I want to get an input from the user like (4,5).But I just want to get the integer values.(4 and 5) I wrote a code for this.But it did not work.How can I fix the problem.
int x, y;
int c;
c = getchar();
while (c != EOF) {
while (c != '(' && c != EOF) {
c = getchar();
}
while (c != ',' && c != EOF) {
c = getchar();
}
x = c;
while (c != ',' && c != EOF) {
c = getchar();
}
while (c != ')' && c != EOF) {
c = getchar();
}
y = c;
}
I get input(1,4). outputs is x= 44 and y= 45 ?
After you find the opening parenthesis, read characters while they are digits (see e.g. isdigit), and create your number.
When the character is not a digit, make sure it's a comma. If it is you read the next number the same as above. Finally make sure you got the closing parenthesis.
The above assumes that there are no whitespace between the parentheses and the numbers, or between the numbers or the comma. Those can be handled by looping and isspace.
As for converting digits to numbers, assuming your system is using ASCII encoding (which is standard on modern PC-like systems) then it's easy since you just subtract '0' from the character to get the digit (see the linked ASCII table to help you understand why). Store the value in a variable, initialized to zero, and multiply by ten as needed.
Depending on the parser you're writing, you could do the handling and recognition of tuples in the parser instead of the lexer. That would make it more flexible.
follow your code flow x value will be character comma.
int c ,x ,y;
c = getchar();
while (c != EOF) {
while (c != '(' && c != EOF) {
c = getchar();
}
c = getchar();
x = c;
while (c != ',' && c != EOF) {
c = getchar();
}
/*x = c;
while (c != ',' && c != EOF) {
c = getchar();
}
*/
c = getchar();
y = c;
while (c != ')' && c != EOF) {
c = getchar();
}
//y = c;
printf("x = %c ,y = %c ",x,y);
}
Hello everyone I just started learning C through THE c PROGRAMMING LANGUAGE Second Edition
by Brian. W.Kernighnan (ISBN-13: 978-8131704943)
So here is a script which counts the characters, line, words
#include <stdio.h>
#define IN 1
#define OUT 0
main()
{
int c, nl, nw, nc, state;
/* c = input, nl = new line, nc = new character, nw = new word, state = (IN/OUT) */
state = OUT;
nl = nw = nc = 0;
while ((c = getchar()) != EOF)
{
++nc;
if (c == '\n')
++nl;
if (c == ' ' || c == '\n' || c == '\t')
state = OUT;
else if (state == OUT)
{
state = IN;
++nw;
}
}
printf(" The number of lines is: %d,\n The number of words is: %d,\n The number of characters is: %d. \n", nl, nw, nc);
}
However I made a script which does the following without the need of defining state IN and OUT
#include <stdio.h>
main()
{
int nw, nl, nc, c ;
nl = nw = nc = c = 0 ;
while ((c = getchar()) != EOF)
{
++nc;
if (c == '\n')
++nl;
else if (c == ' ' || c == '\n' || c == '\t')
++nw;
}
printf("Words:%d \nCharacters:%d \nLines:%d", nw, nc, nl);
}
So what is the difference between these two, why does the author use the state to define IN and OUT ??
[EDIT]
Oh! I see so the script is just to avoid two things :
1. To avoid word count when there are more than one spaces following the word.
2. Secondly my script would count n-2 words I suppose if proper spacing is done.
Which makes the author's script more fullproof.....Is there anything else except these two ??
And thank You for your answes too....
P.S: I'm sorry this is a bit off-topic is it ok to label the question [SOLVED] or is there any other way of doing this ??
Your version is slightly different from his one: in your program, if you have N consecutive spaces, they will be considered as N - 1 words, because for every space you add one to the word count. Also, the last input word won't be considered.
IN literally means "inside a word" and OUT literally means "outside a word". He is tracking the state of the proverbial cursor as he moves through the lines.
"Write a program to copy its input to
its output, replacing each string of
one or more blanks by a single blank."
I'm assuming by this he means input something like...
We(blank)(blank)(blank)go(blank)to(blank)(blank)(blank)the(blank)mall!
... and output it like:
We(blank)go(blank)to(blank)the(blank)mall!
This is probably easier than I'm making it out to be, but still, I can't seem to figure it out. I don't really want the code... more so pseudo code.
Also, how should I be looking at this? I'm pretty sure whatever program I write is going to need at least one variable, a while loop, a couple if statements, and will use both the getchar() and putchar() functions... but besides that I'm at a loss. I don't really have a programmers train of thought yet, so if you could give me some advice as to how I should be looking at "problems" in general that'd be awesome.
(And please don't bring up else, I haven't got that far in the book so right now that's out of my scope.)
Look at your program as a machine that moves between different states as it iterates over the input.
