用递归求和 哪位高手指点指点吧！！

#include<stdio.h>
main()
{long n,m,a,b,p;
   scanf("%d %d",&n,&m);
      p=sum(n,m);
        printf("The sum from %d to %d is %d\n",n,m,p);
}
sum(long n,long m)
{
   if(n==m) return(n);
     else
      return(n+sum(n-1,m));
}
错哪了呢？
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#include <stdio.h>
long int sum(long,long);
int main(void)
{
    long int n,m;
    scanf("%d %d",&n,&m);
    printf("The sum from %d to %d is %d\n",n,m,sum(n,m));
    return 0;
}
long int sum(long n,long m)
{
    if(n==m)
        return m;
    else
        return m+sum(n,m-1);
}
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高手，还是过不去呀
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过不去，你至少把错误说一下吧
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不知道程序具体要实现什么目的,我的理解就是求两个输入数字的和,根据我自己的理解,把程序改了下....在VC下通过了:

#include <stdio.h>
long int sum(long,long);
int main(void)
{
    long int n,m;
    scanf("%d %d",&n,&m);
    printf("The sum from %d to %d is %d\n",n,m,sum(n,m));
    return 0;
}
long int sum(long n,long m)
{
    long temp;
    if(n<m){temp=n;n=m;m=temp;}
    if(n==m)
        return 2*m;
    else
        return 1+sum(n-1,m);
}
----------------解决方案--------------------------------------------------------
#include<stdio.h>
void main()
{long n,m,a,b,p;
   clrscr();
   scanf("%ld %ld",&n,&m);
      p=sum(n,m);
    printf("The sum from %ld to %ld is %ld\n",n,m,p);
}
sum(long n,long m)
{
   if(n==m)
       return(m);
   else
       return(m+sum(n,m-1));
}
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LZ写的没错,只是少了函数声明, 还要注意的是 你输入的两个数,必须是大在前,小的在后, 因为你的递归中 就是那样计算的
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