Copying input to output without unnecessary Spaces in C - c

I'm trying to write a program in C that copies its input to its output while replacing each string of one or more Spaces with a single Space.
My code isn't doing that but is instead taking away every second character.
This is my code:
#include <stdio.h>
main()
{
int c;
int lastc;
lastc = 0;
while(getchar() != EOF){
c = getchar();
if(c == 32 && lastc == 32)
;
else
putchar(c);
lastc = c;
}
}

Your loop should look like:
while((c = getchar()) != EOF){
if(c == 32 && lastc == 32)
;
else
putchar(c);
lastc = c;
}
In your version you get a char with getchar while checking the condition for the while loop and then as a next step you again get a char with getchar. So the first one is not used in your code. Therefore it is taking away every second character.

Keep running in while loop until you get non-space character and print just one space after you get out.
int main()
{
int c;
bool space=false;
while ((c=getchar()) != EOF) {
while (isspace(c)) {
space = true;
c = getchar();
}
if (space) {
putchar(' ');
space = false;
}
putchar(c);
}
return 0;
}

I use fgets() function to getting string from input i.e stdin and store in the scroll string.
Then you must implement a way to analyze string to find spaces in it.
When you find first space, increase index if you face another space.
This is the code.
Code
#include <stdio.h>
int main(void){
char scroll[100];// = "kang c heng junga";
fgets(scroll, 100, stdin);
printf ("Full name: %s\n", scroll);
int flag = 0;
int i=0;
while (scroll[i] != '\0')
{
if (scroll[i] == ' ' )
flag=1;//first space find
printf("%c",scroll[i]);
if (flag==0){
i++;
}else {
while(scroll[i]==' ')
i++;
flag=0;
}
}
return 0;
}
Sample input: Salam be shoma doostane aziz
Program output: Salam be shoma doostane aziz
[Edit]
Use new string st to hold space eliminated string an print as output.
Also this code work for Persian string.
char scroll[100]={0};// = "kang c heng junga";
printf("Enter a string: ");
fgets(scroll, 100, stdin);
printf ("Original string: %s\n", scroll);
char st[100]={0};
int flag = 0;
int i=0;
int j=0;
while (scroll[i] != '\0')
{
if (scroll[i] == ' ' )
flag=1;//first space find
st[j]=scroll[i];
j++;
if (flag==0){
i++;
}else {
while(scroll[i]==' ')
i++;
flag=0;
}
}
printf("Eliminate Spaces: %s", st);

Related

The answer outputs blanks

Program task -
Enter a string, display it word for word on the screen.
The problem is that if you type a lot of spaces between words, they will show up when you check. How can this be fixed?
#include <stdio.h>
int main()
{
int inw = 0, i = 0, count = 0;
char s[10000];
printf("Print string (max 10000 sb):\n");
gets(s);
while (s[i] != '\0') {
if (s[i] != ' ' && s[i] != '\t') {
putchar(s[i]);
}
else if (s[i] == ' ') {
printf("\n");
}
i++;
}
return 0;
}
Ugly, but this gets the job done. Just need a flag to keep track of whether or not you just printed a new line. Also cleaned up unused variables and changed to using fgets
#include <stdio.h>
#include <stdbool.h>
int main()
{
int i = 0;
char s[10000];
bool justPrintedNewline = false;
printf("Print string (max 10000 sb):\n");
fgets(s, sizeof s, stdin);
while (s[i] != '\0') {
if (s[i] != ' ' && s[i] != '\t') {
putchar(s[i]);
justPrintedNewline = false;
}
else if (s[i] == ' ' && justPrintedNewline == false) {
printf("\n");
justPrintedNewline = true;
}
i++;
}
return 0;
}
Demo
You did a great job in the algorithm just fix a little thing.
You can create a flag and after space you increase the flag to 1.
Then you will know you will print just one space.
After printing " " check for a char that isn't " " for update the flag to 0.
When the flag is 1 DONT print anything just wait for another valid char.
Take care,
Ori
Only print a line-feeed when starting a word and after all is done.
Change code to:
If a space
-- print a '\n' when the prior character is a non-white-space.
Else
-- if (prior character is white-space) print a '\n'
-- print it
char prior = 'a';
while (s[i]) {
char ch = s[i];
if (ch != ' ' && ch != '\t') {
if (prior == ' ' || prior == '\t') {
putchar('\n');
}
putchar(ch);
}
prior = ch;
i++;
}
putchar('\n');
There is a bit of a trick to it: use a second, inside loop to skip past spaces and another to print words. The outer loop should only terminate if you have reached the end of the string.
while (s[i] != '\0')
{
// skip all spaces
while ((s[i] != '\0') && isspace( s[i] )) ++i;
// print the word
while ((s[i] != '\0') && !isspace( s[i] ))
{
putchar( s[i] );
}
// print the newline after a word
putchar( '\n' );
}
By the way, gets() is a really, really dangerous function. It should never have been included in the language. You are OK to use it for a homework, but in reality you should use fgets().
char s[1000];
fgets( s, sizeof(s), stdin );
The fgets() function is a bit more fiddly to use than gets(), but the above snippet will work for you.
Your other option for solving this homework is to use scanf() to read a word at a time from user input, and print it each time through the loop. I’ll leave that to you to look up. Don’t forget to specify your max string length in your format specifier. For example, a 100 char array would be a maximum 99-character string, so you would use "%99s" as your format specifier.

