strtok（）帮助 [英] strtok ( ) help

问题描述

我正在使用strtok（）捕获输入行。在我打电话给

" splitCommand"后，我再次调用strtok（）来获取下一行。 Strtok（）

返回NULL（但文件中有更多...）。在'splitCommands''进入图片之前，这并没有发生

。问题是在

splitCommands（）以某种方式修改指针，但我必须调用

函数。有没有办法复制它或什么？


/ *这里是我的代码* /


char * lineOfScript; < br $>
const char * delim =" \ n" ;;


lineOfScript = strstr（scriptFileBuffer，"：preprocess："）; //找到开始

在脚本中放置

lineOfScript = strtok（lineOfScript，delim）; //跳过一行

lineOfScript = strtok（NULL，delim）; //得到下一行...现在我有一个

命令

splitCommand（lineOfScript）; //这可能是我的指针得到的地方

搞砸了......

lineOfScript = strtok（NULL，delim）; //得到下一行，但是strtok返回

NULL


//将命令拆分成单独的单词。

//存储全局数组中的单词

int splitCommand（char * command）{

const char * delimeters =" " ;;

int i = 0;

g_UserCommands [0] = strtok（command，""）;

while（g_UserCommands [i]！= NULL&& i< 5）{

// printf（"％s"，g_UserCommands [i]）; //用于调试...

i + = 1;

g_UserCommands [i] = strtok（NULL，""）;

}

i = 0;

返回1;

}

I''m using strtok( ) to capture lines of input. After I call
"splitCommand", I call strtok( ) again to get the next line. Strtok( )
returns NULL (but there is more in the file...). That didn''t happen
before ''splitCommands'' entered the picture. The problem is in
splitCommands( ) somehow modifying the pointer, but I HAVE to call that
function. Is there a way to make a copy of it or something ?

/* HERE IS MY CODE */

char * lineOfScript;
const char * delim = "\n";

lineOfScript = strstr(scriptFileBuffer,":preprocess:"); //find starting
place in script
lineOfScript = strtok(lineOfScript,delim); //skip a line
lineOfScript = strtok(NULL,delim); //get next line... now I have a
command
splitCommand(lineOfScript); //this is probably where my pointer gets
messed up...
lineOfScript = strtok(NULL,delim); //get next line, but strtok returns
NULL

//Split up command into seperate words.
//Store words in global array
int splitCommand(char * command){
const char * delimeters = " ";
int i = 0;
g_UserCommands[0] = strtok(command, " ");
while(g_UserCommands[i] != NULL && i < 5){
//printf("%s", g_UserCommands[i]);//for debugging...
i+=1;
g_UserCommands[i] = strtok(NULL, " ");
}
i=0;
return 1;
}

推荐答案

ern写道：

我正在使用strtok（）来捕获输入行。在我调用
splitCommand后，我再次调用strtok（）来获取下一行。 Strtok（）
返回NULL（但文件中有更多...）。在splitCommands进入图片之前，这并没有发生。问题在于splitCommands（）以某种方式修改指针，但我必须调用那个
函数。有没有办法复制它或什么？

/ *这里是我的代码* /

char * lineOfScript;
const char * delim = " \\\
";

lineOfScript = strstr（scriptFileBuffer，"：preprocess："）; //在脚本中找到起始位置
lineOfScript = strtok（lineOfScript，delim）; //跳过一行
lineOfScript = strtok（NULL，delim）; //得到下一行...现在我有一个
命令
splitCommand（lineOfScript）; //这可能是我的指针弄乱的地方......
lineOfScript = strtok（NULL，delim）; //获取下一行，但是strtok返回
NULL

//将命令分成单独的单词。
//在全局数组中存储单词
int splitCommand（char * command）{
const char * delimeters =" " ;;
int i = 0;
g_UserCommands [0] = strtok（command，""）;
while（g_UserCommands [i]！= NULL&& i< ; 5）{
// printf（"％s"，g_UserCommands [i]）; //用于调试...
i + = 1;
g_UserCommands [i] = strtok（ NULL，"）;
}
i = 0;
返回1;
}

I''m using strtok( ) to capture lines of input. After I call
"splitCommand", I call strtok( ) again to get the next line. Strtok( )
returns NULL (but there is more in the file...). That didn''t happen
before ''splitCommands'' entered the picture. The problem is in
splitCommands( ) somehow modifying the pointer, but I HAVE to call that
function. Is there a way to make a copy of it or something ?