It reads the input one character at a time. If it sees anything other than a blank, it just prints the character it sees. If it sees a blank, it shifts to a different state. In that state, it prints one blank, and then doesn't print blanks if it sees them. Then, it continues reading the input, but ignores all blanks it sees--until it hits a character that isn't a blank, at which point it shifts back to the first state.
(This concept is called a finite state machine, by the way, and a lot of theoretical computer science work has gone into what they can and can't do. Wikipedia can tell you more, though in perhaps more complicated detail than you're looking for. ;))
Pseudo code
while c = getchar:
if c is blank:
c = getchar until c is not blank
print blank
print c
C
You can substitute use of isblank here if you desire. It is unspecified what characters contrive blank, or what blank value is to be printed in place of others.
After many points made by Matthew in the comments below, this version, and the one containing isblank are the same.
int c;
while ((c = getchar()) != EOF) {
if (c == ' ') {
while ((c = getchar()) == ' ');
putchar(' ');
if (c == EOF) break;
}
putchar(c);
}
Since relational operators in C produce integer values 1 or 0 (as explained earlier in the book), the logical expression "current character non-blank or previous character non-blank" can be simulated with integer arithmetic resulting in a shorter (if somewhat cryptic) code:
int c, p = EOF;
while ((c = getchar()) != EOF) {
if ((c != ' ') + (p != ' ') > 0) putchar(c);
p = c;
}
Variable p is initialized with EOF so that it has a valid non-blank value during the very first comparison.
This is what I got:
while ch = getchar()
if ch != ' '
putchar(ch)
if ch == ' '
if last_seen_ch != ch
putchar(ch)
last_seen_ch = ch
Same explanation with Matt Joiner's, but this code does not use break.
int c;
while ((c = getchar()) != EOF)
{
if (c == ' ') /* find a blank */
{
putchar(' '); /* print the first blank */
while ((c = getchar()) == ' ') /* look for succeeding blanks then… */
; /* do nothing */
}
if (c != EOF) /* We might get an EOF from the inner while-loop above */
putchar(c);
}
I worked really hard at finding a solution that used only the material that has already been covered in the first part of the first chapter of the book. Here is my result:
#include <stdio.h>
/* Write a program to copy its input to its output, replacing each string of one or more blanks by a single blank. */
main()
{
int c;
while ((c = getchar()) != EOF){
if (c == ' '){
putchar(c);
while ((c = getchar()) == ' ')
;
}
if(c != ' ')
putchar(c);
}
}
Here is how I think of the algorithm of this exercise, in pseudo-code:
define ch and bl (where bl is initially defined to be 0)
while ch is = to getchar() which is not = to end of file
do the following:
if ch is = to blank and bl is = 0
--> output ch and assign the value 1 to bl
else --> if ch is = to blank and bl is = 1
--> do nothing
else --> output ch and assign the value 0 to bl
Example implementation in C:
#include <stdio.h>
#include <stdlib.h>
main() {
long ch,bl=0;
while ((ch=getchar()) != EOF)
{
if (ch == ' ' && bl == 0)
{
putchar(ch);
bl=1;
} else if (ch == ' ' && bl == 1) {
// no-op
} else {
putchar(ch);
bl=0;
}
}
return 0;
}
I wrote this and seems to be working.
# include <stdio.h>
int main ()
{
int c,lastc;
lastc=0;
while ((c=getchar()) != EOF)
if (((c==' ')+ (lastc==' '))<2)
putchar(c), lastc=c;
}
#include <stdio.h>
int main()
{
int c;
while( (c = getchar( )) != EOF )
{
if (c == ' ')
{
while ((c = getchar()) == ' ');
putchar(' ');
putchar(c);
}
else
putchar(c);
}
return 0;
}
#include <stdio.h>
main()
{
int c, numBlank=0 ;
while((c= getchar())!=EOF)
{
if(c ==' ')
{
numBlank ++;
if(numBlank <2)
{
printf("character is:%c\n",c);
}
}
else
{
printf("character is:%c\n",c);
numBlank =0;
}
}
}
First declare two variables character and last_character as integers.when you have not reach the end of the file( while(character=getchar() != EOF ) do this;
1. If character != ' ' then
print character
last_character = character
2. If character == ' '
if last_character ==' '
last character = character
else print character
Using the constraints of not using else or and operators. This code only prints a blank when the blank variable is equal to 1 and the only way to reset the counter is by typing something other than a blank. Hope this helps:
include
/* Write a program that replaces strings of blanks with a single blank */
void main(){
int c, bl;
bl = 0;
while((c = getchar()) != EOF){
if(c == ' '){
++bl;
if(bl == 1){
putchar(' ');
}
}
if(c != ' '){
putchar(c);
bl = 0;
}
}
}
Many others have already used the last character logic in their code, but perhaps the following version is easier to read:
int c, prevchar;
while ((c = getchar()) != EOF) {
if (!(c == ' ' && prevchar == ' ')) {
putchar(c);
prevchar = c;
}
}
#include <stdio.h>
main() {
int input, last = EOF;
while ((input = getchar()) != EOF) {
if (input == ' ' && last == ' ') continue;
last = input;
putchar(input);
}
}
I am at the same point in the book. and my solution goes with making a count++ if blank is found and making the count back to zero if anything other than blank is found.
