Linux奇怪 [英] Linux oddity
问题描述
嗨伙计，
将double转换为int时，结果不如我在Linux上所期望的那样：b / b >
#include< stdio.h>
int
main（int argc，char ** argv）
{
double val = 0.24;
int multiplier = 2000;
int result = static_cast< int> （val *乘数）;
printf（" result =％d（应为480）\ n，结果）;
返回0;
}
以上代码在SunOS和Windows上打印480，但在
Linux上使用gcc 3.2打印479.是否有效
对这种差异的解释是什么？
- 基思
Hi Folks,
When converting a double to an int, the result is not as
I''d expect on Linux:
#include <stdio.h>
int
main(int argc, char **argv)
{
double val = 0.24;
int multiplier = 2000;
int result = static_cast<int> (val * multiplier);
printf("result = %d (should be 480)\n", result);
return 0;
}
The above code prints 480 on SunOS and Windows, but on
Linux with gcc 3.2 it prints 479. Is there a valid
explanation for this difference?
- Keith
推荐答案
Keith S.写道：
Keith S. wrote:
嗨伙计，
将double转换为int时，结果不如
我会期待在Linux上：
#include< stdio.h>
int
main（int argc，char ** argv）
{/ / = 0.24;
int multiplier = 2000;
int result = static_cast< int> （val *乘数）;
printf（" result =％d（应为480）\ n，结果）;
返回0;
}
上面的代码在SunOS和Windows上打印480，但在使用gcc 3.2的Linux上打印479.这种差异是否有一个有效的解释？
- Keith
Hi Folks,
When converting a double to an int, the result is not as
I''d expect on Linux:
#include <stdio.h>
int
main(int argc, char **argv)
{
double val = 0.24;
int multiplier = 2000;
int result = static_cast<int> (val * multiplier);
printf("result = %d (should be 480)\n", result);
return 0;
}
The above code prints 480 on SunOS and Windows, but on
Linux with gcc 3.2 it prints 479. Is there a valid
explanation for this difference?
- Keith
嗯，接缝是个数字问题：
#include< cstdio>
int
main（int argc，char ** argv）
{
double val = 0.24;
int multiplier = 2000;
int result = static_cast< int>（val * multiplier）;
printf（" val * multiplier - 480 =％20.20f \ n"， val *乘数 - 480）;
返回0;
}
但这很奇怪。我假设结果只取决于CPU
而不是操作系统...
Hmmm, seams to be a numerical problem:
#include <cstdio>
int
main(int argc, char **argv)
{
double val = 0.24;
int multiplier = 2000;
int result = static_cast<int>(val * multiplier);
printf("val * multiplier - 480 = %20.20f\n", val * multiplier - 480);
return 0;
}
but it''s strange. I assumed that the result would only depend on the CPU
not the Operating System...
Keith S.写道新闻：bm ************ @ ID-169434.news.uni-berlin.de：
Keith S. wrote in news:bm************@ID-169434.news.uni-berlin.de:
嗨伙计，
将double转换为int时，结果不如我在Linux上期望的那样：
#include< stdio.h>
int
main（int argc，char ** argv）
{
double val = 0.24;
int multiplier = 2000;
int result = static_cast< int> （val *乘数）;
printf（" result =％d（应为480）\ n，结果）;
返回0;
}
以上代码在SunOS和Windows上打印480，但在
Linux上使用gcc 3.2打印479.这种差异是否有正确的解释？
Hi Folks,
When converting a double to an int, the result is not as
I''d expect on Linux:
#include <stdio.h>
int
main(int argc, char **argv)
{
double val = 0.24;
int multiplier = 2000;
int result = static_cast<int> (val * multiplier);
printf("result = %d (should be 480)\n", result);
return 0;
}
The above code prints 480 on SunOS and Windows, but on
Linux with gcc 3.2 it prints 479. Is there a valid
explanation for this difference?
漫画者正在使用不同的舍入规则，因为我相信他们可以做的事情是
。请注意，0.24不能在大多数机器上
表示为精确值，实际得到的是
类似于0.23999999999997。所以
给出480的2个编译器向上舍入到最接近的值，并且
linux框是四舍五入的。它的static_cast< int>（）
正在进行四舍五入。
查找函数'sd :: floor（）和std： ：ceil（）在#include< cmath>中定义了
。
HTH
Rob。
-
http：// www .victim-prime.dsl.pipex.com /
Rob Williscroft写道：
Rob Williscroft wrote:
编译器正在使用不同的舍入规则，我相信他们被允许这样做。请注意，0.24不能在大多数机器上表示为精确值，实际得到的是
类似于0.23999999999997。因此，给出480的两个编译器向上舍入到最接近的值，并且
linux盒子正在向下舍入。它的static_cast< int>（）正在进行四舍五入。
The comilers are using diferent rounding rules, as I believe
they are allowed to do. Note that 0.24 can''t on most machines
be represented as an exact value, what you actually get is
something like 0.23999999999997. So the 2 compilers that
give 480 are rounding to up or to the nearest value, and the
linux box is rounding down. Its the static_cast<int>() that
is doing the rounding.
理解，虽然我原以为static_cast< int>'s <
不同平台上的
行为是相同的（关于舍入方法），特别是在使用相同的编译器时...
- 基思
Understood, although I would have expected that static_cast<int>''s
behaviour was the same (with respect to rounding method) on
different platforms, especially when using the same compiler...
- Keith
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