有什么办法比较方便的直接替换宏吗？或者看一下这个程序替换宏后的样子？

Question

有没有什么工具或是编译器、方法之类的能自动替换宏定义？
是有一个程序为了尽可能节省字符数量用了大量宏定义，我想看懂它，但全是宏实在看不明白

#define R { return
#define P P (
#define L L (
#define T S (v, y, c,
#define C ),
#define X x)
#define F );}

int r, a;
P y, X
   R y - ~y << x;
}
Z (X
   R r = x % 2 ? 0 : 1 + Z (x / 2 F
L X
   R x / 2 >> Z (x F
#define U = S(4,13,-4,
T  t)
{
   int
      f = L t C         
      x = r;
   R
         f - 2 ?
         f > 2 ?
         f - v ? t - (f > v) * c : y :
         P f, P T  L X  C 
                          S (v+2, t  U y C  c, Z (X )))
         :
         A (T  L X  C 
                T  Z (X ) F
}
A (y, X
   R L y) - 1
      ? 5 << P y, X 
      : S (4, x, 4, Z (r) F
#define B (x /= 2) % 2 && (
D (X 
{
   int
      f,
      d,
      c = 0,
      t = 7,
      u = 14;
   while (x && D (x - 1 C  B 1))
      d = L L D (X ) C
         f = L r C
         x = L r C
         c - r || (
            L u) || L r) - f ||
            B u = S (4, d, 4, r C 
                   t = A (t, d) C
            f / 2 & B  c = P d, c C 
                              t  U t C 
                              u  U u) )
             C
         c && B
            t = P
               ~u & 2 | B
                  u = 1 << P L c C  u) C 
               P L c C  t) C
            c = r  C
         u / 2 & B 
            c = P t, c C 
            u  U t C 
            t = 9 );
   R a = P P t, P u, P x, c)) C 
                                a F
}
main ()
   R D (D (D (D (D (99)))) F

PS：这是一个在假设无限大内存没有整形上限的情况下，输出一个巨大的数字的程序，这个数字非常大，远远超越想象。所以我想知道生成这个数字的D函数的工作原理

Answer 1

宏展开大概是这样子吧。标准C必须要有返回类型，我加上int为返回类型。
但是貌似逻辑有问题，运行出错！

#include<stdio.h>

int r, a;
int P(y, x)
{
    return y - ~y << x;
}

int Z(x)
{
    return r = x % 2 ? 0 : 1 + Z(x / 2);
}

int L(x)
{
    return x / 2 >> Z(x);
}


int S(v, y, c, t)
{
    int
        f = L(t),
        x = r;
    {
        return f - 2 ? f > 2 ? f - v ? t - (f > v) * c : y : P(f, P(S(v, y, c, L(x)), S(v + 2, t = S(4,13,-4, y), c, Z(x))))
            : A(S(v, y, c, L(x)),
                S(v, y, c, Z(x)));
    }
}

int A(y, x)
{
    return L(y) - 1
               ? 5 << P(y, x)
               : S(4, x, 4, Z(r));
}

int D(x)
{
    int
        f,
        d,
        c = 0,
        t = 7,
        u = 14;
    while (x && D(x - 1), (x /= 2) % 2 && ( 1))
      d = L ( L ( D (x) ) ),
         f = L ( r ),
         x = L ( r ),
         c - r || (
            L ( u) || L ( r) - f ||
            (x /= 2) % 2 && ( u = S (4, d, 4, r ), 
                   t = A (t, d) ),
            f / 2 & (x /= 2) % 2 && (  c = P ( d, c ), 
                              t  = S(4,13,-4, t ), 
                              u  = S(4,13,-4, u) )
             ),
         c && (x /= 2) % 2 && (
            t = P (
               ~u & 2 | (x /= 2) % 2 && (
                  u = 1 << P ( L ( c ),  u) ), 
               P ( L ( c ),  t) ),
            c = r  ),
         u / 2 & (x /= 2) % 2 && ( 
            c = P ( t, c ), 
            u  = S(4,13,-4, t ), 
            t = 9 );
    {
        return a = P(P(t, P(u, P(x, c))),
                     a);
    }
}
int main()
{
    return D(D(D(D(D(99)))));
}