本问题在DEVC++中编译
希望用递归法求解n阶勒让德多项式的值
#include<iostream>
using namespace std;
int main()
{
int f(int,int);
int n,x;
float y;
cin>>n>>x;
y = f(n,x);
cout<<y<<endl;
return 0;
}
float f(int n,int x)
{
float t;
if (n == 0) t = 1;
if (n == 1) t = x;
if (n >= 2) t = ((2*n-1)*x-f(n-1)-(n-1)*f(n-2))/2;
return (t);
}
编译上述程序会出现如下报错
18 35 [Error] too few arguments to function 'float f(int, int)'
求解原因