'Build Your Own Lisp' Solutions: 4.4

Q: What does the \n mean in those strings?


A: '\n' signifies a newline character, so the startup information doesn't appear on the same line as the first prompt.



K&R Solutions - 1.18

Q: Write a program to remove trailing blanks and tabs from each line of input, and to delete entirely blank lines.

A:

#include <stdio.h>
#define MAXLINE 1000

int get_a_line(char s[], int lim);
void chop(char s[], int lim);

int main(void)
{
  int len;
  char line[MAXLINE];

  while ((len = get_a_line(line, MAXLINE)) > 0) {
    if (line[0] == '\n') {
      // Nothing happens. We want to totally ignore strings that are just
      // newlines.
    } else {
      chop(line, MAXLINE);
      printf("%s<---End Of String\n", line);
    }
  }

  return 0;
}

// Same as the previous exercise.
int get_a_line(char s[], int lim) 
{
  int c, i;

  for (i = 0; i < lim-1 && (c=getchar()) != EOF && c != '\n'; ++i) {
    s[i] = c;
  }
  if (c == '\n') {
    s[i] = c;
    ++i;
  }
  s[i] = '\0';
  
  return i;
}

void chop(char s[], int lim) {
  int i;

  // First we find the end of the string.
  // Notice that the the resulting value of 'i' is the length of the string.
  // Also notice that in this context, we could have just moved to the newline
  // before the null char.
  for (i = 0; s[i] != '\0'; ++i) {
  }

  // Now move backwards again, untill we find something that's not whitespace.
  for (i -= 1; s[i] == '\t' || s[i] == '\n' || s[i] == ' '; --i) {
  }

  // And we replace whatever is after with a null char.
  // We're replacing the newline, even though the exercise doesn't tell us
  // to get rid of it. This is becuse it makes the output easier to test.
  s[i+1] = '\0';
}
./trimmer
hi 	  (enter)
hi<---End Of String
how are         you?		(enter)	 
how are 	you?<---End Of String
(enter)
three little birds(enter)
three little birds<---End Of String


K&R Solutions - 1.17

Q: Write a program to print all input lines that are longer than 80 characters.

A:

/* This is really a simpler version of the longest-line program: that already
 * has most of what we need. */
#include <stdio.h>
#define MAXLINE 1000

int get_a_line(char s[], int lim);


int main(void)
{
  int len;
  char line[MAXLINE];

  while ((len = get_a_line(line, MAXLINE)) > 0) {
    if (len > 80) {
      printf("%s", line);
    }
  }

  return 0;
}

int get_a_line(char s[], int lim) 
{
  int c, i;

  for (i = 0; i < lim-1 && (c=getchar()) != EOF && c != '\n'; ++i) {
    s[i] = c;
  }
  if (c == '\n') {
    s[i] = c;
    ++i;
  }
  s[i] = '\0';
  
  return i;
}
This is a short sentence, which will not be repeated.
This, however, is a very, very, very long sentence, which will be repeated back by the program.     
This, however, is a very, very, very long sentence, which will be repeated back by the program.


'Build Your Own Lisp' Solutions: 4.3

Q: Add an extra message to the Version and Exit Information.


A:

...
int main(int argc, char** argv) {

  puts("Lispy Version 0.0.0.0.1, Something Something Something");
  puts("Press Ctrl+c to Exit. Happy Trails!\n");

  while (1) {

    /* Now in either case readline will be correctly defined */
    char* input = readline("lispy> ");
    add_history(input);

    printf("No you're a %s\n", input);
    free(input);

  }

  return 0;
}
...


K&R Solutions - 1.16

Q: Revise the main routine of the longest-line program so it will correctly print the length of arbitrarily long input lines and as much as possible of the text.

A:

...
int main(void)
{
  int len;
  int max;
  char line[MAXLINE];
  char longest[MAXLINE];

  max = 0;
  while ((len = get_a_line(line, MAXLINE)) > 0) {
    if (len > max) {
      max = len;
      copy(longest, line);
    }
    // Following tradition, we will be ignoring the null character that marks
    // the end of every string. The standard library function strlen() also
    // ignores null characters.
    printf("Length: %d\n", len - 1);
    printf("Line: %s\n", line);
  }
  if (max > 0) {
    printf("%s", longest);
  }

  return 0;
}
...
hi
Length: 2
Line: hi

there this is an 
Length: 16
Line: there this is an

arbitrarily long string
Length: 23
Line: arbitrarily long string

moo 
Length: 3
Line: moo

arbitrarily long string


'Build Your Own Lisp' Solutions: 4.2

Q: Change what is echoed back to the user.


A:

...
  while (1) {
    
    /* Now in either case readline will be correctly defined */
    char* input = readline("something of my choice> ");
    add_history(input);

    printf("%s\n", input);
    free(input);
    
  }
...


K&R Solutions - 1.15

Q: Rewrite the temperature conversion program of Section 1.2 to use a function for conversion.

A:

// Compared with the original, this is much easier to read.

#include <stdio.h>
  
float fahrenheit_to_celsius (float fahr) {
  return (5.0/9.0) * (fahr-32.0);
}

int main(void) {
  int lower = 0;
  int upper = 300;
  int step = 20;
  float fahr = lower;

  while (fahr <= upper) {
    printf("%3.0f %6.1f\n", fahr, fahrenheit_to_celsius(fahr));
    fahr = fahr + step;
  }

  return 0;
}
  0  -17.8
 20   -6.7
 40    4.4
 60   15.6
 80   26.7
100   37.8
120   48.9
140   60.0
160   71.1
180   82.2
200   93.3
220  104.4
240  115.6
260  126.7
280  137.8
300  148.9


'Build Your Own Lisp' Solutions: 4.1

Q: Change the prompt from "lispy>" to something of your choice.


A:

...
  while (1) {
    
    char* input = readline("something of my choice> ");
    add_history(input);

    printf("No you're a %s\n", input);
    free(input);
    
  }
...


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 | 
{ | 
| | 
} | 
~ |