For if statement I put another another check to check value of count (if zero) and then print.
Though at this point of learning I shouldn't be concern about efficiency of two methods but which one is efficient a.) Accepted solution here with while inside while or b.) the one I suggested above.
My code goes like below:
#include <stdio.h>
main()
{
int count=0,c;
for( ; (c=getchar())!=EOF; )
{
if(c==' ')
{
if(count==0)
{
putchar(c);
count++;
}
}
if(c!=' ')
{
putchar(c);
count=0;
}
}
}
#include <stdio.h>
main()
{
int CurrentChar, LastChar;
LastChar = '1';
while ((CurrentChar = getchar()) != EOF)
{
if (CurrentChar != ' ')
{
putchar(CurrentChar);
LastChar = '1';
}
else
{
if (LastChar != ' ')
{
putchar(CurrentChar);
LastChar = ' ';
}
}
}
}
a way to make it easier for the new people are stuck on this book
(by not knowing any thing then what brought up until page 22 in K&R).
credits to #Michael , #Mat and #Matthew to help me to understand
#include <stdio.h>
main()
{
int c;
while ((c = getchar()) != EOF) /* state of "an input is not EOF" (1) */
{
if (c == ' ') /* "blank has found" -> change the rule now */
{
while ((c = getchar ()) == ' '); /* as long you see blanks just print for it a blank until rule is broken (2) */
putchar(' ');
}
putchar(c); /* either state (2) was broken or in state (1) no blanks has been found */
}
}
1.Count the number of blanks.
2.Replace the counted number of blanks by a single one.
3.Print the characters one by one.
<code>
main()
{
int c, count;
count = 0;
while ((c = getchar()) != EOF)
{
if (c == ' ')
{
count++;
if (count > 1)
{
putchar ('\b');
putchar (' ');
}
else putchar (' ');
}
else
{
putchar (c);
count = 0;
}
}
return;
}
</code>
#include <stdio.h>
int main(void)
{
long c;
long nb = 0;
while((c = getchar()) != EOF) {
if(c == ' ' || c == '\t') {
++nb;
} else {
if(nb > 0) {
putchar(' ');
nb = 0;
}
putchar(c);
}
}
return 0;
}
To do this using only while loops and if statements, the trick is to add a variable which remembers the previous character.
Loop, reading one character at a time, until EOF:
If the current character IS NOT a space:
Output current character
If the current character IS a space:
If the previous character WAS NOT a space:
Output a space
Set previous character to current character
In C code:
#include <stdio.h>
main()
{
int c, p;
p = EOF;
while ((c = getchar()) != EOF) {
if (c != ' ')
putchar(c);
if (c == ' ')
if (p != ' ')
putchar(' ');
p = c;
}
}
I am also starting out with the K&R textbook, and I came up with a solution that uses only the material which had been covered up until that point.
How it works:
First, set some counter 'blanks' to zero. This is used for counting blanks.
If a blank is found, increase the counter 'blanks' by one.
If a blank is not found, then first do a sub-test: is the counter 'blanks' equal or bigger than 1? If yes, then first, print a blank and after that, set the counter 'blanks' back to zero.
After this subtest is done, go back and putchar whatever character was not found to be a blank.
The idea is, before putcharing a non-blank character, first do a test to see, if some blank(s) were counted before. If there were blanks before, print a single blank first and then reset the counter of blanks. That way, the counter is zero again for the next round of blank(s). If the first character on the line is not a blank, the counter couldn't have increased, hence no blank is printed.
One warning, I haven't gone very far into the book, so I'm not familiar with the syntax yet, so it's possible that the {} braces might be written in different places, but my example is working fine.
#include <stdio.h>
/* Copy input to output, replacing each string of one or more blanks by a single blank. */
main()
{
int c, blanks;
blanks = 0;
while ((c = getchar()) != EOF) {
if (c != ' ') {
if (blanks >= 1)
printf(" ");
blanks = 0;
putchar(c); }
if (c == ' ')
++blanks;
}
}
Like many other people, I am studying this book as well and found this question very interesting.
I have come up with a piece of code that only uses what has been explained before the exercice (as I am not consulting any other resource but just playing with the code).
There is a while loop to parse the text and one if to compare the current character to the previous one.
Are there any edge cases where this code would not work ?