/* HERE IS MY CODE */

char * lineOfScript;
const char * delim = "\n";

lineOfScript = strstr(scriptFileBuffer,":preprocess:"); //find starting
place in script
lineOfScript = strtok(lineOfScript,delim); //skip a line
lineOfScript = strtok(NULL,delim); //get next line... now I have a
command
splitCommand(lineOfScript); //this is probably where my pointer gets
messed up...
lineOfScript = strtok(NULL,delim); //get next line, but strtok returns
NULL

//Split up command into seperate words.
//Store words in global array
int splitCommand(char * command){
const char * delimeters = " ";
int i = 0;
g_UserCommands[0] = strtok(command, " ");
while(g_UserCommands[i] != NULL && i < 5){
//printf("%s", g_UserCommands[i]);//for debugging...
i+=1;
g_UserCommands[i] = strtok(NULL, " ");
}
i=0;
return 1;
}

In splitCommand（）你开始一个新的strtok（）" session"，所以当

你返回并接下来调用它时，实际上它会从''
剩下的地方搜索''命令''。这是
splitCommand（）的局部变量，并且没有告诉我们曾经有什么内容

splitCommand（）返回（它实际上/不是/还有）。


C按值传递/所有/参数（甚至指针），即副本

用于被调用函数。函数返回后，

对象不再存在。试图访问它会调用Undefined

行为。


干杯


弗拉基米尔

-

我的电子邮件地址是真的，我看了。

In splitCommand() you start a new strtok() "session", so when
you return and invoke it next it actually searches from where it
left off in ''command''. This was a local variable to
splitCommand(), and there''s no telling what''s there once
splitCommand() returns (it actually /is not/ there anymore).

C passes /all/ parameters by value (even pointers), i.e. a copy
is made for the called function. After the function returns the
object ceases to exist. Trying to access it invokes Undefined
Behaviour.

Cheers

Vladimir
--
My e-mail address is real, and I read it.

Vladimir S. Oka写道：

Vladimir S. Oka wrote:

ern写道：

我正在使用strtok（）来捕获输入行。在我调用
splitCommand后，我再次调用strtok（）来获取下一行。 Strtok（）
返回NULL（但文件中有更多...）。在splitCommands进入图片之前，这并没有发生。问题在于splitCommands（）以某种方式修改指针，但我必须调用那个
函数。有没有办法复制它或什么？

/ *这里是我的代码* /

char * lineOfScript;
const char * delim = " \\\
";

lineOfScript = strstr（scriptFileBuffer，"：preprocess："）; //在脚本中找到起始位置
lineOfScript = strtok（lineOfScript，delim）; //跳过一行
lineOfScript = strtok（NULL，delim）; //得到下一行...现在我有一个
命令
splitCommand（lineOfScript）; //这可能是我的指针弄乱的地方......
lineOfScript = strtok（NULL，delim）; //获取下一行，但是strtok返回
NULL

//将命令分成单独的单词。
//在全局数组中存储单词
int splitCommand（char * command）{
const char * delimeters =" " ;;
int i = 0;
g_UserCommands [0] = strtok（command，""）;
while（g_UserCommands [i]！= NULL&& i< ; 5）{
// printf（"％s"，g_UserCommands [i]）; //用于调试...
i + = 1;
g_UserCommands [i] = strtok（ NULL，"）;
}
i = 0;
返回1;
}

I''m using strtok( ) to capture lines of input. After I call
"splitCommand", I call strtok( ) again to get the next line. Strtok( )
returns NULL (but there is more in the file...). That didn''t happen
before ''splitCommands'' entered the picture. The problem is in
splitCommands( ) somehow modifying the pointer, but I HAVE to call that
function. Is there a way to make a copy of it or something ?

