{"id":6583,"date":"2022-09-04T03:29:34","date_gmt":"2022-09-03T21:59:34","guid":{"rendered":"https:\/\/jassweb.com\/solved\/solved-how-to-print-cities-correctly-using-recursive-closed\/"},"modified":"2022-09-04T03:29:34","modified_gmt":"2022-09-03T21:59:34","slug":"solved-how-to-print-cities-correctly-using-recursive-closed","status":"publish","type":"post","link":"https:\/\/jassweb.com\/solved\/solved-how-to-print-cities-correctly-using-recursive-closed\/","title":{"rendered":"[Solved] How to print cities correctly using recursive? [closed]"},"content":{"rendered":"

[ad_1]
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very probably a reasonable definition of cities<\/em> (which was much better named City<\/em>, no reason so have it plural) is<\/p>\n

typedef struct cities {\n  char name[64];\n  struct cities * e;\n  struct cities * n;\n  struct cities * s;\n  struct cities * w;\n} cities;\n<\/code><\/pre>\n
and your problem is to write all the links without entering in an infinite loop<\/p>\n
A first thing to do is to immediately modify your program because you indicate the west etc of a city but you do not indicate for which city, so for instance :<\/p>\n
  if(cityHead->w != NULL && w != 1){\n      printf(\"--\");\n      printf(\"%s West: %s\\n\", cityHead->name, cityHead->w->name);\n      printcity(cityHead->w,1,0,0,0);\n  }else{\n      printf(\"--\");\n      printf(\"%s West: None\\n\", cityHead->name);\n  }\n<\/code><\/pre>\n
and so on for the other cases<\/p>\n
Let see with 2 cities with the change :<\/p>\n
int main()\n{\n  cities * c1 = newcity(\"c1\");\n  cities * c2 = newcity(\"c2\");\n\n  c1->e = c2;\n  c2->w = c1;\n\n  printcity(c1, 0, 0, 0, 0);\n}\n<\/code><\/pre>\n
Execution :<\/p>\n
pi@raspberrypi:\/tmp $ .\/a.out\n--c1 West: None\n--c1 East: c2\n--c2 West: None\n--c2 East: None\n--c2 South: None\n--c2 North: None\n--c1 South: None\n--c1 North: None\n<\/code><\/pre>\n
and we see the program does not indicate the east of c2<\/em> is c1<\/em><\/p>\n
The reason is simple, for instance in<\/p>\n
\n
 if(cityHead->w != NULL && w != 1){\n     printf(\"--\");\n     printf(\"%s West: %s\\n\", cityHead->name, cityHead->w->name);\n     printcity(cityHead->w,1,0,0,0);\n }else{\n     printf(\"--\");\n     printf(\"%s West: None\\n\", cityHead->name);\n }\n<\/code><\/pre>\n<\/blockquote>\n
w<\/em> is clearly used to not enter in an infinite, but it also blocks the  printf<\/em><\/p>\n
So let modify it to have :<\/p>\n
  if(cityHead->w != NULL){\n      printf(\"--\");\n      printf(\"%s West: %s\\n\", cityHead->name, cityHead->w->name);\n      if (w != 1)\n        printcity(cityHead->w,1,0,0,0);\n  }else{\n      printf(\"--\");\n      printf(\"%s West: None\\n\", cityHead->name);\n  }\n<\/code><\/pre>\n
There is also an error when you want to manage the north, the recursive call is :<\/p>\n
\n
printcity(cityHead->s,0,0,1,0);\n<\/code><\/pre>\n<\/blockquote>\n
but it must be<\/p>\n
printcity(cityHead->n,0,0,1,0);\n<\/code><\/pre>\n
So finaly :<\/p>\n
void printcity(cities *cityHead,int e,int w,int s,int n){\n if(cityHead){\n\n    if(cityHead->w != NULL){\n        printf(\"--\");\n        printf(\"%s West: %s\\n\", cityHead->name, cityHead->w->name);\n        if (w != 1)\n          printcity(cityHead->w,1,0,0,0);\n    }else{\n        printf(\"--\");\n        printf(\"%s West: None\\n\", cityHead->name);\n    }\n\n    if(cityHead->e != NULL){\n        printf(\"--\");\n        printf(\"%s East: %s\\n\", cityHead->name,cityHead->e->name);\n        if (e != 1)\n          printcity(cityHead->e,0,1,0,0);\n\n    }else{\n        printf(\"--\");\n        printf(\"%s East: None\\n\", cityHead->name);\n    }\n\n    if(cityHead->s != NULL){\n        printf(\"--\");\n        printf(\"%s South: %s\\n\", cityHead->name,cityHead->s->name);\n        if (s != 1)\n          printcity(cityHead->s,0,0,0,1);\n\n    }else{\n        printf(\"--\");\n        printf(\"%s South: None\\n\", cityHead->name);\n    }\n\n    if(cityHead->n != NULL){\n        printf(\"--\");\n        printf(\"%s North: %s\\n\", cityHead->name,cityHead->n->name);\n        if (n != 1)\n          printcity(cityHead->n,0,0,1,0);\n\n    }else{\n        printf(\"--\");\n        printf(\"%s North: None\\n\", cityHead->name);\n    }\n }\n}\n<\/code><\/pre>\n
