问题描述
我有一个这样的字符串:
I have a string like this:
a;b;c;d;e
f;g;h;i;j
1;2;3;4;5
我想逐个元素解析它.我使用了嵌套的 strtok 函数,但它只是拆分了第一行并使令牌指针为空.我怎样才能克服这个?代码如下:
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, "
");
while(token != NULL && *token != EOF)
{
char a[128], b[128];
strcpy(a,token);
strcpy(b,a);
printf("a:%s
",a);
char *token2 = strtok(a,";");
while(token2 != NULL)
{
printf("token2 %s
",token2);
token2 = strtok(NULL,";");
}
strcpy(token,b);
token = strtok(NULL, "
");
if(token == NULL)
{
printf("its null");
}
}
输出:
token 2 a
token 2 b
token 2 c
token 2 d
token 2 e
推荐答案
你不能用 strtok()
;使用 strtok_r()
来自 POSIX 或 strtok_s()
(如果可用),或重新考虑您的设计.
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);
这两个函数可以互换.
注意一个变种 strtok_s()
在 C11 的可选部分中指定(ISO/IEC 9899:2011 中的附录 K).但是,除了 Microsoft 之外,很少有供应商实现了该部分标准中的接口.附件 K 中指定的 strtok_s()
版本与 Microsoft 的 strtok_s()
具有不同的接口——类似的问题困扰着附件 K 中指定的许多其他函数.>
使用 strtok_r()
Note that a variant strtok_s()
is specified in an optional part of C11 (Annex K in ISO/IEC 9899:2011). However, few suppliers other than Microsoft have implemented the interfaces in that section of the standard. The version of strtok_s()
specified in Annex K has a different interface from Microsoft's strtok_s()
— similar problems bedevil a number of the other functions specified in Annex K.
#include <string.h>
#include <stdio.h>
int main(void)
{
char str[] = "a;b;c;d;e
f;g;h;i;j
1;2;3;4;5
";
char *end_str;
char *token = strtok_r(str, "
", &end_str);
while (token != NULL)
{
char *end_token;
printf("a = %s
", token);
char *token2 = strtok_r(token, ";", &end_token);
while (token2 != NULL)
{
printf("b = %s
", token2);
token2 = strtok_r(NULL, ";", &end_token);
}
token = strtok_r(NULL, "
", &end_str);
}
return 0;
}
结果
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
没有 strtok_r()
这适用于上下文 - 前提是数据以换行符结尾.
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
f;g;h;i;j
1;2;3;4;5
";
char *string = data;
char *token = strchr(string, '
');
while (token != NULL)
{
/* String to scan is in string..token */
*token++ = ' ';
printf("a = %s
", string);
char *token2 = strtok(string, ";");
while (token2 != NULL)
{
printf("b = %s
", token2);
token2 = strtok(NULL, ";");
}
string = token;
token = strchr(string, '
');
}
return 0;
}
输出
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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