nrand48源代码看起来很奇怪？ [英] nrand48 source code looks weird ?
问题描述
这是来自gnuwin32 libgw32c的nrand48的源代码
long int nrand48（xsubi）unsigned short int xsubi [3];
{
long int result;
（void）__nrand48_r（xsubi，& __ libc_drand48_data，& result）;
返回结果;
}
// long int nrand48（xsubi）unsigned short int xsubi [3];
''unsigned short int xsubi [3];''部分意味着''（xsubi）''后面是什么？
来自二进制文件的stdlib.h中的原型是:(由于某种原因stdlib.h
不在源代码发布中？）
extern long int nrand48（unsigned short int __xsubi [3]）__ THROW;
所以乍一看我会猜测它与抛出异常有关...但是现在我再看一遍它我认为它是'b $ b'这是xsubi的类型。
原始看起来很奇怪：
1. long int nrand48（xsubi）unsigned short i nt xsubi [3];
这不应该是这样的：
2. long int nrand48（unsigned short int xsubi [3]）;
为什么他们把它写成1.？
这只是他们写作的首选方式。 ..还是有一些
的功能？
然后原型再次看起来像2.！？嗯？ :)
Euhm ......难道这是一个古怪的程序员的案例吗？呃。
再见，
Skybuck。
Skybuck Flying<否**** @ hotmail.com>写道：这是来自gnuwin32的nrand48的源代码libgw32c
long int nrand48（xsubi）unsigned short int xsubi [3];
{
long int result ;
（void）__nrand48_r（xsubi，& __ libc_drand48_data，& result）;
返回结果;
}
// long int nrand48（xsubi）unsigned short int xsubi [3];
''（xsubi）''''unsigned short int xsubi [3];''部分意味着什么？
这是函数定义的旧方式，在没有C标准的情况下，K& R天写回了
。这基本上相当于
借给
long int nrand48（unsigned short int xsubi [3]）
在那段时间的代码中你通常有函数定义
开头
my_func（a，b，c）
int a，
char * b，
unsigned short c
{
现在将写成
int my_func（int a，char * b，unsigned short c）
{
来自二进制文件的stdlib.h中的原型是:(由于某种原因，stdlib.h
不在源代码分发中？）
extern long int nrand48（unsigned short int __xsubi [3]）__ THROW;
乍一看，我猜它与抛出异常有关......但是现在我再次看它，我认为它是xsubi的类型。
最好不要看头文件，除非你有很好的理由
（或只是出于好奇）。很多系统依赖的魔法在那里上$
。
问候，Jens
-
\ Jens Thoms Toerring ___ Je***********@physik.fu- berlin.de
\ __________________________ http：// www。 toerring.de
< Je *********** @ physik.fu-berlin &由Matchi.com提供回到GT;在消息中写道
新闻：2l ************ @ uni-berlin.de ...Skybuck Flying< no ** **@hotmail.com>写道：这是来自gnuwin32 libgw32c的nrand48的源代码
long int nrand48（xsubi）unsigned short int xsubi [3];
{
long int result;
（void）__nrand48_r（xsubi，& __ libc_drand48_data，& result）;< br>
返回结果;
}
// long int nrand48（xsubi）unsigned short int xsubi [3] ;
''unsigned short int xsubi [3];''部分意味着''（xsubi）''后面是什么？
这是功能定义的旧方式，在没有C标准的K& R天中写回。这基本等同于
long int nrand48（unsigned short int xsubi [3]）
在从那时起你的典型代码中有函数定义
my_func（a，b，c）
int a，
char * b，
unsigned short c
{
现在将写成
int my_func（int a，char * b，unsigned short c）
{
是的，我开始认为C真的是＃^＃@
语言:)
Visual C / C ++ 6.0并不支持旧方式...
<<< ---叹气--->>>
所以现在我必须重新写一遍...
