将字符串解析为数组 [英] parsing string into an array
问题描述
我想将字符串解析为数组。我在网上找到了以下代码来解析字符串并将其打印出来。
结果
正是我想要的：
char * pch;
pch = strtok（缓冲区，"" ）;
while（pch！= NULL）
{
printf（"％s\ n，pch）;
pch = strtok（NULL，&,;，。"）;
}
要将输出存储到字符串数组中，我创建了一个二维的
数组并将输出复制到数组：
char输出[5] [55];
char * pch;
int i = 0;
pch = strtok（str，""）;
while（pch ！= NULL）
{
strcpy（output [i]，pch）;
printf（"％s \ n"， pch）;
printf（" output [％d] is％s。\ n" ;; i，output [i ++]）;
pch = strtok（NULL ，，，;
}
这不起作用。输出数组中的内容与
输出（屏幕上）完全不同。应该怎么做？
I would like to parse a string into an array. I found on the net
the following codes which parse a string and print it. The result
is exactly what I want:
char * pch;
pch = strtok (buffer," ");
while (pch != NULL)
{
printf ("%s\n",pch);
pch = strtok (NULL, " ,.");
}
To store the output to a string array, I created a two-dimensional
array and copied the output to the array:
char output[5][55];
char * pch;
int i=0;
pch = strtok (str," ");
while (pch != NULL)
{
strcpy(output[i],pch);
printf ("%s\n",pch);
printf("output[%d] is %s.\n";i,output[i++]);
pch = strtok (NULL, " ,.");
}
This doesn''t work. What''s in the output array is nothing like the
output (on screen). How should it be done?
推荐答案
在文章< lb ****************** ***@twister.southeast.rr.com> ，
John Smith< js **** @ company.com>写道：
In article <lb*********************@twister.southeast.rr.com> ,
John Smith <js****@company.com> wrote:
printf（" output [％d] is％s。\ n" ;; i，output [i ++]）;
printf("output[%d] is %s.\n";i,output[i++]);
如果我没记错，这会使用未定义的行为。
后期增量可以在
序列点之间的任何时间完成，所以它不是定义对i的第一个
引用是否在
增量之前或之后取值。
它也是一种语法错误将分号放在中间
语句中。该行根本不应该编译。
-
''ignorandus（拉丁语）：值得不知道''
- 自我指涉期刊
If I recall correctly, this uses undefined behaviour.
The post increment may be done at any time between the
sequence points, so it is not defined whether the first
reference to i has its value taken before or after the
increment.
It also is a syntax error to have the semicolon in the middle
of the statement. That line should not compile at all.
--
''ignorandus (Latin): "deserving not to be known"''
-- Journal of Self-Referentialism
Walter Roberson写道：
Walter Roberson wrote:
文章< lb ********** ***********@twister.southeast.rr.com> ，
John Smith< js **** @ company.com>写道：
In article <lb*********************@twister.southeast.rr.com> ,
John Smith <js****@company.com> wrote:
printf（" output [％d] is％s。\ n" ;; i，output [i ++]）;
printf("output[%d] is %s.\n";i,output[i++]);
如果我没记错，这会使用未定义的行为。
后期增量可能在
序列点之间的任何时间完成，所以没有定义是否第一个
对...的引用在
增量之前或之后取值。
在语句的中间部分使用分号也是语法错误。该行根本不应该编译。
If I recall correctly, this uses undefined behaviour.
The post increment may be done at any time between the
sequence points, so it is not defined whether the first
reference to i has its value taken before or after the
increment.
It also is a syntax error to have the semicolon in the middle
of the statement. That line should not compile at all.
感谢您的回复。那个分号是一个错字。
实际上，我也尝试了以下（单独的i ++）并且
也没用。
char输出[5] [55];
char * pch;
int i = 0;
pch = strtok（str，""）;
while（pch！= NULL）
{
strcpy（output [i] ，pch）;
printf（"％s \ nn"，pch）;
printf（" output [％d] is％s。\ n" ;，i，输出[i]）;
i ++;
pch = strtok（NULL，"，。"）;
}
Thanks for the reply. That semicolon was a typo.
Actually, I had also tried the following (a separate i++) and it
did not work either.
char output[5][55];
char * pch;
int i=0;
pch = strtok (str," ");
while (pch != NULL)
{
strcpy(output[i],pch);
printf ("%s\n",pch);
printf("output[%d] is %s.\n",i,output[i]);
i++;
pch = strtok (NULL, " ,.");
}
John Smith写道：
John Smith wrote:
其实我也试过以下（单独的i ++）它也没有用。
char输出[5] [55];
char * pch;
int i = 0;
pch = strtok（str，""）;
while（pch！= NULL）
{
strcpy（output [i]，pch）;
printf（"％s \ n"，pch）;
printf（" output [％d] i s％s。\ n"，i，output [i]）;
i ++;
pch = strtok（NULL，" ，。;
}
Actually, I had also tried the following (a separate i++) and it
did not work either.
char output[5][55];
char * pch;
int i=0;
pch = strtok (str," ");
while (pch != NULL)
{
strcpy(output[i],pch);
printf ("%s\n",pch);
printf("output[%d] is %s.\n",i,output[i]);
i++;
pch = strtok (NULL, " ,.");
}
这里提供的代码似乎没问题。你是什么
的意思是什么都不起作用。
-
Al Bowers
坦帕，Fl USA
mailto： xa******@myrapidsys.com （删除x发送电子邮件）
http://www.geocities.com/abowers822 /
The code provided here seems to be ok. What do you
mean by "did not work either".
--
Al Bowers
Tampa, Fl USA
mailto: xa******@myrapidsys.com (remove the x to send email)
http://www.geocities.com/abowers822/
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