C中功能编程的意外结果

Question

在C编程中尝试函数风格时,我尝试将以下Haskell代码转换为C.

f (0, 0, 0, 1) = 0
f (0, 0, 1, 0) = f (0, 0, 0, 1) + 1
f (0, 1, 0, 0) = f (0, 0, 1, 1) + 1
f (1, 0, 0, 0) = f (0, 1, 1, 1) + 1
f (a, b, c, d) = (p + q + r + s) / (a + b + c + d)
    where
    p
        | a > 0 = a * f (a - 1, b + 1, c + 1, d + 1)
        | otherwise = 0
    q
        | b > 0 = b * f (a, b - 1, c + 1, d + 1)
        | otherwise = 0
    r
        | c > 0 = c * f (a, b, c - 1, d + 1)
        | otherwise = 0
    s
        | d > 0 = d * f (a, b, c, d - 1)
        | otherwise = 0

main = print (f (1, 1, 1, 1))

至

#include <stdio.h>
#include <stdlib.h>

#define int const int
#define double const double

double f(int a, int b, int c, int d)
{
    if (a == 0 && b == 0 && c == 0 && d == 1)
    {
        return 0.0;
    }
    else if (a == 0 && b == 0 && c == 1 && d == 0)
    {
        return f(0, 0, 0, 1) + 1.0;
    }
    else if (a == 0 && b == 1 && c == 0 && d == 0)
    {
        return f(0, 0, 1, 1) + 1.0;
    }
    else if (a == 1 && b == 0 && c == 0 && d == 0)
    {
        return f(0, 1, 1, 1) + 1.0;
    }
    else
    {
        int p = a > 0 ? a * f(a - 1, b + 1, c + 1, d + 1) : 0;
        int q = b > 0 ? b * f(a, b - 1, c + 1, d + 1) : 0;
        int r = c > 0 ? c * f(a, b, c - 1, d + 1) : 0;
        int s = d > 0 ? d * f(a, b, c, d - 1) : 0;
        return (double)(p + q + r + s) / (double)(a + b + c + d);
    }
}

int main(void)
{
    printf("%f\n", f(1, 1, 1, 1));
    return EXIT_SUCCESS;
}

我期望完全相同的行为,但C程序总是输出0.0.随着f(0, 0, 1, 1)他们两个输出0.5,但每当数得到一个垃圾越大,C版根本不起作用.出了什么问题？

Answer 1

int p = a > 0 ? a * f(a - 1, b + 1, c + 1, d + 1) : 0;
int q = b > 0 ? b * f(a, b - 1, c + 1, d + 1) : 0;
int r = c > 0 ? c * f(a, b, c - 1, d + 1) : 0;
int s = d > 0 ? d * f(a, b, c, d - 1) : 0;

这里递归调用的结果f在存储在int变量中时被截断为整数.因此,例如,如果a是1并且f(a-1, b+1, c+1, c+1)是0.5,p则将为0而不是0.5因为您无法存储0.5在int中.

在Haskell代码中,所有变量都是双精度(或者说是小数),因此您应该在C版本中执行相同操作并将所有变量和参数声明为double.