用C嵌套的strtok函数问题 [英] Nested strtok function problem in C
问题描述
我有一个这样的字符串:
A; B,C,D组;ê
F;克; H;我;Ĵ
1; 2; 3; 4; 5
和我想通过元素分析它的元素。我用嵌套的strtok的功能,但它只是分裂第一线,使空令牌指针。我该如何克服这个问题?这里是code:
令牌= strtok的(STR,\\ n);而(令牌= NULL&放大器;!&安培;!*记号= EOF)
{
烧焦一个[128],B [128];
的strcpy(一,令牌);
的strcpy(B,A);
的printf(A:%S \\ n,一);
字符* token2 = strtok的(A,);
而(token2!= NULL)
{
的printf(token2%S \\ n,token2);
token2 =的strtok(NULL,);
}
的strcpy(令牌,B);
令牌= strtok的(NULL,\\ n);
如果(令牌== NULL)
{
的printf(空的);
}
}
输出:
标记2
令牌2 B
令牌2 C
令牌2天
令牌2ê
您不能做到这一点与的strtok()
;使用 strtok_r()
一>从POSIX或 strtok_s()
从微软(如果有),或重新考虑你的设计。
的char * strtok_r(字符*限制S,为const char *限制九月,
焦炭**限制持续);
字符* strtok_s(字符* strToken,为const char * strDelimit,焦炭**背景);
这两个功能可以互换。虽然 strtok_s()
是C11的可选部分(附件K在ISO / IEC 9899:2011),比微软其他一些供应商已经实现了标准的那部分接口
随着strtok_r()
的#include<&string.h中GT;
#包括LT&;&stdio.h中GT;INT主要(无效)
{
烧焦海峡[] =一个; B,C,D组;ê\\ NF;克; H,I; J□\\ n1的; 2; 3; 4; 5 \\ N的;
字符* end_str;
char *之令牌= strtok_r(STR,\\ n,&安培; end_str); 而(令牌!= NULL)
{
字符* end_token;
的printf(A =%S \\ N标记);
字符* token2 = strtok_r(令牌,;,&放大器; end_token);
而(token2!= NULL)
{
的printf(B =%S \\ n,token2);
token2 = strtok_r(NULL,;,&放大器; end_token);
}
令牌= strtok_r(NULL,\\ n,&安培; end_str);
} 返回0;
}
结果
A = A,B,C,D组;ê
B = A
B =
B = C
B =ð
B =Ë
A = F;克; H;我;Ĵ
B =˚F
B =摹
B = ^ h
B =
B =Ĵ
一个= 1; 2; 3; 4; 5
B = 1
B = 2
B = 3
B = 4
B = 5
无strtok_r()
这工作在背景 - 提供的数据以换行符结束
。 的#include<&string.h中GT;
#包括LT&;&stdio.h中GT;INT主要(无效)
{
CHAR数据[] =一个; B,C,D组;ê\\ NF;克; H,I; J□\\ n1的; 2; 3; 4; 5 \\ N的;
字符*字符串=数据;
字符*记号=和strchr(字符串,'\\ n'); 而(令牌!= NULL)
{
/ *字符串进行扫描是string..token * /
*记号++ ='\\ 0';
的printf(A =%S \\ n,字符串);
字符* token2 = strtok的(字符串,);
而(token2!= NULL)
{
的printf(B =%S \\ n,token2);
token2 =的strtok(NULL,);
}
字符串=记号。
令牌=和strchr(字符串,'\\ n');
} 返回0;
}
输出
A = A,B,C,D组;ê
B = A
B =
B = C
B =ð
B =Ë
A = F;克; H;我;Ĵ
B =˚F
B =摹
B = ^ h
B =
B =Ĵ
一个= 1; 2; 3; 4; 5
B = 1
B = 2
B = 3
B = 4
B = 5
I have a string like this:
a;b;c;d;e
f;g;h;i;j
1;2;3;4;5
and i want to parse it element by element. I used nested strtok function but it just splits first line and makes null the token pointer. How can i overcome this? Here is the code:
token = strtok(str, "\n");
while(token != NULL && *token != EOF)
{
char a[128], b[128];
strcpy(a,token);
strcpy(b,a);
printf("a:%s\n",a);
char *token2 = strtok(a,";");
while(token2 != NULL)
{
printf("token2 %s\n",token2);
token2 = strtok(NULL,";");
}
strcpy(token,b);
token = strtok(NULL, "\n");
if(token == NULL)
{
printf("its null");
}
}
Output:
token 2 a
token 2 b
token 2 c
token 2 d
token 2 e
You cannot do that with strtok()
; use strtok_r()
from POSIX or strtok_s()
from Microsoft if they are available, or rethink your design.
char *strtok_r(char *restrict s, const char *restrict sep,
char **restrict lasts);
char *strtok_s(char *strToken, const char *strDelimit, char **context);
These two functions are interchangeable. Although strtok_s()
is an optional part of C11 (Annex K in ISO/IEC 9899:2011), few suppliers other than Microsoft have implemented the interfaces in that section of the standard.
With strtok_r()
#include <string.h>
#include <stdio.h>
int main(void)
{
char str[] = "a;b;c;d;e\nf;g;h;i;j\n1;2;3;4;5\n";
char *end_str;
char *token = strtok_r(str, "\n", &end_str);
while (token != NULL)
{
char *end_token;
printf("a = %s\n", token);
char *token2 = strtok_r(token, ";", &end_token);
while (token2 != NULL)
{
printf("b = %s\n", token2);
token2 = strtok_r(NULL, ";", &end_token);
}
token = strtok_r(NULL, "\n", &end_str);
}
return 0;
}
Results
a = a;b;c;d;e
b = a
b = b
b = c
b = d
b = e
a = f;g;h;i;j
b = f
b = g
b = h
b = i
b = j
a = 1;2;3;4;5
b = 1
b = 2
b = 3
b = 4
b = 5
Without strtok_r()
This works in context - provided that the data ends with a newline.
#include <string.h>
#include <stdio.h>
int main(void)
{
char data[] = "a;b;c;d;e\nf;g;h;i;j\n1;2;3;4;5\n";
char *string = data;
char *token = strchr(string, '\n');
while (token != NULL)
{
/* String to scan is in string..token */
*token++ = '\0';
printf("a = %s\n", string);
char *token2 = strtok(string, ";");
while (token2 != NULL)
{
printf("b = %s\n", token2);
token2 = strtok(NULL, ";");
}
string = token;
token = strchr(string, '\n');
}
return 0;
}
Output
a = a;b;c;d;e
b = a
b = b
b = c
b = d
b = e
a = f;g;h;i;j
b = f
b = g
b = h
b = i
b = j
a = 1;2;3;4;5
b = 1
b = 2
b = 3
b = 4
b = 5
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