【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= 0;
    unsigned int iCharCnt = 0;
    const char* p = pStr;

    if (NULL == pStr)
    {
        return 0;
    }

    while ('\0' != *p)
    {
        if (('.' == *p)
            || (',' == *p)
            || (';' == *p)
            || ('?' == *p)
            || (' ' == *p)
            || ('_' == *p)
            || ('"' == *p))
        {
            printf("[%c]",*p);
            iCharCnt=0;
        }  
        else if (!((*p >= 'a'&& *p<='z')
            || (*p >='A' && *p<='Z')
            || (*p >= '0' && *p <='9')))
        {
            printf("[%c]",*p);
            iCharCnt = 0;
        }
        else if (0 == 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 = 0;
    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 0;
}

/**
    显示结果如下：
[ ][.][.][.][.]4
[ ][.][,][!][@][!][$][!][$][!][!][!][!][!][!][!][!][!][!][!][!][.]5
[ ][.][)][(][*][&][&][&][*][(][?][>][>][<][>][,][.][.][.]4
[.][ ][.]3
Press any key to continue
**/