/* HERE IS MY CODE */

char * lineOfScript;
const char * delim = "\n";

lineOfScript = strstr(scriptFileBuffer,":preprocess:"); //find starting
place in script
lineOfScript = strtok(lineOfScript,delim); //skip a line
lineOfScript = strtok(NULL,delim); //get next line... now I have a
command
splitCommand(lineOfScript); //this is probably where my pointer gets
messed up...
lineOfScript = strtok(NULL,delim); //get next line, but strtok returns
NULL

//Split up command into seperate words.
//Store words in global array
int splitCommand(char * command){
const char * delimeters = " ";
int i = 0;
g_UserCommands[0] = strtok(command, " ");
while(g_UserCommands[i] != NULL && i < 5){
//printf("%s", g_UserCommands[i]);//for debugging...
i+=1;
g_UserCommands[i] = strtok(NULL, " ");
}
i=0;
return 1;
}

在splitCommand（）中你开始一个新的strtok（）会话，所以当你返回
并接下来调用它时，它实际上是从它停止的位置搜索
''command''。这是splitCommand（）的局部变量，并且没有
告诉splitCommand（）返回时它是什么（它实际上/不是/
那里）。

In splitCommand() you start a new strtok() "session", so when you return
and invoke it next it actually searches from where it left off in
''command''. This was a local variable to splitCommand(), and there''s no
telling what''s there once splitCommand() returns (it actually /is not/
there anymore).

有关strtok（）如何工作的详细说明，请参阅系统帮助/手册。 C标准在7.21.5.8中对此进行了描述。

我很久以前就用它来禁止释义。


干杯


弗拉基米尔


-

我的电子邮件地址是真的，我看了。

Consult your system help/manuals for a detailed description of
how strtok() works. The C Standard describes it in 7.21.5.8.
I''ve last used it too long ago to dare paraphrase either.

Cheers

Vladimir

--
My e-mail address is real, and I read it.

ern写道：

我正在使用strtok（）来捕获输入行。在我调用
splitCommand后，我再次调用strtok（）来获取下一行。 Strtok（
）返回NULL（但文件中有更多...）。在splitCommands进入图片之前，这并没有发生。问题在于splitCommands（）以某种方式修改指针，但我必须调用
该函数。有没有办法复制它或什么？

/ *这里是我的代码* /

char * lineOfScript;
const char * delim = " \\\
";

lineOfScript = strstr（scriptFileBuffer，"：preprocess："）; //找到
脚本中的起始位置
lineOfScript = strtok（lineOfScript，delim）; //跳过一行
lineOfScript = strtok（NULL，delim）; //得到下一行...现在我有一个
命令
splitCommand（lineOfScript）; //这可能是我的指针弄乱的地方......
lineOfScript = strtok（NULL，delim）; //获取下一行，但是strtok
返回NULL

//将命令分成单独的单词。
//在全局数组中存储单词
int splitCommand（char * command）{
const char * delimeters =" " ;;
int i = 0;
g_UserCommands [0] = strtok（command，""）;
while（g_UserCommands [i]！= NULL&& i< ; 5）{
// printf（"％s"，g_UserCommands [i]）; //用于调试...
i + = 1;
g_UserCommands [i] = strtok（ NULL，"）;
}
i = 0;
返回1;
}

I''m using strtok( ) to capture lines of input. After I call
"splitCommand", I call strtok( ) again to get the next line. Strtok(
) returns NULL (but there is more in the file...). That didn''t happen
before ''splitCommands'' entered the picture. The problem is in
splitCommands( ) somehow modifying the pointer, but I HAVE to call
that function. Is there a way to make a copy of it or something ?