My program does not detect the last character while printing the line with its line number in c language?

Write a program that reads lines from the standard input. Each line is printed on the standard output preceded by its line number. Try to write the program so that it has no built in limit on how long a line it can handle.
#include <stdio.h>
int main()
{
int ch;
int pos = 1;
printf("Enter the line :\n");
while ((ch = getchar()) != EOF)
{
if (pos == 1)
{
printf("%d\t", pos);
pos++;
}
putchar(ch);
if (ch == '\n')
printf("%d\t", pos++);
}
}
OP's code is almost there.
Keep track of line number and column position.
User input might not end with a '\n'. Better to increment and print the line number when in column position 0 and data has arrived.
To avoid numeric limits, code could use a wider type than int.
#include <stdio.h>
int main(void) {
int ch;
long long line_count = 0;
long long column_position = 0;
printf("Enter the line :\n");
while ((ch = getchar()) != EOF) {
if (column_position == 0) {
printf("%lld\t", ++line_count);
}
column_position++;
putchar(ch);
if (ch == '\n') {
column_position = 0;
}
}
fflush(stdout); // insure any last line without a \n is printed before quitting
}
Is very easy (and faster) using fgets.
Just print the line, then search for a trailing newline (strchr can help) and if you find it, print the number of line.
#include <stdio.h>
#include <string.h>
int main(void)
{
char str[2048];
int ln = 0;
printf("%5d) ", ++ln);
while (fgets(str, sizeof str, stdin)) {
printf("%s", str);
if (strchr(str, '\n')) {
printf("%5d) ", ++ln);
}
}
puts("EOF");
return 0;
}

C- While loop not working

#include <stdio.h>
#include <string.h>
int main(void)
{
char str1[1000];
int i, letter, space = 0;
char ch = str1[i];
printf("Enter a sentence: ");
scanf("%[^\n]s", str1);
printf("you enter %s\n", str1);
while (i != strlen(str1)) {
if (ch != ' ') {
letter++;
} else if (ch = ' ') {
space++;
}
i++;
}
printf("%d %d", letter, space);
}
My while loop isn't working and I can't seem to locate the problem. I am using the terminal in ubuntu and after printing the user string, I get a blank line. I have to use Ctrl-Z to stop the script.
Mistakes I see: using uninitialised variables - local variables do not get initialised automatically.
Another is you do not read a character from the string within the loop.
The third is the unecessary and syntactically incorrect if (ch=' ') which should have been if (ch==' ')
#include<stdio.h>
#include<string.h>
int main(void){
char str1[1000];
int i = 0, letter = 0, space = 0; // initialise all to 0;
printf("Enter a sentence: ");
scanf("%[^\n]s",str1);
printf("you enter %s\n",str1);
while (i!=strlen(str1)){
char ch = str1[i]; // move this inside the loop
if (ch!= ' '){
letter++;
}else { // unnecessary - you already checked space
space++;
}
i++;
}
printf("%d %d\n", letter, space);
}
Program session:
Enter a sentence: hallo my friend
you enter hallo my friend
13 2
You need to initialize i to 0 at the beginning of the program.
int i,letter,space = 0;
the above line will only set space to 0 and not i.
This shall work :
int i = 0;
int letter = 0;
int space = 0;
char ch = ' ';
printf("Enter a sentence: ");
scanf("%[^\n]s",str1);
printf("you enter %s\n",str1);
while (i!=strlen(str1)){
ch = str1[i];
if (ch!= ' '){
letter++;
} else {
space++;
}
i++;
}
You need to initialise i to 0 before entering the loop. Also if (ch=' ') should be if (ch==' '). Next time, try to compile with warnings enabled (-Wall -Wextra).
Also, while probably not related to the issue here, using strlen() in a condition of a while statement can be a performance disaster if the string is too long. Furthermore, do note that the scanf() function may write beyond the end of the buffer if the input string is too long. For this specific case, I recommend getting the string using getline() (which dynamically allocates the necessary buffer).

