不工作 [英] cos not working

问题描述

HI


我是C的新手，这不起作用。我们都知道cos Pi是-1

但这里答案是1.000000 !!!!我使用MS编译所以请帮助!!!

extern void main（）;

extern int printf（）;

extern double cos（）;

void main（）{

#define Pi 22/7

printf（"％lf"，cos （Pi））;

返回;

}


回答

1.000000< - - 不对!!!!!


谢谢大家:)

解决方案

Miles Sorenson写道：

HI


我是C的新手，这是行不通的。我们都知道cos Pi是-1

但这里答案是1.000000 !!!!我使用MS编译所以请帮助!!!

extern void main（）;

extern int printf（）;

extern double cos（）;

那里有什么用的？


让我们再试一次，好吗？


#include< math.h>

#include< stdio.h>


int main（void）

{

const double pi = 22.0 / 7;


printf（"％lf \ n"，cos（pi） ）;


返回0;

}


c99 -lm test.c


结果：


-0.999999

-

Ian Collins

在文章< Ra ***************** @ newsfe01.iad>，

Miles Sorenson< ; ms ************* @ aol.comwrote：

> extern int printf（）;
extern double COS（）;

为什么你自己声明这些功能？这样做非常容易发生错误。改为包括标准标题。

> #define Pi 22/7

22/7是3.这就是你想要的吗？

> printf（"％lf"，cos（Pi））;

cos（）采用什么样的参数？一双。但是你是
传递一个整数。由于你的声明很差，

不是原型，它不会被转换为双倍，所以你是完全垃圾传递给函数的
。


- 理查德

-

请记得提及我/你留下的录音带。

Ian Collins写道：

Miles Sorenson写道：

> HI

我是C的新手，这不起作用。我们都知道cos Pi是-1
但这里答案是1.000000 !!!!我使用MS编译所以请帮助!!!

extern void main（）;
extern int printf（）;
extern double cos（）;

那里有什么用的？


让我们再试一次，好吗？


#include< math.h>

#include< stdio.h>


int main（void）

{

const double pi = 22.0 / 7;


printf（"％lf \ n"，cos（pi） ）;


返回0;

}


c99 -lm test.c


结果：


-0.999999

％f用于输出C89和C89中的类型double值C99。


中不作任何内容，作为C99中的printf格式指定符。


％lf使该程序在C89中未定义。


-

pete

HI

I''m newby with C and this is not working. we all know that cos Pi is -1
but answer here is 1.000000 !!!! i use MS compile so please help!!!

extern void main();
extern int printf();
extern double cos();
void main(){
#define Pi 22/7
printf("%lf",cos(Pi));
return;
}

answer
1.000000 <--- not right!!!!!

thanks everyone :)

解决方案

Miles Sorenson wrote:

HI

I''m newby with C and this is not working. we all know that cos Pi is -1
but answer here is 1.000000 !!!! i use MS compile so please help!!!

extern void main();
extern int printf();
extern double cos();

What are all those there for?

Let''s try again, shall we?

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

int main(void)
{
const double pi = 22.0/7;

printf("%lf\n",cos(pi));

return 0;
}

c99 -lm test.c

result:

-0.999999

--
Ian Collins

In article <Ra*****************@newsfe01.iad>,
Miles Sorenson <ms*************@aol.comwrote:

>extern int printf();
extern double cos();

Why are you declaring these functions yourself? Doing so is extremely
error prone. Include the standard headers instead.

>#define Pi 22/7

22/7 is 3. Is that what you want?

>printf("%lf",cos(Pi));

What kind of argument does cos() take? A double. But you''re
passing it an integer. Because of your poor declaration, which
isn''t a prototype, it won''t be converted to a double, so you are
passing complete rubbish to the function.

-- Richard
--
Please remember to mention me / in tapes you leave behind.

Ian Collins wrote:

Miles Sorenson wrote:

>HI

I''m newby with C and this is not working. we all know that cos Pi is -1
but answer here is 1.000000 !!!! i use MS compile so please help!!!

extern void main();
extern int printf();
extern double cos();

What are all those there for?

Let''s try again, shall we?

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

int main(void)
{
const double pi = 22.0/7;

printf("%lf\n",cos(pi));

return 0;
}

c99 -lm test.c

result:

-0.999999

%f is for outputting type double values in both C89 and C99.

The letter ''l'' is allowed, but does nothing, in "%lf"
as a printf format specifer in C99.

The %lf makes that program undefined in C89.

--
pete

这篇关于不工作的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！