18-11-11 Weekly Links


K&R Solutions - 1.14

Q: Write a program to print a histogram of the frequencies of different characters in its input.

A:

#include <stdio.h>

int main(void) 
{
  int FREQUENCY_LENGTH = 256; // We'll just count all ASCII values. Why not?
  int frequencies[FREQUENCY_LENGTH];
  int i, j, c;
  int tally = 0;

  for (i = 0; i < FREQUENCY_LENGTH; ++i) {
    frequencies[i] = 0;
  }

  // This part gets a lot more simple.
  while ((c = getchar()) != EOF) {
    ++frequencies[c];
  }

  printf("\n");
  printf("Character Frequencies\n");
  printf("---------------------\n");
  /* For the sake of brevity, we will only print visible characters.
   * The ASCII values for visible characters are all between 33 and 126,
   * a fact I pulled off www.asciitable.com
   */
  for (i = 33; i <= 126; ++i) {
    printf("%c | ", i);

    for (j = 0; j < frequencies[i]; ++j) {
      printf("-");
    }

    printf("\n");
  }

  return 0;
}
America! Land of my dreams, home of the Whopper!

(ctrl+d)

Character Frequencies
---------------------
! | --
" | 
# | 
$ | 
% | 
& | 
' | 
( | 
) | 
* | 
+ | 
, | -
- | 
. | 
/ | 
0 | 
1 | 
2 | 
3 | 
4 | 
5 | 
6 | 
7 | 
8 | 
9 | 
: | 
; | 
< | 
= | 
> | 
? | 
@ | 
A | -
B | 
C | 
D | 
E | 
F | 
G | 
H | 
I | 
J | 
K | 
L | -
M | 
N | 
O | 
P | 
Q | 
R | 
S | 
T | 
U | 
V | 
W | -
X | 
Y | 
Z | 
[ | 
\ | 
] | 
^ | 
_ | 
` | 
a | ---
b | 
c | -
d | --
e | -----
f | --
g | 
h | ---
i | -
j | 
k | 
l | 
m | ----
n | -
o | ----
p | --
q | 
r | ---
s | -
t | -
u | 
v | 
w | 
x | 
y | -
z | 
{ | 
| | 
} | 
~ | 


18-10-28 Weekly Links




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+ 


'Build Your Own Lisp' Solutions: 3.12

Q: What does the 'typedef' keyword do exactly?


A: 'typedef' lets you define a type, which is to say, it lets you give a new name to a pre-existing type (which can be useful for organizational/readability purposes), or to define a struct as a new type, unique to your program.



K&R Solutions - 1.13, Part 1

Q: Write a program to print a histogram of the lengths of words in its input. It is easy to draw the histogram with the bars horizontal; a vertical orientation is more challenging.

A:

#include <stdio.h>

int main(void) 
{
  /* Because of screen space and the limitations of terminal output, words
   * longer than ten are going to be categorized as greater than 10.
   * 
   * In this model, frequencies[0] is a one-character word, and frequencies[1]
   * is a two-character word. This 'off-by-one' behaviour seems unintuitive
   * and certainly leads to bugs, but there is a very good, low-level reason
   * that becomes very obvious when you learn a bit of assembly. You could find
   * a work-around (like leaving frequencies[0] unused), but it's better to get
   * use to thinking of things as starting with the 0th element because, in the
   * long run, there's no real avoiding it in programming.
   */
  int FREQUENCY_LENGTH = 10; 
  // It's tradition to capitalize this we won't be changing.
  int frequencies[FREQUENCY_LENGTH];
  int i, j, c;
  int tally = 0;

  /* Initialize all frequencies to 0 */
  /* The book hasn't told us this yet, but the easiest way to initiate all
   * elements in an array to 0 is simply to go:
   * int frequencies[10] = {0};
   * BUT doing it with what the book has tought us gives us a better feel for
   * the way that arrays work.
   */
  for (i = 0; i < FREQUENCY_LENGTH; ++i) {
    frequencies[i] = 0;
  }

  /* Count word-length frequencies */
  /* I'm missing colons and semi-colons here for the sake of space, but you get
   * the drift. In a professional setting, you would probably want to create 
   * a separate subroutine called is_word_barrier() to keep your code cleaner
   * and more readable, but we haven't got to that part of the book yet. 
   */
  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;
    }
  }

  /* Print horizontal-barred histogram */
  printf("\n");
  printf("Word Frequencies: Horizontal\n");
  printf("----------------------------\n");
  for (i = 0; i < FREQUENCY_LENGTH; ++i) {
    if (i != 0) {
      if (i == FREQUENCY_LENGTH - 1) {
        printf("%3d+ | ", i);
      } else {
        printf("%3d  | ", i);
      } 
      for (j = 0; j < frequencies[i]; ++j) {
        printf("-");
      }
      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)

Word Frequencies: Horizontal
----------------------------
  1  | -
  2  | ----
  3  | -------
  4  | -----
  5  | ---
  6  | ------
  7  | ---
  8  | ---
  9+ | -


K&R Solutions - 1.12

Q: Write a program that prints its input one word per line.

A:

#include <stdio.h>

int main(void)
{
  char c;

  while ((c = getchar()) != EOF) {
    if (c == '\t' || c == '\n' || c == ' ') {
      printf("\n");
    } else {
      putchar(c);
    }
  }

  return 0;
}
> muh muh muh     moo cows
muh
muh
muh
moo
cows


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