Pascal - > C [英] Pascal -> C
问题描述
Hallo allemaal，
在我的Pascal程序中，我在一个函数中收集了所有消息。所有
消息都有自己的号码。这种方法的优点之一
是如果有人想要更改消息，他只需要在一个地方查看
。
该函数声明如下：
函数zin（by：byte）：string;
调用函数的行看起来像简单如下
这个：
writeln（zin（12））;
到目前为止，我找到了这个函数的替代品但我不满足于b
。我定义了一个''char pst [255]''和函数''void
zin（int by）''填充pst和想要的消息。但这意味着我现在
需要两行而不是一行：
zin（12）;
printf（"％s"，pst ）;
当然我试图把''zin（int by）''变成一个函数，
返回一个值，所以我可以把它们结合起来两行就像
''printf（"％s"，zin（12））;''但是我得到了编译错误或只有
废话出现了在屏幕上。
同时我找到了一个解决方案，形式为
char * zin [20] = {" .."， "" ..."等等。$
printf（"％s"，zin [12]）;
但这适用于固定邮件但是那些邮件
是由动态函数生成的吗？
欢迎任何帮助！
-
___
/ __ | __
/ / | _ / Groetjes，Ruud Baltissen
\ \__ | _ \
\ ___ | http://Ruud.C64.org
Hallo allemaal,
In my Pascal program I gathered all messages in one function. All
messages have their own number. One of the advantages of this method
is that if someone wants to change a messages, he only has to look in
one place.
The function is declared as follow:
function zin(by : byte) : string;
A line calling the function can look like as simple as
this:
writeln(zin(12));
So far I have found a replacement for this function but I''m not
satisfied with it. I defined a ''char pst[255]'' and the function ''void
zin(int by)'' fills pst with the wanted message. But that means I now
need two lines instead of one:
zin(12);
printf("%s", pst);
Of course I tried to turn ''void zin(int by)'' into a funtion that
returns a value so I could combine those two line to something like
''printf("%s", zin(12));'' but I either got an compilation error or only
nonsense appeared on the screen.
In the meanwhile I found a solution in the form of
char *zin[20] = { "..", ""..." etc. }
printf("%s", zin[12]);
But that works fine for fixed messages but what about messages that
are generated by a function on the fly?
Any help is welcome!
--
___
/ __|__
/ / |_/ Groetjes, Ruud Baltissen
\ \__|_\
\___| http://Ruud.C64.org
推荐答案
Ruud说：
< snip>
Ruud said:
<snip>
当然我试过将''zin（int by）''转换成一个函数，
返回一个值，这样我就可以将这两行合并为类似
''printf（" ％s"，zin（12））;''但是我得到了一个编译错误或只有
废话出现在屏幕上。
Of course I tried to turn ''void zin(int by)'' into a funtion that
returns a value so I could combine those two line to something like
''printf("%s", zin(12));'' but I either got an compilation error or only
nonsense appeared on the screen.
为什么不修改zin返回（指向第一个字符的指针）
字符串？
const char * zin（int msg）
{
static char pst [MAX_LENGTH]; / *在您方便时定义MAX_LENGTH * /
const char * msglist [] =
{
" Hello"，
Goodbye，
Whatever
};
size_t count = sizeof msglist / sizeof msglist [0];
if（msg> = 0&& msg< count）
{
strcpy（pst， msglist [msg]）;
}
其他
{
/ *大概这是你的机器 - 生成的消息是
生成，并加载到pst
* /
}
返回pst ;
}
现在printf（"％s"，zin（12））变得合法 - 但要注意！静态缓冲区
表示您只有一个存储区域用于当前消息，因此
printf（"％s％s"，zin） （12），zin（1））很可能不会做你想要的东西
来做。如果这是一个问题，你需要考虑将缓冲区传递给
函数。
-
Richard Heathfield< http://www.cpax.org.uk>
电子邮件：-http：// www。 + rjh @
谷歌用户：< http://www.cpax.org.uk/prg/writings/googly.php>
Usenet是一个奇怪的放置" - dmr 1999年7月29日
Why not modify zin to return (a pointer to the first character of) the
string?
const char *zin(int msg)
{
static char pst[MAX_LENGTH]; /* define MAX_LENGTH at your convenience */
const char *msglist[] =
{
"Hello",
"Goodbye",
"Whatever"
};
size_t count = sizeof msglist / sizeof msglist[0];
if(msg >= 0 && msg < count)
{
strcpy(pst, msglist[msg]);
}
else
{
/* presumably this is where your machine-generated messages are
generated, and loaded into pst
*/
}
return pst;
}
Now printf("%s", zin(12)) becomes legal - but beware! The static buffer
means that you''ve only got one storage area for "the current message", so
printf("%s %s", zin(12), zin(1)) will very likely not do what you want it
to do. If that''s a problem, you need to think about passing a buffer into
the function.
