write a recursive version of function which returns the position of the first appearance of a specified digit in a positive number.count from right and starts from 1.
i.e:input 1234 and 3,output 2,input 123241 and 2,output 3.
以上都不难，可以实现，但是他又规定如果那个digit不在那个数中的话，输出0，例如：input 234 and 1,output 0.这可把我难倒了。
注意：该程序要用recursive version写。
请高手指教，很急，谢谢！

int fun (int n, int d)
{
int n;

if (n <= 9) return (n == d) ? 1 : 0;
if (n%9 == d) return 1;
tmp = fun(n/10, d);
return tmp ? (tmp+1) ? 0;
}

?好像不能当双目运算符吧……

if (n%9 == d) return 1;
should be
if (n%10 == d) return 1;

但是比我的程序短了一半，厉害厉害！

int rposition(int number,int digit)
{
int static i=0;
int temp;
if(number%10==digit)
{
i=0;
return 1;
}
else if(number==0)
{
i++;
return 1+rposition(number/10,digit);
}
}

#include "stdio.h"
int position(unsigned long number, int digit)
{
int i=0, remainder;
unsigned long number0=number;
while (number0!=0)
{
number0=number/10;
remainder=number-number0*10;
number=number0;
if (remainder==digit)
{
i++;break;
}
else
i++;
}
if (number0==0)
return(0);

else
return(i);
}
void main()
{
unsigned long number=324263555;
int digit=8;
printf("(%ld,%d)=%d\n",number,digit,position(number,digit));
}