{"content":{"id":318,"title":"K\u0026R Solutions - 1.13, Part 2","body":"\u003cp\u003e\u003cb\u003eQ:\u003c/b\u003e The same as part one, but vertical.\u003c/p\u003e\r\n\u003cp\u003e\u003cb\u003eA:\u003c/b\u003e\u003c/p\u003e \r\n\u003cpre\u003e\u003ccode class=\"cpp\"\u003e#include \u0026ltstdio.h\u0026gt\r\n\r\nint main(void) \r\n{\r\n int FREQUENCY_LENGTH = 10; \r\n int frequencies[FREQUENCY_LENGTH];\r\n int i, j, c;\r\n int tally = 0;\r\n int highest = 0;\r\n\r\n // We're tallying in the exact same way.\r\n for (i = 0; i \u003c FREQUENCY_LENGTH; ++i) {\r\n frequencies[i] = 0;\r\n }\r\n\r\n while ((c = getchar()) != EOF) {\r\n if (c == '.' || c == '\\n' || c == '\\t' || c == ' ' || c == ',') {\r\n if (tally \u003e FREQUENCY_LENGTH) {\r\n tally = FREQUENCY_LENGTH - 1;\r\n }\r\n ++frequencies[tally];\r\n tally = 0;\r\n } else {\r\n ++tally;\r\n }\r\n }\r\n\r\n /* First, we need the height of the highest bar.\r\n * At the top of the function, temp is set to zero. Normally a function\r\n * wouldn't be this long, but we haven't got to that part fo the book yet.\r\n */\r\n for (i = 0; i \u003c FREQUENCY_LENGTH; i++) {\r\n if (frequencies[i] \u003e highest) {\r\n highest = frequencies[i];\r\n }\r\n }\r\n\r\n /* Now we keep drawing the bars untill our 'highest' variable, and everything\r\n * in our tallies is zeroed out.\r\n */\r\n while (highest) {\r\n // Remember there are no zero-length words so we skip 0. Feel free to play\r\n // around with this, because some junk data is getting stored in\r\n // frequencies[0]. Why?\r\n for(i = 1; i \u003c FREQUENCY_LENGTH; i++) {\r\n if (frequencies[i] == highest) {\r\n printf(\" | \");\r\n frequencies[i]--;\r\n } else {\r\n printf(\" \");\r\n }\r\n }\r\n printf(\"\\n\");\r\n highest--;\r\n }\r\n\r\n /* Now we print a handy guide at the bottom.\r\n */\r\n for (i = 1; i \u003c FREQUENCY_LENGTH; i++) {\r\n printf(\"---\");\r\n }\r\n printf(\"-\\n\");\r\n for (i = 1; i \u003c FREQUENCY_LENGTH; i++) {\r\n if (i == FREQUENCY_LENGTH - 1) {\r\n printf(\" %d+\", i);\r\n } else {\r\n printf(\" %d \", i);\r\n }\r\n }\r\n printf(\"\\n\");\r\n\r\n return 0;\r\n}\r\n\u003c/code\u003e\u003c/pre\u003e\r\n\u003cpre\u003e\u003ccode class=\"lisp\"\u003eThe worst of misery\r\nIs when a nature framed for noblest things\r\nCondemns itself in youth to petty joys,\r\nAnd, sore athirst for air, breathes scanty life\r\nGasping from out the shallows.\r\nSupercalifragilisticexpialidocious.\r\n\r\n(ctrl+d)\r\n\r\n | \r\n | | \r\n | | | \r\n | | | | \r\n | | | | | | | \r\n | | | | | | | \r\n | | | | | | | | | \r\n----------------------------\r\n 1 2 3 4 5 6 7 8 9+ \r\n\u003c/code\u003e\u003c/pre\u003e\r\n\u003cp\u003eYou can play around with this quite a bit. Removing one line and then changing one character, and we get a nice little graph:\u003c/p\u003e\r\n\u003cpre\u003e\u003ccode class=\"lisp\"\u003e . \r\n . \r\n . \r\n . \r\n . . . \r\n \r\n . . \r\n----------------------------\r\n 1 2 3 4 5 6 7 8 9+ \r\n\u003c/code\u003e\u003c/pre\u003e","publication_date":"2018-10-14T01:00:00.000Z","created_at":"2018-10-13T17:15:36.000Z","updated_at":"2018-10-13T19:47:42.000Z","user_id":1,"rating":null},"tags":"\u003ca class=\"changeable-title\" href=\"/q?tag=k%26r\"\u003ek\u0026amp;r\u003c/a\u003e \u003ca class=\"changeable-title\" href=\"/q?tag=programming\"\u003eprogramming\u003c/a\u003e \u003ca class=\"changeable-title\" href=\"/q?tag=answers\"\u003eanswers\u003c/a\u003e \u003ca class=\"changeable-title\" href=\"/q?tag=c\"\u003ec\u003c/a\u003e \u003ca class=\"changeable-title\" href=\"/q?tag=unix\"\u003eunix\u003c/a\u003e"}

