Specifically, the code is a solution to Exercise 1-9 in K&R C Programming Language 2nd Edition. I already solved it, but I have a question.
Write a program to copy its input to its output, replacing each string of one or more blanks by a single blank.
This code works and returns the desired output
int ch, last;
for (last = 0; (ch = getchar()) != EOF; last = ch)
if (ch == ' ' && last == ' ')
;
else
putchar(ch);
This version of the code doesn't work and instead prints literally the same input with excess spaces included.
int ch, last;
last = 0;
while ((ch = getchar()) != EOF)
if (ch == ' ' && last == ' ')
;
else
putchar(ch);
last = ch;
Could somebody tell me the difference between these two versions of code and why the latter version doesn't work?
last = ch; is not inside the loop, so you're not setting it until after you've read the entire file. So you're comparing with 0 every time.
If you want to have multiple statements in a loop, you have to put {} around them.
while ((ch = getchar()) != EOF) {
if (ch == ' ' && last == ' ') {
} else {
putchar(ch);
}
last = ch;
}
This is a perfect example of Why is it considered a bad practice to omit curly braces?
In the 2nd version you need to place the body of the while loop in a {} as it consist of two statements (if and last= assignment). Otherwise only the first statement is executed in the loop, and the 2nd after the loop terminates.
int ch, last;
last = 0;
while ((ch = getchar()) != EOF) {
if (ch == ' ' && last == ' ')
;
else
putchar(ch);
last = ch;
}
I would invert the condition:
if (ch != ' ' || last != ' ')
putchar(ch);
The answers about curly braces are correct.
There is, however, another way to transform the for() expression into a while() loop without an abundance of filigree in the code.
Below traversing a string stands in for getting characters from stdin.
#include <stdio.h>
int main() {
char *str = " Once upon a time in a land far away ";
int i = 0, ch, last = 0;
while( (ch = str[i++]) != '\0' )
if ( ch != ' ' || last != ' ' )
putchar( last = ch ); // NB: copy value
return 0;
}
Output
" Once upon a time in a land far away " // quotes added to show SPs
(Shameless duplication of #Alan Wind's observation about boolean logic.)
I have written two codes for splitting sentences to words, one of which worked while other one didn't.
Only difference is in the if condition under for loop.
First one gave no output. While second one is giving expected output. Please explain logically, don't teach DeMorgan Laws. I have purely applied my logic in both of them which I think is also correct in first code.
NOT WORKING ONE:
#include <stdio.h>
void main(void){
int i=0, m=0, n=0, j, l=0;
char sen[500], wrd[500][500];
printf("Input a sentence:\n");
gets(sen);
while(sen[i]!='\0'){
l++;
i++;
}
for(i=0;i<=l;i++){
if(sen[i] != ' '||sen[i] != '\0'){ //Here is the difference.
wrd[m][n]=sen[i];
n++;
}
else{
wrd[m][n]='\0';
puts(wrd[m]);
printf("\n");
m++;
n=0;
}
}
}
WORKING ONE:
#include <stdio.h>
void main(void){
int i=0, m=0, n=0, j, l=0;
char sen[500], wrd[500][500];
printf("Input a sentence:\n");
gets(sen);
while(sen[i]!='\0'){
l++;
i++;
}
for(i=0;i<=l;i++){
if(sen[i] == ' '||sen[i] == '\0'){ //Here is the difference.
wrd[m][n]='\0';
puts(wrd[m]);
printf("\n");
m++;
n=0;
}
else{
wrd[m][n]=sen[i];
n++;
}
}
}
Only difference is in the if condition of for loop.
First one have: if(sen[i] != ' '||sen[i] != '\0')
Second have: if(sen[i] == ' '||sen[i] == '\0')
You seem to be under the impression that the condition:
(sen[i] != ' '||sen[i] != '\0')
is the logical negation of:
(sen[i] == ' '||sen[i] == '\0')
That is not the case. DeMorgan's Laws dictate the logical equivalence of boolean statements.
The general case of what you want is !(A || B) == (!A && !B). So you can move a negation inside of a grouped expression, but a logical OR turns into a logical AND, and vice versa.
In your specific case, the inverse of the latter condition is:
!(sen[i] == ' ' || sen[i] == '\0')
Which becomes:
(sen[i] != ' ' && sen[i] != '\0')
Talking through this, you want to append to wrd if both sen[i] is not a space AND sen[i] is not a newline. You instead said you want to append to wrd if either sen[i] is not a space OR sen[i] is not a newline. Any character is either not a space or not a newline, so this condition will always be true.
First one gave no output because if(sen[i] != ' '||sen[i] != '\0') always evaluates to True.
Ok, I will try to explain you logically without going through DeMorgan Laws.
Suppose a string is entered:
This is a string.
Control will go through the loop.
Consider i=3, at indexing 3 character is 's' (sen[3]='s').
