怎么写?? [英] how to write??

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问题描述

大家好!

我的问题:

怎么写(在Builder C ++中)

10 ^( - b * z)

因为我不知道:(

感谢您的帮助..

johnnyB

解决方案

janek写道:

大家好!
我的问题:
如何写(在Builder C ++中)
10 ^( - b * z)
因为我不知道:(




如果我没有误会它是pow(10.0,b * z);你需要包括math.h


-

WW aka Attila


" janek"< si ****@poczta.gazeta.pl>写在消息中

news:bk ********** @ inews.gazeta.pl ...

大家好!
我的问题:
如何写(在Builder C ++中)
10 ^( - b * z)
因为我不知道:(
感谢您的帮助..
johnnyB




我猜,但试试''std :: pow(10, - b * z) ''。

你有到#include< cmath>首先。


Buster写道:

我猜,但试试''std :: pow(10, - b * z) ''。
你必须#include< cmath>第一个。




在我们再次讨论cmath / std :: vs. math / nostd之前我有

提出了后者有两个原因。一:不幸的是我已经看到了

编译器,其中包含一个cxxx标题确定是disater的配方。另一个

(不那么重要)是它也适用于C.


-

WW aka Attila


Hello all!
My question:
How to write (in Builder C++)
10^(-b*z)
cause i have no idea :(
Thanks for any help..
johnnyB

解决方案

janek wrote:

Hello all!
My question:
How to write (in Builder C++)
10^(-b*z)
cause i have no idea :(



If I am not mistaking it is pow(10.0, b*z); You need to include math.h

--
WW aka Attila


"janek" <si****@poczta.gazeta.pl> wrote in message
news:bk**********@inews.gazeta.pl...

Hello all!
My question:
How to write (in Builder C++)
10^(-b*z)
cause i have no idea :(
Thanks for any help..
johnnyB



I''m guessing, but try ''std::pow (10, - b * z)''.
You have to #include <cmath> first.


Buster wrote:

I''m guessing, but try ''std::pow (10, - b * z)''.
You have to #include <cmath> first.



Before we go into a debate again on the cmath/std:: vs. math/nostd I have
proposed the latter for two reasons. One: unfortunately I have seen
compilers where include a cxxx header is sure recipe for disater. The other
(way less important) is that it also works in C.

--
WW aka Attila


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