幸运的是我...它只是一点点代码......幸运的是我
很少在C中编程：P = D
Thx的描述......大声笑......它必须是痛苦的C代码：D
再见，
Skybuck。
This is the source code for nrand48 from gnuwin32 libgw32c
long int nrand48 (xsubi) unsigned short int xsubi[3];
{
long int result;
(void) __nrand48_r (xsubi, &__libc_drand48_data, &result);
return result;
}
// long int nrand48 (xsubi) unsigned short int xsubi[3];
What does the ''unsigned short int xsubi[3];'' part mean after ''(xsubi)'' ?
The prototype in stdlib.h from the binaries is: (for some reason stdlib.h
was not in source distribution ?)
extern long int nrand48 (unsigned short int __xsubi[3]) __THROW;
So at first sight I would guess it has to do with throwing exceptions... but
now that I look at it again it I think it''s the type for xsubi.
The original looks really weird:
1. long int nrand48 (xsubi) unsigned short int xsubi[3];
Isn''t that supposed to look like this:
2. long int nrand48 (unsigned short int xsubi[3] );
Why did they write it as 1. ?
Is that just their preferred way of writing it down... or does it have some
sort of functionality ?
The prototype then again looks like 2. !? huh ?! :)
Euhm... Could this be a case of wacky c programmers ? eh.
Bye,
Skybuck.
Skybuck Flying <no****@hotmail.com> wrote:This is the source code for nrand48 from gnuwin32 libgw32c long int nrand48 (xsubi) unsigned short int xsubi[3];
{
long int result; (void) __nrand48_r (xsubi, &__libc_drand48_data, &result); return result;
} // long int nrand48 (xsubi) unsigned short int xsubi[3]; What does the ''unsigned short int xsubi[3];'' part mean after ''(xsubi)'' ?
That''s the very old way function definitions where written back in
the K&R days when there wasn''t a C standard. It''s basically equiva-
lent to
long int nrand48( unsigned short int xsubi[ 3 ] )
In code dating back from that time you typically had function definitions
starting with
my_func( a, b, c )
int a,
char *b,
unsigned short c
{
which nowadays would be written as
int my_func( int a, char *b, unsigned short c )
{
The prototype in stdlib.h from the binaries is: (for some reason stdlib.h
was not in source distribution ?) extern long int nrand48 (unsigned short int __xsubi[3]) __THROW; So at first sight I would guess it has to do with throwing exceptions... but
now that I look at it again it I think it''s the type for xsubi.
Better don''t look at the header files unless you have very good reasons
to (or just out of curiosity). Lots of system-dependend magic is going
on there.
Regards, Jens
--
\ Jens Thoms Toerring ___ Je***********@physik.fu-berlin.de
\__________________________ http://www.toerring.de
<Je***********@physik.fu-berlin.de> wrote in message
news:2l************@uni-berlin.de...Skybuck Flying <no****@hotmail.com> wrote:This is the source code for nrand48 from gnuwin32 libgw32c
long int nrand48 (xsubi) unsigned short int xsubi[3];
{
long int result;
(void) __nrand48_r (xsubi, &__libc_drand48_data, &result);
return result;
}
// long int nrand48 (xsubi) unsigned short int xsubi[3];
What does the ''unsigned short int xsubi[3];'' part mean after ''(xsubi)'' ?
That''s the very old way function definitions where written back in
the K&R days when there wasn''t a C standard. It''s basically equiva-
lent to
long int nrand48( unsigned short int xsubi[ 3 ] )
In code dating back from that time you typically had function definitions
starting with
my_func( a, b, c )
int a,
char *b,
unsigned short c
{
which nowadays would be written as
int my_func( int a, char *b, unsigned short c )
{
Yes, I was starting to think that C was a really #^#@
language :)
Visual C/C++ 6.0 doesn''t the support the old way...
<<<--- sigh --->>>
So now I have to re-write it all...
Fortunately for me... It''s only a little bit of code... and fortunately I
rarely program in C :P =D
Thx for the description... Lol... it must be painfull to code in C :D
Bye,
Skybuck.
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