out of that let notice the initial test<\/p>\n
\n
if(cityHead){\n<\/code><\/pre>\n<\/blockquote>\n
is only useful if printcity<\/em> is initially called with NULL, the function itself never recur with a NULL<\/p>\n
Now the execution is :<\/p>\n
pi@raspberrypi:\/tmp $ .\/a.out\n--c1 West: None\n--c1 East: c2\n--c2 West: c1\n--c2 East: None\n--c2 South: None\n--c2 North: None\n--c1 South: None\n--c1 North: None\n<\/code><\/pre>\n
If we have 4 cities making a square :<\/p>\n
int main()\n{\n  cities * c1 = newcity(\"c1\");\n  cities * c2 = newcity(\"c2\");\n  cities * c3 = newcity(\"c3\");\n  cities * c4 = newcity(\"c4\");\n\n  c1->e = c2;\n  c2->w = c1;\n\n  c1->s = c3;\n  c3->n = c1;\n\n  c2->s = c4;\n  c4->n = c2;\n\n  c4->w = c3;\n  c3->e = c4;\n\n  printcity(c1, 0, 0, 0, 0);\n}\n<\/code><\/pre>\n
the execution write a verrrrrrrry long times before to stop after too many recursions and a too large stack.<\/p>\n
The reason is the int<\/em> in parameters do not have enough extend in the graph.<\/p>\n
A way to solve is to add a field in the struct<\/em> to mark if the city was already managed. But out of modifying the struct<\/em> that requests at the end to again go through the graph to clear the mark.<\/p>\n
An other way is to save the already managed cities in a list, giving that list in parameter :<\/p>\n
#include <stdio.h>\n#include <stdlib.h>\n#include <string.h>\n\ntypedef struct cities {\n  char name[64];\n  struct cities * e;\n  struct cities * n;\n  struct cities * s;\n  struct cities * w;\n} cities;\n\ntypedef struct CityList {\n  cities * city;\n  struct CityList * next;\n} CityList;\n\nint isPresent(cities * c, CityList * l)\n{\n  while (l) {\n    if (l->city == c)\n      return 1;\n    l = l->next;\n  }\n  return 0;\n}\n\nvoid add(cities * city, CityList ** l)\n{\n  CityList * cell = malloc(sizeof(CityList));\n\n  cell->city = city;\n  cell->next = *l;\n  *l = cell;\n}\n\nvoid clear(CityList ** l)\n{\n  while (*l) {\n    CityList * c = *l;\n\n    *l = (*l)->next;\n    free(c);\n  }\n}\n\n\nvoid printcity(cities *cityHead, CityList ** l){\n if(cityHead && !isPresent(cityHead, *l)) {\n    add(cityHead, l);\n    if(cityHead->w != NULL){\n        printf(\"--\");\n        printf(\"%s West: %s\\n\", cityHead->name, cityHead->w->name);\n        printcity(cityHead->w, l);\n    }else{\n        printf(\"--\");\n        printf(\"%s West: None\\n\", cityHead->name);\n    }\n\n    if(cityHead->e != NULL){\n        printf(\"--\");\n        printf(\"%s East: %s\\n\", cityHead->name,cityHead->e->name);\n        printcity(cityHead->e, l);\n\n    }else{\n        printf(\"--\");\n        printf(\"%s East: None\\n\", cityHead->name);\n    }\n\n    if(cityHead->s != NULL){\n        printf(\"--\");\n        printf(\"%s South: %s\\n\", cityHead->name,cityHead->s->name);\n        printcity(cityHead->s, l);\n\n    }else{\n        printf(\"--\");\n        printf(\"%s South: None\\n\", cityHead->name);\n    }\n\n    if(cityHead->n != NULL){\n        printf(\"--\");\n        printf(\"%s North: %s\\n\", cityHead->name,cityHead->n->name);\n        printcity(cityHead->n, l);\n\n    }else{\n        printf(\"--\");\n        printf(\"%s