#include <stdio.h>
main() {
// c current character
// pc previous character
int c, pc;
while ((c = getchar()) != EOF) {
// A truthy evaluation implies 1
// (learned from chapter 1, exercice 6)
// Avoid writing a space when
// - the previous character is a space (+1)
// AND
// - the current character is a space (+1)
// All the other combinations return an int < 2
if ((pc == ' ') + (pc == c) < 2) {
putchar(c);
}
// update previous character
pc = c;
}
}
for(nb = 0; (c = getchar()) != EOF;)
{
if(c == ' ')
nb++;
if( nb == 0 || nb == 1 )
putchar(c);
if(c != ' ' && nb >1)
putchar(c);
if(c != ' ')
nb = 0;
}
Here is my answer, I am currently in the same spot you were years ago.
I used only the syntax taught until this point in the books and it reduces the multiple spaces into one space only as required.
#include<stdio.h>
int main(){
int c
int blanks = 0; // spaces counter
while ((c = getchar()) != EOF) {
if (c == ' ') { // if the character is a blank
while((c = getchar()) == ' ') { //check the next char and count blanks
blanks++;
// if(c == EOF){
// break;
// }
}
if (blanks >= 0) { // comparing to zero to accommodate the single space case,
// otherwise ut removed the single space between chars
putchar(' '); // print single space in all cases
}
}
putchar(c); //print the next char and repeat
}
return 0;
}
I removed the break part as it was not introduced yet in the book,hope this help new comers like me :)
This is a solution using only the techniques described so far in K&R's C. In addition to using a variable to achieve a finite state change for distinguishing the first blank space from successive blank spaces, I've also added a variable to count blank spaces along with a print statement to verify the total number. This helped me to wrap my head around getchar() and putchar() a little better - as well as the scope of the while loop within main().
// Exercise 1-9. Write a program to copy its input to its output, replacing
// each string of one or more blanks by a single blank.
#include <stdio.h>
int main(void)
{
int blank_state;
int c;
int blank_count;
printf("Replace one or more blanks with a single blank.\n");
printf("Use ctrl+d to insert an EOF after typing ENTER.\n\n");
blank_state = 0;
blank_count = 0;
while ( (c = getchar()) != EOF )
{
if (c == ' ')
{
++blank_count;
if (blank_state == 0)
{
blank_state = 1;
putchar(c);
}
}
if (c != ' ')
{
blank_state = 0;
putchar(c);
}
}
printf("Total number of blanks: %d\n", blank_count);
return 0;
}
#include<stdio.h>
#include<stdlib.h>
int main(void)
{
int c, flag=0;
while((c=getchar()) != EOF){
if(c == ' '){
if(flag == 0){
flag=1;
putchar(c);
}
}else{
flag=0;
putchar(c);
}
}
return 0;
}
I hope this will help.
/*a program that copies its input to its output, replacing each string of one or more blanks by a single blank*/
#include <stdio.h>
#include<stdlib.h>
int main(void)
{
double c;
char blank = ' ';
while((c = getchar()) != EOF)
{
if(c == ' ')
{
putchar(c);
while(( c = getchar() )== ' ')
{
if(c != ' ')
break;
}
}
if(c == '\t')
{
putchar(blank);
while((c = getchar()) == '\t')
{
if(c != '\t')
break;
}
}
putchar(c);
}
return 0;
}
// K & R Exercise 1.9
// hoping to do this with as few lines as possible
int c = 0, lastchar = 0;
c = getchar();
while (c != EOF) {
if (lastchar != ' ' || c != ' ') putchar(c);
lastchar=c;
c=getchar();
}
Considering what's asked in the question, I have also made sure that the program runs smooth in case of various tabs, spaces as well as when they're clubbed together to form various combinations!
Here's my code,
int c, flag = 1;
printf("Enter the character!\n");
while ((c = getchar()) != EOF) {
if (c == ' '||c == '\t') {
c=getchar();
while(c == ' '|| c == '\t')
{
c = getchar();
}
putchar(' ');
if (c == EOF) break;
}
putchar(c);
}
Feel free to run all test cases using various combinations of spaces and tabs.
Solution1: as per topics covered in the k&R book:
#include <stdio.h>
int main()
{
int c;
while ((c = getchar()) != EOF)
{
if (c == ' ')
{while ( getchar() == ' ' )
; // ... we take no action
}
putchar(c);
}
return 0;
}
Solution2 : using program states:
int main()
{
int c, nblanks = 0 ;
while ((c = getchar()) != EOF)
{
if (c != ' ')
{ putchar(c);
nblanks = 0;}
else if (c==' ' && nblanks == 0) // change of state
{putchar(c);
nblanks++;}
}
return 0;
}
Solution3 : based on last seen char
int main()
{
int c, lastc = 0;
while ((c = getchar()) != EOF)
{
if ( c != ' ')
{putchar(c);}
if (c == ' ')
{
if (c==lastc)
;
else putchar(c);
}
lastc = c;
}
return 0;
}