【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
 **/