关于字符串查找的问题

求子串在父串中出现的次数

#include <stdio.h>
#include <string.h>
int main()
{

char * pstr1 = "hehelalalahehe";
char * psub = "hehe";
printf("%d\n",caculate(pstr1,psub));

}


int caculate(char *str,char *substr)
{

char * pstr = str;
char * psub = substr;
int times = 0;

while(*pstr != '\0')
{
if(*pstr == *psub)
{
pstr++;
psub++;
if(*psub == '\0')
{
psub = substr;
times++;
}
}
else
{
pstr++;
}

return times;
}



}


请问为何我在调试的时候，执行到if(*psub == '\0')
就直接return times了？问题出在哪里
（正确输出结果是2，但实际为0）






[解决办法]

    while(*pstr != '\0')
    {
        if(*pstr == *psub) //以你的为例 这里成立进入if
        {
           pstr++;
           psub++;
           if(*psub == '\0')  //此时*psub不为'\0' 跳过这个if
           { 
               psub = substr;
               times++;
           }
                                 //跳过上面的if就到这 
 
        }            //继续向下到这外面的if结束
        else         //if else 结构执行if，不用在再执行else
        {
            pstr++;
        }
                     //跳过else到这
        return times;  //然后你就return了times的值还是初值0
    }

//"hehehe"包含1个"hehe"的代码
int caculate(char *str, char *substr)
{
    char * pstr = str;
    char * psub = substr;
    int times = 0;

    while (*pstr != '\0')
    {
        if (*pstr == *psub)
        {
           pstr++;
           psub++;
           if(*psub == '\0')
           { 
               psub = substr;
               times++;
           }
        }
        else
        {
            psub = substr;//这里
            pstr++;
        }
    }
    
    return times;//这里
}

//"hehehe"包含2个"hehe"的代码
int caculate(const char *str, const char *substr)
{
    int times = 0;

    while (*str != '\0')
    {
        const char *pstr = str;
        const char *psub = substr;
        do
        {
            if (!*psub)
            {
                ++times;
            }
        } while (*pstr++ == *psub++);
        
        ++str;
    }
    
    return times;
}