Salida de programas en C | conjunto 11

Preguntado por Shobhit

#include<stdio.h>
int fun(int n, int *fg)
{
   int t, f;
   if(n <= 1)
   {
     *fg = 1;
      return 1;
   }
   t = fun(n-1, fg);
   f = t + *fg;
   *fg = t;
   return f;
}
int main( )
{
  int x = 15;
  printf ( "%d\n", fun (5, &x));
  getchar();
  return 0;
}

En el programa anterior, habrá llamadas recursivas hasta que n no sea menor o igual a 1.

fun(5, &x) 
       \
         \
       fun(4, fg)
           \
             \
            fun(3, fg)
                 \
                   \
                   fun(2, fg)
                      \
                        \
                          fun(1, fg)

fun(1, fg) ya no llama a fun() porque n es 1 ahora, y va dentro de la parte if. Cambia el valor en la dirección fg a 1 y devuelve 1.

Diversión interior (2, fg)

   t = fun(n-1, fg); --> t = 1
  /* After fun(1, fg) is called, fun(2, fg) does following */
   f = t + *fg;      -->  f = 1 + 1 (changed by fun(1, fg)) = 2
   *fg = t;           --> *fg = 1
  return f (or return 2)

Diversión interior (3, fg)

   t = fun(2, fg); --> t = 2
  /* After fun(2, fg) is called, fun(3, fg) does following */
   f = t + *fg;       -->  f = 2 + 1 = 3
   *fg = t;            --> *fg = 2
  return f (or return 3)

Diversión interior (4, fg)

   t = fun(3, fg);   --> t = 3
  /* After fun(3, fg) is called, fun(4, fg) does following */
   f = t + *fg;       -->  f = 3 + 2 = 5
   *fg = t;            --> *fg = 3
   return f (or return 5)

Diversión interior (5, fg)

   t = fun(4, fg);   -->  t = 5
  /* After fun(4, fg) is called, fun(5, fg) does following */
   f = t + *fg;       -->  f = 5 + 3 = 8
   *fg = t;            --> *fg = 5
   return f (or return 8 )

Finalmente, se imprime el valor devuelto por fun(5, &x), por lo que se imprime 8 en la pantalla