North: None\\n\", cityHead->name);\n    }\n }\n}\n\ncities * newcity(char name[])\n{\n   cities *temp = (cities*)malloc(sizeof(cities));\n\n   strcpy(temp->name,name);\n   temp->e = temp->n = temp->s = temp->w = NULL;\n   return temp;\n } \n\n\nint main()\n{\n  cities * c1 = newcity(\"c1\");\n  cities * c2 = newcity(\"c2\");\n  cities * c3 = newcity(\"c3\");\n  cities * c4 = newcity(\"c4\");\n  CityList * l = NULL;\n\n  c1->e = c2;\n  c2->w = c1;\n\n  c1->s = c3;\n  c3->n = c1;\n\n  c2->s = c4;\n  c4->n = c2;\n\n  c4->w = c3;\n  c3->e = c4;\n\n  puts(\"[]\");\n  printcity(c1, &l);\n  clear(&l);\n\n  \/* to make a 8 *\/\n\n  cities * c5 = newcity(\"c5\");\n  cities * c6 = newcity(\"c6\");\n\n  c3->s = c5;\n  c5->n = c3;\n\n  c4->s = c6;\n  c6->n = c4;\n\n  c5->e = c6;\n  c6->w = c5;\n\n  puts(\"8\");\n  printcity(c1, &l);\n  clear(&l);\n\n  free(c1);\n  free(c2);\n  free(c3);\n  free(c4);\n  free(c5);\n  free(c6);\n\n  return 0;\n}\n<\/code><\/pre>\n
Execution :<\/p>\n
pi@raspberrypi:\/tmp $ .\/a.out\n[]\n--c1 West: None\n--c1 East: c2\n--c2 West: c1\n--c2 East: None\n--c2 South: c4\n--c4 West: c3\n--c3 West: None\n--c3 East: c4\n--c3 South: None\n--c3 North: c1\n--c4 East: None\n--c4 South: None\n--c4 North: c2\n--c2 North: None\n--c1 South: c3\n--c1 North: None\n8\n--c1 West: None\n--c1 East: c2\n--c2 West: c1\n--c2 East: None\n--c2 South: c4\n--c4 West: c3\n--c3 West: None\n--c3 East: c4\n--c3 South: c5\n--c5 West: None\n--c5 East: c6\n--c6 West: c5\n--c6 East: None\n--c6 South: None\n--c6 North: c4\n--c5 South: None\n--c5 North: c3\n--c3 North: c1\n--c4 East: None\n--c4 South: c6\n--c4 North: c2\n--c2 North: None\n--c1 South: c3\n--c1 North: None\npi@raspberrypi:\/tmp $ \n<\/code><\/pre>\n
\n
Note your definition of newcity<\/em> is dangerous because of the line<\/p>\n
\n
strcpy(temp->name,name);\n<\/code><\/pre>\n<\/blockquote>\n
where you can write out of the field name<\/em> in cities<\/em> in case the name is too long, producing an undefined behavior<\/p>\n<\/p><\/div>\n
<\/div>\n<\/div>\n
            <\/span> <\/p><\/div>\n<\/div>\n
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solved How to print cities correctly using recursive? [closed] <\/p>\n","protected":false},"excerpt":{"rendered":"
[ad_1] very probably a reasonable definition of cities (which was much better named City, no reason so have it plural) is typedef struct cities { char name[64]; struct cities * e; struct cities * n; struct cities * s; struct cities * w; } cities; and your problem is to write all the links without … Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[320],"tags":[324,1826],"class_list":["post-6583","post","type-post","status-publish","format-standard","hentry","category-solved","tag-c","tag-recursive-datastructures"],"yoast_head":"\n[Solved] How to print cities correctly using recursive? 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Read more","og_url":"https:\/\/jassweb.com\/solved\/solved-how-to-print-cities-correctly-using-recursive-closed\/","og_site_name":"JassWeb","article_published_time":"2022-09-03T21:59:34+00:00","author":"Kirat","twitter_card":"summary_large_image","twitter_misc":{"Written by":"Kirat","Est. reading time":"7 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Article","@id":"https:\/\/jassweb.com\/solved\/solved-how-to-print-cities-correctly-using-recursive-closed\/#article","isPartOf":{"@id":"https:\/\/jassweb.com\/solved\/solved-how-to-print-cities-correctly-using-recursive-closed\/"},"author":{"name":"Kirat","@id":"https:\/\/jassweb.com\/solved\/#\/schema\/person\/65c9c7b7958150c0dc8371fa35dd7c31"},"headline":"[Solved] How to print cities correctly using recursive? 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