圆的问题 [英] problem with round

问题描述

全部，

i我试图使用圆形（）函数，我通过google

在math.h中声明（
http://www.gnu.org/software/ libc / man ... unctions.html）。

此函数不起作用，我得到一个未引用的符号错误。当包含math.h时，

函数rint（）有效。

是否在所有math.h文件中找不到round（）？如果是这样的话可能是一个

替代方案。我在一台运行sparc-sun-solaris的机器上运行

系统，gcc版本为2.95.2。

谢谢

all,
i am trying to use the function round() which I found through google
to be declared in math.h (
http://www.gnu.org/software/libc/man...unctions.html).
this function does not work and i get an unreferenced symbol error. the
function rint() works when math.h is included.
is round() not found in all math.h files? if so what might be an
alternative. i work on a machine that runs a sparc-sun-solaris operating
system with gcc version 2.95.2.
thanks

推荐答案

m< ma ** @ mschumacher.com>写道：

m <ma**@mschumacher.com> writes:

我正在尝试使用函数round（），我通过google发现它在math.h中声明（
http://www.gnu.org/software/libc/男人... unctions.html）。这个
功能不起作用，我得到一个未引用的符号错误。
函数rint（）在包含math.h时有效。

i am trying to use the function round() which I found through
google to be declared in math.h (
http://www.gnu.org/software/libc/man...unctions.html). this
function does not work and i get an unreferenced symbol error. the
function rint() works when math.h is included.

这是在C FAQ中。


14.3：我正在尝试做一些简单的触发，我正在#including< math.h>，

但是我一直得到undefined：sin编译错误。


答：确保你实际上是在连接数学库。对于

实例，在Unix下，在编译/链接时，通常需要使用-lm选项，命令行的* end * * * *。另请参阅

问题13.25,13.26和14.2。

-

int main（void）{char p [] =" ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuv wxyz. \

\ n"，* q =" kl BIcNBFr.NKEzjwCIxNJC" ;; int i = sizeof p / 2; char * strchr（）; int putchar（\

）; while（* q）{i + = strchr（p，* q ++） - p; if（i> =（int）sizeof p）i- = sizeof p-1; putchar（p [i] \

）;}返回0;}

This is in the C FAQ.

14.3: I''m trying to do some simple trig, and I am #including <math.h>,
but I keep getting "undefined: sin" compilation errors.

A: Make sure you''re actually linking with the math library. For
instance, under Unix, you usually need to use the -lm option, at
the *end* of the command line, when compiling/linking. See also
questions 13.25, 13.26, and 14.2.
--
int main(void){char p[]="ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuv wxyz.\
\n",*q="kl BIcNBFr.NKEzjwCIxNJC";int i=sizeof p/2;char *strchr();int putchar(\
);while(*q){i+=strchr(p,*q++)-p;if(i>=(int)sizeof p)i-=sizeof p-1;putchar(p[i]\
);}return 0;}

Ben Pfaff写道：

Ben Pfaff wrote:

m< ma ** @ mschumacher.com>写道：

m <ma**@mschumacher.com> writes:

我正在尝试使用函数round（），我通过google发现它在math.h中声明（
< a rel =nofollowhref =http://www.gnu.org/software/libc/manual/html_node/Rounding-Functions.html)\"target =_ blank> http://www.gnu.org/软件/的libc /人...... unctions.html）。这个
功能不起作用，我得到一个未引用的符号错误。
函数rint（）在包含math.h时有效。

i am trying to use the function round() which I found through
google to be declared in math.h (
http://www.gnu.org/software/libc/man...unctions.html). this
function does not work and i get an unreferenced symbol error. the
function rint() works when math.h is included.

这是在C FAQ中。

14.3：我是试图做一些简单的触发，我是#including< math.h>，
但我一直得到undefined：sin编译错误。
答：确保你实际上是在连接数学库。对于
实例，在Unix下，在编译/链接时，通常需要在命令行的* end *处使用-lm选项。另见
问题13.25,13.26和14.2。

This is in the C FAQ.

14.3: I''m trying to do some simple trig, and I am #including <math.h>,
but I keep getting "undefined: sin" compilation errors.

A: Make sure you''re actually linking with the math library. For
instance, under Unix, you usually need to use the -lm option, at
the *end* of the command line, when compiling/linking. See also
questions 13.25, 13.26, and 14.2.

抱歉没有先阅读常见问题解答。谢谢回复。我使用

#define round（x）（（x）> = 0）工作

？（int）（（x）+0.5）:( int ）（（x）-0.5）

对C来说是新手，但看起来很奇怪。这种链接只是使用像头文件这样的东西来敲打

。我相信我错过了一些东西。


谢谢

sorry for not having read the FAQ first. thanks for the reply. i worked
around that using
#define round(x) ((x)>=0)?(int)((x)+0.5):(int)((x)-0.5)
am new to C but seems mighty strange.this kind of linking just beats
using something like a header file. i am sure i am missing something.

thanks

m< ma ** @ mschumacher.com>写道：

m <ma**@mschumacher.com> writes:

抱歉没有先阅读常见问题解答。谢谢回复。我使用
#define round（x）（（x）> = 0）来解决这个问题？（int）（（x）+0.5）:( int）（（x）-0.5）

sorry for not having read the FAQ first. thanks for the reply. i
worked around that using
#define round(x) ((x)>=0)?(int)((x)+0.5):(int)((x)-0.5)

对于它的价值，它应该可行，但只有当'x''

四舍五入到范围'int''。

-

int main（void）{char p [] =" ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuv wxyz。\

\ n"，* q =" kl BIcNBFr.NKEzjwCIxNJC" ;; int i = sizeof p / 2; char * strchr（）; int putchar（\

）; while（* q ）{i + = strchr（p，* q ++） - p; if（i> =（int）sizeof p）i- = sizeof p-1; putchar（p [i] \

） ;}返回0;}

For what it''s worth, that should work okay, but only if `x''
rounds to a value in the range of `int''.
--
int main(void){char p[]="ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuv wxyz.\
\n",*q="kl BIcNBFr.NKEzjwCIxNJC";int i=sizeof p/2;char *strchr();int putchar(\
);while(*q){i+=strchr(p,*q++)-p;if(i>=(int)sizeof p)i-=sizeof p-1;putchar(p[i]\
);}return 0;}

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