实现`strtok`，其分隔符有多个字符

代码段：

char str[] = "String1::String2:String3:String4::String5"; char *deli = "::"; char *token = strtok(str,deli); while(token != NULL) { printf("Token= \"%s\"\n", token); token=strtok(NULL,deli); } 

上面的代码片段产生输出：

 Token="String1" Token="String2" Token="String3" Token="String4" Token="String5" 

但我希望输出为：

 Token="String1" Token="String2:String3:String4" Token="String5" 

我知道我没有得到预期的输出，因为strtok的第二个参数中的每个字符都被视为分隔符。

为了获得预期的输出，我编写了一个程序，它使用strstr （以及其他东西）将给定的字符串拆分为标记，以便获得预期的输出。 这是程序：

 #include  #include  #include  #include  int myStrtok(char* str,char* deli) { if(str==NULL || deli==NULL) return -1; int tokens=0; char *token; char *output=str; while((token=strstr(output,deli))!=NULL) { bool print=true; if(output != token) { printf("Token = \""); tokens++; print=false; } while(output != token) { putchar(*output); output++; } if(print==false) printf("\"\n"); output+=strlen(deli); } if(strlen(output)>0) { printf("Token = \"%s\"",output); tokens++; } printf("\n\n"); return tokens; } int main(void) { char str[]="One:1:Two::Three::::"; char *deli="::"; int retval; printf("Original string=\"%s\"\n\n",str); if((retval=myStrtok(str,deli))==-1) printf("The string or the delimeter is NULL\n"); else printf("Number of tokens=%d\n", retval); return(EXIT_SUCCESS); } 

上述程序产生预期的输出。

我想知道是否有更容易/更简单的方法来做到这一点。 有没有？

Answer 1

一个字符串分隔符函数，它使用strtok的原型并模仿其用法：

 char *strtokm(char *str, const char *delim) { static char *tok; static char *next; char *m; if (delim == NULL) return NULL; tok = (str) ? str : next; if (tok == NULL) return NULL; m = strstr(tok, delim); if (m) { next = m + strlen(delim); *m = '\0'; } else { next = NULL; } return tok; } 

Answer 2

如果你不关心与strtok相同的用法，我会用这个：

 // "String1::String2:String3:String4::String5" with delimiter "::" will produce // "String1\0\0String2:String3:String4\0\0String5" // And words should contain a pointer to the first S, the second S and the last S. char **strToWordArray(char *str, const char *delimiter) { char **words; int nwords = countWords(str, delimiter); //I let you decide how you want to do this words = malloc(sizeof(*words) * (nwords + 1)); int w = 0; int len = strlen(delimiter); words[w++] = str; while (*str != NULL) { if (strncmp(str, delimiter, len) == 0) { for (int i = 0; i < len; i++) { *(str++) = 0; } if (*str != 0) words[w++] = str; else str--; //Anticipate wrong str++ down; } str++; } words[w] = NULL; return words; } 

Answer 3

代码来自strsep https://code.woboq.org/userspace/glibc/string/strsep.c.html

 char *strsepm( char **stringp, const char *delim ) { char *begin, *end; begin = *stringp; if ( begin == NULL ) return NULL; /* Find the end of the token. */ end = strstr( begin , delim ); if ( end != NULL ) { /* Terminate the token and set *STRINGP past NUL character. */ *end = '\0'; end += strlen( delim ); *stringp = end; } else { /* No more delimiters; this is the last token. */ *stringp = NULL; } return begin; } int main( int argc , char *argv [] ) { char *token_ptr; char *token; const char *delimiter = "&&"; char buffer [ 256 ]; strcpy( buffer , " && Hello && Bernd && waht's && going && on &&"); token_ptr = buffer; while ( ( token = strsepm( &token_ptr , delimiter ) ) != NULL ) { printf( "\'%s\'\n" , token ); } } 

结果：

 ' ' ' Hello ' ' Bernd ' ' waht's ' ' going ' ' on ' ''