--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999
10月21日，09：38，Ruud< Ruud.Baltis ... @ apg.nlwrote：
On 21 Oct, 09:38, Ruud <Ruud.Baltis...@apg.nlwrote:
在我的Pascal程序中，我在一个函数中收集了所有消息。所有
消息都有自己的号码。这种方法的优势之一
是如果有人想要更改消息，他只需要在一个地方查看
。
In my Pascal program I gathered all messages in one function. All
messages have their own number. One of the advantages of this method
is that if someone wants to change a messages, he only has to look in
one place.
如果你想知道什么是消息，它永远不会出现在你看的
的地方...... />
非常长的宏名称可以提供帮助。
错误（27）;
和某处
{27，通信链接丢失}
因此更好的宏可能是
错误（LINK_LOST）;
我参与了一个错误宏看起来像这样的项目
错误（SYSTEM_ERROR_COMMUNICATIONS_LINK_IS_LOST）;
但是这个是国际化（I18Ning）实际输出
字符串。
and if you want to know what a message is it''s never in the
place where you look...
Very long macro names can help.
error (27);
and somewhere is
{27, "communications link is lost"}
hence a better macro might be
error (LINK_LOST);
I worked on a project where the error macros looked like this
error (SYSTEM_ERROR_COMMUNICATIONS_LINK_IS_LOST);
But this was internationalising (I18Ning) the actual output
strings.
该函数声明如下：
函数zin（by：byte）：string;
调用函数的行看起来就像
一样简单：
writeln（zin（12））;
到目前为止，我已经找到了这个功能的替代品，但我不满足于b
$ b。我定义了一个''char pst [255]''和函数''void
zin（int by）''填充pst和想要的消息。但这意味着我现在
需要两行而不是一行：
zin（12）;
printf（"％s"，pst ）;
The function is declared as follow:
function zin(by : byte) : string;
A line calling the function can look like as simple as
this:
writeln(zin(12));
So far I have found a replacement for this function but I''m not
satisfied with it. I defined a ''char pst[255]'' and the function ''void
zin(int by)'' fills pst with the wanted message. But that means I now
need two lines instead of one:
zin(12);
printf("%s", pst);
你可以返回静态字符串
const char * zin（int n）
{
开关（n）
{
案例1：返回链接丢失;
休息;
案例2：返回O2超压;
休息;
默认：
返回坏错误代码;
}
}
printf（"％s"，zin（12））;
you could return static strings
const char* zin (int n)
{
switch (n)
{
case 1: return "link lost";
break;
case 2: return "O2 over pressure";
break;
default:
return "bad error code";
}
}
printf ("%s", zin(12));
当然我试图把''zin（int by）''变成一个函数，
返回一个值，所以我可以将这些结合起来两行就像
''printf（"％s"，zin（12））;''但是我得到了编译错误或只有
废话出现了屏幕上。
Of course I tried to turn ''void zin(int by)'' into a funtion that
returns a value so I could combine those two line to something like
''printf("%s", zin(12));'' but I either got an compilation error or only
nonsense appeared on the screen.
显示你的代码
show your code
同时我找到了一个解决方案
char * zin [20] = {" .."，"" ..."等等。
printf（"％s"，zin [12]）;
In the meanwhile I found a solution in the form of
char *zin[20] = { "..", ""..." etc. }
printf("%s", zin[12]);
是的。或者
struct String_tab_entry
{
int n;
const char * str;
};
String_tab_entry string_tab [] =
{
{12，" tile lost" }，
{99，" intruder"}，
（101，fire}，
};
然后搜索数组。字符串然后不必连续
甚至订购。（订购可以使搜索更快）
yep. Or
struct String_tab_entry
{
int n;
const char *str;
};
String_tab_entry string_tab [] =
{
{12, "tile lost"},
{99, "intruder"},
(101, "fire"},
};
Then search the array. The strings then don''t have to be contiguous
or even ordered. (tho'' ordered can make the search faster)
但这适用于固定消息但是动态函数生成的消息是什么？
But that works fine for fixed messages but what about messages that
are generated by a function on the fly?