Now, if condition will be checked.
sen[3] != ' ' == True
sen[3] != '\0' == True
Now,
if(sen[i] != ' '||sen[i] != '\0') >>> True||True >>> True.
So, body of if block will be executed. Hence, no output.
Now i=4, at indexing 4 character is ' ' (sen[3]=' ').
Now, if condition will be checked.
sen[4] != ' ' == False
sen[4] != '\0' == True
Now, if(sen[i] != ' '||sen[i] != '\0') >>> False||True >>> True.
Again, body of if block will be executed. Hence, no output.
else block will never be executed which contains statements for output.
More logically, You want to append to wrd if either sen[i] is not a space OR sen[i] is not a null-character. Every character is either not a space or not a newline, so this condition will always be true. In your logic, else blok will be executed only when such character is encountered which is both space and \0(null). Since such character does not exist else block in your code will never execute. Hence, NO OUTPUT.
Here are the equivalences:
if(sen[i] == ' '||sen[i] == '\0'){
// code block #1
} else {
// code block #2
}
is logically equivalent to the following:
if(sen[i] != ' ' && sen[i] != '\0'){
// code block #2
} else {
// code block #1
}
Basically when you inverted individual conditions you neglected to also change the OR into AND.
About the logic in your code, in the code which is NOT working, in the sentence:
if(sen[i] != ' '||sen[i] != '\0'){
If an empty space was found you get the first parameter as a false, but the second one compares against the end of the string, but it's NOT equal! So this comparison becomes true... false or true gives you true... So the code do what you are expecting to do with a char or a word
I'm creating an absurdly simple program in C to mess around with getchar(). The program will print out what you input until you press enter and it will guarantee that your lines are no more than 80 chars each. To do this, I keep a running count of how many characters have been entered. Once the char count hits 70, the next space encountered will cause a line break. If no space is encountered between 70-80, a line break will occur regardless. I realize this is a super naive implementation and could be optimized left and right, but remember, I'm just messing around:
while ((c = getchar()) != '\n') {
if (lineLengthCount < 70) {
putchar(c);
lineLengthCount++;
}
else if (lineLengthCount < 80 && (c == ' ')) {
printf("%c\n", c);
lineLengthCount = 0;
}
else {
printf("%c\n", c);
lineLengthCount = 0;
}
}
The problem is the c == ' ' conditional doesn't seem to be actually checking for a space. I get output like this:
fitter happier more productive comfortable not drinking too much regula
r exercise at the gym three days a week getting on better with your ass
ociate employee contemporaries at ease eating well no microwaved dinner
where I was hoping that the lines would be truncated when a space was encountered. Instead, no matter what character is entered after line 70, a new line is created. am I missing something? Does ' ' really mean any character?
while ((c = getchar()) != '\n') {
if (lineLengthCount < 70) {
putchar(c);
lineLengthCount++;
}
else if (lineLengthCount < 80 && (c == ' ')) {
printf("%c\n", c);
lineLengthCount = 0;
}
else if (lineLengthCount >= 80){
printf("%c\n", c);
lineLengthCount = 0;
}
else{
putchar(c);
lineLengthCount++;
}
}
I think this should work. That should prevent the else from executing when there are less than 80 characters but the character isn't a space.
EDIT: I realized now that instead if lineLengthCount is less than 80 but the character isn't a space it wouldn't get printed at all, so I added another else at the end to fix it.
Wouldn't this be shorter and more concise?
while ((c = getchar()) != '\n') {
putchar(c);
if((c == ' ' && lineLengthCount >= 70) || lineLengthCount >= 80){
printf("\n");
lineLengthCount = 0;
}
else
++lineLengthCount;
}
There's a problem with your conditions: if lineLengthCount is > 70 but the next character is not a space, the last else will be hit, breaking line and resetting the counter.
If you're at all unsure of what's going on, I would recommend breaking up the "if" conditional into three explicit checks:
while ((c = getchar()) != '\n') {
lineLengthCount++;
if (lineLengthCount < 70) {
putchar(c);
}
if (lineLengthCount < 80 && (c == ' ')) {
printf("%c\n", c);
lineLengthCount = 0;
}
if (lineLengthCount == 80) {
printf("%c\n", c);
lineLengthCount = 0;
}
}
If you want to see what's happening, write some debugging output in each "if" to notice when it is called.
Once it works, and you understand why, you can edit it down and combine the "ifs" ...
Using ' ' is completely valid. You could also try using the C standard library function isspace() to check if the character is a space. This function returns a boolean expression, As in:
char ch = '0';
if (isspace(ch))
//the char is a space...
By 'is space' this function actually means is any 'whitespace' character, so that includes '\n' or any other character that prints as empty space.
You could also the decimal value 32 which means the same as a space:
if (ch==32)
However for readability I would rather use the first version!
"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;
}