奇怪的算术 [英] Weird arithmetic

问题描述

嗨新闻组，


我是C的新手并且想知道以下代码：


#include< stdio.h>


int main（无效）

{

double m = 100.45;

double n = 200.45;

double o = 200.45;


int i = m * 100;

int j = n * 100;


o = n * 100;

int k = o;


printf（"％lf =％i \ n"

"％lf =％i \ n"

"％lf =％i \ n"，m * 100，i，n * 100，j，o，k）;


返回0;

}

编译为GCC-3.4.3 on一个Linux盒子（Fedora Core 3 x86）输出

看起来像这样：

$ gcc -Wall -pedantic -std = c99 test.c -o测试

$ ./test

10045.000000 = 10045

20045.000000 = 20044

20045.000000 = 20045


怎么会这样？


提前致谢！

- Pascal

解决方案

gcc -Wall -pedantic -std = c99 test.c -o test

./ test

10045.000000 = 10045

20045.000000 = 20044

20045.000000 = 20045


怎么会这样？


提前致谢！

- Pascal

" Pascal Gallois < [e-mail]>在消息中写道

news：pa **************************** @ nowhere.net ... < blockquote class =post_quotes>嗨新闻组，

我是C的新手并且想知道以下代码：

#include< stdio.h>

int main（void）
{
double m = 100.45;
double n = 200.45;
double o = 200.45;

int i = m * 100;
int j = n * 100;

o = n * 100;
int k = o;

printf（ "％lf =％i \ n"
"％lf =％i \ n"
"％lf =％i \ n"，m * 100，i，n * 100，j，o，k）;

返回0;
}

在Linux机器上用GCC-3.4.3编译（Fedora Core 3 x86）输出
看起来像这样：

Hi newsgroup,

I am new to C and wonder about the following code:

#include <stdio.h>

int main(void)
{
double m = 100.45;
double n = 200.45;
double o = 200.45;

int i = m * 100;
int j = n * 100;

o = n * 100;
int k = o;

printf("%lf = %i\n"
"%lf = %i\n"
"%lf = %i\n", m * 100, i, n * 100, j, o, k);

return 0;
}
Compiled with GCC-3.4.3 on a Linux box (Fedora Core 3 x86) the output
looks like this:

$ gcc -Wall -pedantic -std=c99 test.c -o test
$ ./test
10045.000000 = 10045
20045.000000 = 20044
20045.000000 = 20045

How could this be?

Thanks in advance!
- Pascal

解决方案

gcc -Wall -pedantic -std=c99 test.c -o test

./test
10045.000000 = 10045
20045.000000 = 20044
20045.000000 = 20045

How could this be?

Thanks in advance!
- Pascal

"Pascal Gallois" <[e-mail]> wrote in message
news:pa****************************@nowhere.net...

Hi newsgroup,

I am new to C and wonder about the following code:

#include <stdio.h>

int main(void)
{
double m = 100.45;
double n = 200.45;
double o = 200.45;

int i = m * 100;
int j = n * 100;

o = n * 100;
int k = o;

printf("%lf = %i\n"
"%lf = %i\n"
"%lf = %i\n", m * 100, i, n * 100, j, o, k);

return 0;
}
Compiled with GCC-3.4.3 on a Linux box (Fedora Core 3 x86) the output
looks like this:

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