在这种情况下你需要某种缓冲。这是相当的标准C问题。解决方案是
1。使用静态缓冲区
const char * zin（int n，const char * arg）
{
static char buffer [ 256];
开关（n）
{
案例1：
sprintf（缓冲区，链接丢失％s，arg）;
返回缓冲区;
}
}
多次通话可能有问题
printf（"％s％s"，zin（12），zin（13））;
2.传递一个缓冲区
const char * zin（int n，char * buffer，const char * arg）
{switch（n）
{
案例1：
sprintf（缓冲区，链接丢失％s，arg）;
返回缓冲区;
}
}
3. malloc缓冲区（但有人必须清理
之后）。
4.让zin（）更加聪明（未经测试的代码！）
#include< stdarg.h>
typedef void PrintFunc（const char *）;
void zin（int n，PrintFunc * print，const char * format，...）
{
va_list args;
va_start（args，format）;
char buffer [256];
vsprintf（buffer，format，args）;
prin t（缓冲区）;
}
void print（const char * s）
{
printf（"％s"，s）;
}
zin（23，print，link lost％d，link_id）;
这有点不整洁。但关键是你可以使用stdarg功能将仲裁
参数传递给zin（）。你可以通过函数参数来实现实际的i / o.
如果不了解你的应用程序的更多信息，我就不能再进一步了。< br $> b $ b希望我已经给了你一些想法。
-
Nick Keighley
In that case you need a buffer of some sort. This is fairly
standard C problem. Solutions are
1. use a static buffer
const char* zin (int n, const char* arg)
{
static char buffer [256];
switch (n)
{
case 1:
sprintf (buffer, "link lost %s", arg);
return buffer;
}
}
Multiple calls may be a problem
printf ("%s %s", zin(12), zin(13));
2. pass a buffer
const char* zin (int n, char *buffer, const char* arg)
{ switch (n)
{
case 1:
sprintf (buffer, "link lost %s", arg);
return buffer;
}
}
3. malloc a buffer (but then someone has to clean up
afterwards).
4. Make zin() a lot cleverer (Untested code!)
#include <stdarg.h>
typedef void PrintFunc (const char*);
void zin (int n, PrintFunc *print, const char *format, ...)
{
va_list args;
va_start (args, format);
char buffer [256];
vsprintf (buffer, format, args);
print (buffer);
}
void print (const char *s)
{
printf ("%s", s);
}
zin (23, print, "link lost %d", link_id);
This is a bit untidy. But the point is you can pass arbitary
arguments to zin() using stdarg functionality. You can
also pass function arguments to do the actual i/o.
Without knowing more about your app I can''t go any furthur.
Hope I''ve given you some ideas.
--
Nick Keighley
Nick Keighley< ni ****************** @ hotmail.comwrote：
Nick Keighley <ni******************@hotmail.comwrote:
10月21日，09：38，Ruud< Ruud.Baltis ... @ apg.nlwrote：
On 21 Oct, 09:38, Ruud <Ruud.Baltis...@apg.nlwrote:
在我的Pascal程序中，我收集了所有消息在一个功能。所有
消息都有自己的号码。这种方法的优势之一
是如果有人想要更改消息，他只需要在一个地方查看
。
In my Pascal program I gathered all messages in one function. All
messages have their own number. One of the advantages of this method
is that if someone wants to change a messages, he only has to look in
one place.
如果你想知道什么是消息，它就永远不会出现在你看的
的地方...... />
非常长的宏名称可以提供帮助。
错误（27）;
和某处
{27，通信链接丢失}
因此更好的宏可能是
错误（LINK_LOST）;
我参与了一个错误宏看起来像这样的项目
错误（SYSTEM_ERROR_COMMUNICATIONS_LINK_IS_LOST）;
and if you want to know what a message is it''s never in the
place where you look...
Very long macro names can help.
error (27);
and somewhere is
{27, "communications link is lost"}
hence a better macro might be
error (LINK_LOST);
I worked on a project where the error macros looked like this
error (SYSTEM_ERROR_COMMUNICATIONS_LINK_IS_LOST);
或者在这种情况下，甚至更好，使用枚举。
Richard
Or in this case, even better, use an enum.
Richard
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