K&R Solutions - 1.13, Part 2

Q: The same as part one, but vertical.

A:

#include <stdio.h>

int main(void) 
{
  int FREQUENCY_LENGTH = 10; 
  int frequencies[FREQUENCY_LENGTH];
  int i, j, c;
  int tally = 0;
  int highest = 0;

  // We're tallying in the exact same way.
  for (i = 0; i < FREQUENCY_LENGTH; ++i) {
    frequencies[i] = 0;
  }

  while ((c = getchar()) != EOF) {
    if (c == '.' || c == '\n' || c == '\t' || c == ' ' || c == ',') {
      if (tally > FREQUENCY_LENGTH) {
        tally = FREQUENCY_LENGTH - 1;
      }
      ++frequencies[tally];
      tally = 0;
    } else {
      ++tally;
    }
  }

  /* First, we need the height of the highest bar.
  *  At the top of the function, temp is set to zero. Normally a function
  *  wouldn't be this long, but we haven't got to that part fo the book yet.
  */
  for (i = 0; i < FREQUENCY_LENGTH; i++) {
    if (frequencies[i] > highest) {
      highest = frequencies[i];
    }
  }

  /* Now we keep drawing the bars untill our 'highest' variable, and everything
   * in our tallies is zeroed out.
   */
  while (highest) {
    // Remember there are no zero-length words so we skip 0. Feel free to play
    // around with this, because some junk data is getting stored in
    // frequencies[0]. Why?
    for(i = 1; i < FREQUENCY_LENGTH; i++) {
      if (frequencies[i] == highest) {
        printf(" | ");
        frequencies[i]--;
      } else {
        printf("   ");
      }
    }
    printf("\n");
    highest--;
  }

  /* Now we print a handy guide at the bottom.
   */
  for (i = 1; i < FREQUENCY_LENGTH; i++) {
    printf("---");
  }
  printf("-\n");
  for (i = 1; i < FREQUENCY_LENGTH; i++) {
    if (i == FREQUENCY_LENGTH - 1) {
      printf(" %d+", i);
    } else {
      printf(" %d ", i);
    }
  }
  printf("\n");

  return 0;
}
The worst of misery
Is when a nature framed for noblest things
Condemns itself in youth to petty joys,
And, sore athirst for air, breathes scanty life
Gasping from out the shallows.
Supercalifragilisticexpialidocious.

(ctrl+d)

       |                   
       |        |          
       |  |     |          
    |  |  |     |          
    |  |  |  |  |  |  |    
    |  |  |  |  |  |  |    
 |  |  |  |  |  |  |  |  | 
----------------------------
 1  2  3  4  5  6  7  8  9+ 

You can play around with this quite a bit. Removing one line and then changing one character, and we get a nice little graph:

       .                   
                .          
          .                
    .                      
             .     .  .    
                           
 .                       . 
----------------------------
 1  2  3  4  5  6  7  8  9+