C 混淆的科学计算器。请解释一下它是如何工作的

我试图理解下面的程序是如何工作的。它是一个命令行科学计算器。来源于。对于IOCCC条目来说，它看起来相当可读，但显然不是

#include <stdio.h>
#include <math.h>
#define clear 1;if(c>=11){c=0;sscanf(_,"%lf%c",&r,&c);while(*++_-c);}\
  else if(argc>=4&&!main(4-(*_++=='('),argv))_++;g:c+=
#define puts(d,e) return 0;}{double a;int b;char c=(argc<4?d)&15;\
  b=(*_%__LINE__+7)%9*(3*e>>c&1);c+=
#define I(d) (r);if(argc<4&&*#d==*_){a=r;r=usage?r*a:r+a;goto g;}c=c
#define return if(argc==2)printf("%f\n",r);return argc>=4+
#define usage main(4-__LINE__/26,argv)
#define calculator *_*(int)
#define l (r);r=--b?r:
#define _ argv[1]
#define x

double r;
int main(int argc,char** argv){
  if(argc<2){
    puts(
      usage: calculator 11/26+222/31
      +~~~~~~~~~~~~~~~~~~~~~~~~calculator-\
      !                          7.584,367 )
      +~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~+
      ! clear ! 0 ||l   -x  l   tan  I (/) |
      +~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~+
      ! 1 | 2 | 3 ||l  1/x  l   cos  I (*) |
      +~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~+
      ! 4 | 5 | 6 ||l  exp  l  sqrt  I (+) |
      +~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~+
      ! 7 | 8 | 9 ||l  sin  l   log  I (-) |
      +~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(0
    );
  }
  return 0;
}

---

有人能解释一下它是如何工作的吗？

它扩展成这样：

double r;
int
main (int argc, char **argv)
{
  if (argc < 2)
    {
      if (argc == 2)
    printf ("%f\n", r);
      return argc >= 4 + 0;
    }
  {
    double a;
    int b;
    char c = (argc < 4 ? main (4 - 21 / 26,
                   argv) : *argv[1] * (int) 11 / 26 + 222 / 31 +
          ~~~~~~~~~~~~~~~~~~~~~~~~*argv[1] * (int) -!7.584) & 15;
    b = (*argv[1] % 21 + 7) % 9 * (3 * 367 >> c & 1);
    c += +~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~+!1;
    if (c >= 11)
      {
    c = 0;
    sscanf (argv[1], "%lf%c", &r, &c);
    while (*++argv[1] - c);
      }
    else if (argc >= 4 && !main (4 - (*argv[1]++ == '('), argv))
      argv[1]++;
  g:c += !0 || (r);
    r = --b ? r : -(r);
    r = --b ? r : tan (r);
    if (argc < 4 && *"/" == *argv[1])
      {
    a = r;
    r = main (4 - 23 / 26, argv) ? r * a : r + a;
    goto g;
      }
    c = c | +~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~+!1 | 2 | 3 || (r);
    r = --b ? r : 1 / (r);
    r = --b ? r : cos (r);
    if (argc < 4 && *"*" == *argv[1])
      {
    a = r;
    r = main (4 - 25 / 26, argv) ? r * a : r + a;
    goto g;
      }
    c = c | +~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~+!4 | 5 | 6 || (r);
    r = --b ? r : exp (r);
    r = --b ? r : sqrt (r);
    if (argc < 4 && *"+" == *argv[1])
      {
    a = r;
    r = main (4 - 27 / 26, argv) ? r * a : r + a;
    goto g;
      }
    c = c | +~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~+!7 | 8 | 9 || (r);
    r = --b ? r : sin (r);
    r = --b ? r : log (r);
    if (argc < 4 && *"-" == *argv[1])
      {
    a = r;
    r = main (4 - 29 / 26, argv) ? r * a : r + a;
    goto g;
      }
    c = c | +~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(0);
  }
  if (argc == 2)
    printf ("%f\n", r);
  return argc >= 4 + 0;
}

它仍然不是完全可读的，但至少现在没有什么是对你隐藏的。你应该能够做你的简化，而不会被咬得那么厉害

---

@cmaster说：“出于同样的原因，我不打算进行完整的解释。

研究这段源代码的预处理形式，也许可以使用gcc-C-E obfusc.C | grep-v'^'| indent>obfusc。然后我查看了里面的̀obfusc.I肯定不会。IOCCC条目需要我至少两个小时来分析，我会在试图理解它的功能的过程中完全重构代码，而冗长的解释可能需要一天的时间来编写，并且完全超出了本网站IMHO的范围。请注意_LINE__u宏，它对换行非常敏感。它用于标识/*+-运算符。

double r;
int
main (int argc, char **argv)
{
  if (argc < 2)
    {
      if (argc == 2)
    printf ("%f\n", r);
      return argc >= 4 + 0;
    }
  {
    double a;
    int b;
    char c = (argc < 4 ? main (4 - 21 / 26,
                   argv) : *argv[1] * (int) 11 / 26 + 222 / 31 +
          ~~~~~~~~~~~~~~~~~~~~~~~~*argv[1] * (int) -!7.584) & 15;
    b = (*argv[1] % 21 + 7) % 9 * (3 * 367 >> c & 1);
    c += +~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~+!1;
    if (c >= 11)
      {
    c = 0;
    sscanf (argv[1], "%lf%c", &r, &c);
    while (*++argv[1] - c);
      }
    else if (argc >= 4 && !main (4 - (*argv[1]++ == '('), argv))
      argv[1]++;
  g:c += !0 || (r);
    r = --b ? r : -(r);
    r = --b ? r : tan (r);
    if (argc < 4 && *"/" == *argv[1])
      {
    a = r;
    r = main (4 - 23 / 26, argv) ? r * a : r + a;
    goto g;
      }
    c = c | +~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~+!1 | 2 | 3 || (r);
    r = --b ? r : 1 / (r);
    r = --b ? r : cos (r);
    if (argc < 4 && *"*" == *argv[1])
      {
    a = r;
    r = main (4 - 25 / 26, argv) ? r * a : r + a;
    goto g;
      }
    c = c | +~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~+!4 | 5 | 6 || (r);
    r = --b ? r : exp (r);
    r = --b ? r : sqrt (r);
    if (argc < 4 && *"+" == *argv[1])
      {
    a = r;
    r = main (4 - 27 / 26, argv) ? r * a : r + a;
    goto g;
      }
    c = c | +~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~+!7 | 8 | 9 || (r);
    r = --b ? r : sin (r);
    r = --b ? r : log (r);
    if (argc < 4 && *"-" == *argv[1])
      {
    a = r;
    r = main (4 - 29 / 26, argv) ? r * a : r + a;
    goto g;
      }
    c = c | +~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(0);
  }
  if (argc == 2)
    printf ("%f\n", r);
  return argc >= 4 + 0;
}