C programming getchar()

I have two problems writing my code. The first problem I have is getting my getchar() to work if the user enters no text and just hits enter. I need to print an error if they do so and prompt the user to reenter the text in a loop until they do enter text. Is there any way to do so because everything I have tried has failed.
Here is the code I have for that section:
printf("Enter a text message: ");
while((c=getchar()) != '\n' && c != EOF)
{
text[i]= c;
i++;
}
I am new to C so I am limited on ideas to fix my dilemma. As you can see I am setting the input equal to an array. This leads to my second problem, I need to limit the input to no more than 100 characters. But, instead of giving the user an error I need to just chop off the extra characters and just read the first 100.
The simplest solution to your problem is to use fgets. We can give limit to the input so that it doesn't read the extra characters after the given limit.
Refer this sample code. Here I am printing the string if the user is not pressing Enter key:
#include <stdio.h>
int main()
{
char str[100];
fgets(str, 100, stdin);
if(str[0] != '\n')
{
puts(str);
}
return 0;
}
#include <stdio.h>
#define MAXSIZE 100
int main() {
char text[MAXSIZE+1]; // one extra for terminating null character
int i = 0;
int c;
while (1) {
printf("Enter a text message: ");
i = 0;
while ((c = getchar()) != '\n' && c != '\r' && c != EOF) {
if (i < MAXSIZE) {
text[i]= c;
i++;
}
}
if (i > 0 || c == EOF)
break;
printf("Empty string not allowed.\n");
}
text[i] = '\0';
printf("You entered: %s\n", text);
return 0;
}
Test code to detect non-compliant system:
#include <stdio.h>
int main() {
int c;
printf("Just hit enter: ");
c = getchar();
if (c == '\r')
printf("\\r detected!!!\n");
else if (c == '\n')
printf("\\n detected.\n");
else
printf("Yikes!!!\n");
return 0;
}
First of all getchar() can take only one character an input. It cannot take more than one character.
char c;
int total_characters_entered = 0;
do
{
printf ("Enter a text message: ");
c = getchar();
if (c != '\n')
{
total_characters_entered++;
}
} while (total_characters_entered <= 100);
I have written some code that will iterate in while loop until user has entered 100 characters excluding "Simple Enter without any text"
Please let me know if it does not satisfy your requirement. We will work on that.

