-mangleling - am I wrong here?) I do not know how dobbles looks in
bits
and if it is the same on all machines, but I think it could be made
fast ....

If I had a fast command like this (and gcc/g++ with -O3 does not make
the
comparison x>=0 fast - maybe the answer is that that is the real
problem
,but if it is possible to make a fast function without a compare the
language would signal that this works without a compare and therefore
is faster)

it would be possible to write something like :
Depending on how fast the function is this might he faster than a
binary
search on v.

Does anybody has some comments on this ?

double cos(double v)
{
:rerun
v=fabs(v)-(2.0*PI);
int part = 0;

part=part + not_negative(v); // v was bigger than 2PI
v=v+PI/4.0;
part=part + not_negative(v);
v=v+PI/4.0;
part=part + not_negative(v);
v=v+PI/4.0;
part=part + not_negative(v);
v=v+PI/4.0;
part=part + not_negative(v);
v=v+PI/4.0;
part=part + not_negative(v);
v=v+PI/4.0;
part=part + not_negative(v);
v=v+PI/4.0;
part=part + not_negative(v);

switch (part)
{
case 0 : return something for case0;
case 1 : return something for case1;
case 2 : return something for case2;
case 3 : return something for case3;
case 4 : return something for case4;
case 5 : return something for case5;
case 6 : return something for case6;
case 7 : return something for case7;
case 8 : x is not in [0;2PI].
// This was not a "part" of the deal but
// fix x and goto rerun
}
}

推荐答案

tmartsum写道：

tmartsum wrote:

我不是C ++（/ C）的专家，但也不是一个完整的初学者。

当我发现从double到int非常昂贵时，我正在编写一个更快的cos（和sin）近似基于
预先计算的数据（有一些成功），所以可能是cos / sin近似
然而我（仍然）需要找出单位上的位置 - 角度
（一个双（v）从0到2PI）是。我找了一个像

int not_negative（double d）的命令//如果d ==（+）0.0则返回1，d == 1.2 aso else 0
（和int not_positive（ double d）//如果d == - 返回1 = 0.0 d == - 1.6（否则
0））

我的猜测是这样的命令可以快速实现（有些
bit-mangleling - 我在这里错了吗？）我不知道在所有机器上看起来是多么粗糙，但我认为它可以制作成br？快速....

I am not an expert on C++(/C), but also not a complete beginner.

I was writing a faster cos (and sin) approximation based on
precalculated data (with some success) when I discovered that going
from double to int is quite expensive, so maybe a cos/sin approximation
without precalculation would be better (faster).

Nevertheless I (still) needed to find out where on the unit-cirkle the
angle
(a double (v) from 0 to 2PI) was. And I looked for a command like

int not_negative(double d) // return 1 if d==(+)0.0, d==1.2 aso else 0
(and int not_positive(double d) // return 1 if d==-0.0 d==-1.6 (else
0))

My guess is that such a command could be made real fast (with some
bit-mangleling - am I wrong here?) I do not know how dobbles looks in
bits
and if it is the same on all machines, but I think it could be made
fast ....

解决方案不可移植。但是，无论如何，看一下你的双重格式的b / b
编译器实现细节，它将是由指数和尾数组成的
，在尾数中找到符号位，

使非便携式实现依赖于小心方法

来测试它。

准备将其更改为当你发现更好的方法时（比如

分区来发现PI / 4.0包含多少倍）


Zara

The solution would not be portable. But, anyhow, take a look at your
compiler implementation details about the double format, which will be
formed by exponent and mantissa, locate the sign bit in the mantissa,
make a non-portable implementation dependent be-careful-with-it method
to test it.
An be ready to change it as soon as you discover better ways (such as
division to discover how many times PI/4.0 is contained in a double)

Zara

Zara skrev：

Zara skrev:

该解决方案不可移植。但是，无论如何，看一下你的双格式的编译器实现细节，它将由指数和尾数组成，在尾数中找到符号位，
做一个非便携式实现依赖于小心的方法来测试它。
一旦你找到更好的方法（例如分区发现PI多少次）就准备好改变它/4.0包含在一个双））
Zara

The solution would not be portable. But, anyhow, take a look at your
compiler implementation details about the double format, which will be
formed by exponent and mantissa, locate the sign bit in the mantissa,
make a non-portable implementation dependent be-careful-with-it method
to test it.
An be ready to change it as soon as you discover better ways (such as
division to discover how many times PI/4.0 is contained in a double) Zara

感谢您的回答 - 我的猜测是它不便携。

（我想编写可移植的C ++（或者在这种情况下它只是C）


我的主题也是：C ++中的sign_command missing（？） ;因此

编译器可能知道如何快速为给定平台做这件事，

但我不能做出快速算法，除非我编写创建C ++的C ++

（我不喜欢。）


PS：我应该先写的东西：

int part = st atic_cast< INT> （楼层（v *（1 /（2.0 * PI）* 8）））;


不是（我记得它）表现良好。演员和地板都很昂贵。我考虑不使用2

原因的预先计算数据。

1）查找可能的缓存未命中

2）演员来自double to int

Thanks for the answer - It was my guess that it would not be portable.
(And I want to write portable C++ (or in this case it is just C)

My subject was also : "sign_command missing(?) in C++" hence the
compiler would probably know how to do this quick for a given platform,
but I can make no fast algoritm unless I write C++ that creates C++
(Which I am not fond of.)

PS : Something I should have written first :
int part = static_cast<int> (floor(v*(1/(2.0*PI)*8)));

is not (as I remember it) performing good. Both the cast and floor are
expensive. I was considering not using pre_calculated data of 2
reasons.
1) The lookup with a possible cache-miss
2) The cast from double to int

tmartsum写道：

tmartsum wrote:

我的主题还有：C ++中的sign_command missing（？） ;因此
编译器可能知道如何快速地为给定的平台做这个，但是除非我编写创建C ++的C ++（我不喜欢它），否则我不能做出快速的算法。 ）


sign命令应该是


inline bool not_negative（double d）{return d> = 0.0;}


如果我们尊重便携性，这是最有效的方式。

PS：我应该首先编写的东西：
int part = static_cast< int> （楼层（v *（1 /（2.0 * PI）* 8）））;

不是（我记得它）表现良好。

My subject was also : "sign_command missing(?) in C++" hence the
compiler would probably know how to do this quick for a given platform,
but I can make no fast algoritm unless I write C++ that creates C++
(Which I am not fond of.)
The sign command should be

inline bool not_negative(double d) {return d>=0.0;}

This is the most efficient way, if we honour portability.

PS : Something I should have written first :
int part = static_cast<int> (floor(v*(1/(2.0*PI)*8)));

is not (as I remember it) performing good.

这个是次优的，实际上是多余的。 *如果您保证

v> = 0.0，则以下表达式就足够了：


int part = static_cast< int> （v *（1 /（2.0 * PI）* 8））;


因为将实体类型转换为整数类型，所以将
向零舍入。

This one is sub-optimal, it is in fact redundant. *IF* you guarantee
that v >=0.0, then the following expression is enough:

int part = static_cast<int> (v*(1/(2.0*PI)*8));

Because casting form real types to integer types is done rounding
towards zero.

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