/* HERE IS MY CODE */

char * lineOfScript;
const char * delim = "\n";

lineOfScript = strstr(scriptFileBuffer,":preprocess:"); //find
starting place in script
lineOfScript = strtok(lineOfScript,delim); //skip a line
lineOfScript = strtok(NULL,delim); //get next line... now I have a
command
splitCommand(lineOfScript); //this is probably where my pointer gets
messed up...
lineOfScript = strtok(NULL,delim); //get next line, but strtok
returns NULL

//Split up command into seperate words.
//Store words in global array
int splitCommand(char * command){
const char * delimeters = " ";
int i = 0;
g_UserCommands[0] = strtok(command, " ");
while(g_UserCommands[i] != NULL && i < 5){
//printf("%s", g_UserCommands[i]);//for debugging...
i+=1;
g_UserCommands[i] = strtok(NULL, " ");
}
i=0;
return 1;
}

strtok （lineOfScript，delim）;


初次调用后，strtok（）必须记住你要求的数据

to解析。要做到这一点，它会复制任何指向

的lineOfScript，并将其存储在一些内部缓冲区中[它维护，并且你不能直接使用
访问]。

strtok（NULL，delim）;


当你再次调用它时，将NULL作为第一个参数传递，它只是继续

从以前停止的地方进行解析 - 即，它继续解析内部缓冲区的内部缓冲区，由最初指向的lineOfScript设置。

strtok（" BOO，delim）;


现在，如果再次使用非NULL初始参数调用它，它会忘记以前存储的任何

数据/正在处理并重置其内部缓冲区，以便将您刚刚传入的任何数据复制到一个BOO的副本中。在这种情况下。所以，

无论你还没有解析 - 最初由lineOfScript引用 -

现在已经丢失并被遗忘。

底线，你不能做你想做的事情......


p = strtok（p1，p2）;


while（strtok（NULL，p2））

{

p3 = strtok（p4，...）;
< br $> b $ b ...

}

-

============= ===================================== =============

试图减少信号上的不必要的噪音......


免责声明：

我提供的任何评论/代码可能= = = 100％便携，也不是

语义正确[读 - ''不是100％迂腐正确'']。

我不太关心这一点，而且我认为这是与大多数访客相同的

。但是，请放心，任何

''必要''（？）更正几乎肯定会出现v.soon

[读 - ''加上他们认为合适的噪音，一个学生将会很快就会接受

。'


警告：请务必阅读标签。没有旁边的细节

过滤器提供。远离儿童。不要点燃。

====================================== ============ =============

strtok(lineOfScript,delim);

After the initial call, strtok() has to ''remember'' the data you''ve asked it
to parse. To do that here, it makes a copy of whatever lineOfScript pointed
to, and stores it in some internal buffer [that it maintains, and you can''t
directly access].
strtok(NULL,delim);

When you call it again, passing NULL as the first param, it simply continues
parsing from wherever it previously left off - i.e., it continues to parse
its internal buffer as set by whatever lineOfScript originally pointed to.
strtok("BOO",delim);

Now, if you call it again with a non-NULL initial param, it forgets whatever
data it was previously storing/working on and resets its internal buffer to
whatever data you''ve just passed in - a copy of "BOO" in this case. So,
whatever you didn''t yet parse - that was originally ref''ed by lineOfScript -
is now lost and forgotten.

Bottom line, you can''t do what you''re trying to do with ...

p = strtok(p1, p2);

while(strtok(NULL, p2))
{
p3 = strtok(p4, ...);

...
}
--
================================================== =============
In an attempt to reduce ''unwanted noise'' on the ''signal'' ...

Disclaimer:

Any comment/code I contribute might =NOT= be 100% portable, nor
semantically correct [read - ''not 100% pedantically correct''].
I don''t care too much about that though, and I reckon it''s the
same with most ''visitors'' here. However, rest assured that any
''essential'' (?) corrections WILL almost certainly appear v.soon
[read - ''to add noise as they see fit, a pedant will be along
shortly''].

WARNINGS: Always read the label. No beside-the-point minutiae
filter supplied. Keep away from children. Do not ignite.
================================================== =============

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