【c实现，vc6调试通过】给出一字符串指针，计算出字符串指针中单词数

 #include <stdio.h>

 /*

     给出一字符串指针，计算出字符串指针中单词数，

     单词不包括'.',',',';','?','_','"'，由0-9数字或26个字母组成

     by zww @ 2013.09.26

     vc 6.0编译通过

 */

 unsigned int get_string_word_cnt(const char* pStr)

 {

     unsigned int iCnt= ;

     unsigned int iCharCnt = ;

     const char* p = pStr;

     if (NULL == pStr)

     {

         return ;

     }

     while ('\0' != *p)

     {

         if (('.' == *p)

             || (',' == *p)

             || (';' == *p)

             || ('?' == *p)

             || (' ' == *p)

             || ('_' == *p)

             || ('"' == *p))

         {

             printf("[%c]",*p);

             iCharCnt=;

         }

         else if (!((*p >= 'a'&& *p<='z')

             || (*p >='A' && *p<='Z')

             || (*p >= '' && *p <='')))

         {

             printf("[%c]",*p);

             iCharCnt = ;

         }

         else if ( == iCharCnt)

         {

             iCharCnt++;

             iCnt++;

         }

         p++;

     }

     return iCnt;

 }

 int main()

 {

     char * str1 = "hello world.1...2";

     char * str2 = "hello world.sd,fg!@!$!$!!!!!!!!!!!!hj.";

     char * str3 = "hello world.fdfg)(*&&&*(hjdfh?>><>,...";

     char * str4 = ".hello world.12";

     int i = ;

     i = get_string_word_cnt(str1);

     printf(" %d\r\n",i);

     i = get_string_word_cnt(str2);

     printf(" %d\r\n",i);

     i = get_string_word_cnt(str3);

     printf(" %d\r\n",i);

     i = get_string_word_cnt(str4);

     printf(" %d\r\n",i);

     return ;

 }

 /**

     显示结果如下：

 [ ][.][.][.][.]4

 [ ][.][,][!][@][!][$][!][$][!][!][!][!][!][!][!][!][!][!][!][!][.]5

 [ ][.][)][(][*][&][&][&][*][(][?][>][>][<][>][,][.][.][.]4

 [.][ ][.]3

 Press any key to continue

 **/