counting the number of sentences in a paragraph in c

As part of my course, I have to learn C using Turbo C (unfortunately).
Our teacher asked us to make a piece of code that counts the number of characters, words and sentences in a paragraph (only using printf, getch() and a while loop.. he doesn't want us to use any other commands yet). Here is the code I wrote:
#include <stdio.h>
#include <conio.h>
void main(void)
{
clrscr();
int count = 0;
int words = 0;
int sentences = 0;
char ch;
while ((ch = getch()) != '\n')
{
printf("%c", ch);
while ((ch = getch()) != '.')
{
printf("%c", ch);
while ((ch = getch()) != ' ')
{
printf("%c", ch);
count++;
}
printf("%c", ch);
words++;
}
sentences++;
}
printf("The number of characters are %d", count);
printf("\nThe number of words are %d", words);
printf("\nThe number of sentences are %d", sentences);
getch();
}
It does work (counts the number of characters and words at least). However when I compile the code and check it out on the console window I can't get the program to stop running. It is supposed to end as soon as I input the enter key. Why is that?
Here you have the solution to your problem:
#include <stdio.h>
#include <conio.h>
void main(void)
{
clrscr();
int count = 0;
int words = 0;
int sentences = 0;
char ch;
ch = getch();
while (ch != '\n')
{
while (ch != '.' && ch != '\n')
{
while (ch != ' ' && ch != '\n' && ch != '.')
{
count++;
ch = getch();
printf("%c", ch);
}
words++;
while(ch == ' ') {
ch = getch();
printf("%c", ch);
}
}
sentences++;
while(ch == '.' && ch == ' ') {
ch = getch();
printf("%c", ch);
}
}
printf("The number of characters are %d", count);
printf("\nThe number of words are %d", words);
printf("\nThe number of sentences are %d", sentences);
getch();
}
The problem with your code is that the innermost while loop was consuming all the characters. Whenever you enter there and you type a dot or a newline it stays inside that loop because ch is different from a blank. However, when you exit from the innermost loop you risk to remain stuck at the second loop because ch will be a blank and so always different from '.' and '\n'. Since in my solution you only acquire a character in the innermost loop, in the other loops you need to "eat" the blank and the dot in order to go on with the other characters.
Checking these conditions in the two inner loops makes the code work.
Notice that I removed some of your prints.
Hope it helps.
Edit: I added the instructions to print what you type and a last check in the while loop after sentences++ to check the blank, otherwise it will count one word more.
int ch;
int flag;
while ((ch = getch()) != '\r'){
++count;
flag = 1;
while(flag && (ch == ' ' || ch == '.')){
++words;//no good E.g Contiguous space, Space at the beginning of the sentence
flag = 0;;
}
flag = 1;
while(flag && ch == '.'){
++sentences;
flag=0;
}
printf("%c", ch);
}
printf("\n");
I think the problem is because of your outer while loop's condition. It checks for a newline character '\n', as soon as it finds one the loop terminates. You can try to include your code in a while loop with the following condition
while((c=getchar())!=EOF)
this will stop taking input when the user presses Ctrl+z
Hope this helps..
You can implement with ease an if statement using while statement:
bool flag = true;
while(IF_COND && flag)
{
//DO SOMETHING
flag = false;
}
just plug it in a simple solution that uses if statements.
For example:
#include <stdio.h>
#include <conio.h>
void main(void)
{
int count = 0;
int words = 1;
int sentences = 1;
char ch;
bool if_flag;
while ((ch = getch()) != '\n')
{
count++;
if_flag = true;
while (ch==' ' && if_flag)
{
words++;
if_flag = false;
}
if_flag = true;
while (ch=='.' && if_flag)
{
sentences++;
if_flag = false;
}
}
printf("The number of characters are %d", count);
printf("\nThe number of words are %d", words);
printf("\nThe number of sentences are %d", sentences);
getch();
}
#include <stdio.h>
#include <ctype.h>
int main(void){
int sentence=0,characters =0,words =0,c=0,inside_word = 0,temp =0;
// while ((c = getchar()) != EOF)
while ((c = getchar()) != '\n') {
//a word is complete when we arrive at a space after we
// are inside a word or when we reach a full stop
while(c == '.'){
sentence++;
temp = c;
c = 0;
}
while (isalnum(c)) {
inside_word = 1;
characters++;
c =0;
}
while ((isspace(c) || temp == '.') && inside_word == 1){
words++;
inside_word = 0;
temp = 0;
c =0;
}
}
printf(" %d %d %d",characters,words,sentence);
return 0;
}
this should do it,
isalnum checks if the letter is alphanumeric, if its an alphabetical letter or a number, I dont expect random ascii characters in my sentences in this program.
isspace as the name says check for space
you need the ctype.h header for this. or you could add in
while(c == ' ') and whie((c>='a' && c<='z') || (c >= 'A' && c<='Z')
if you don't want to use isalpace and isalnum, your choice, but it will be less elegant :)
The trouble with your code is that you consume the characters in each of your loops.
a '\n' will be consumed either by the loop that scans for words of for sentences, so the outer loop will never see it.
Here is a possible solution to your problem:
int sentences = 0;
int words = 0;
int characters = 0;
int in_word = 0; // state of our parser
int ch;
do
{
int end_word = 1; // consider a word wil end by default
ch = getch();
characters++; // count characters
switch (ch)
{
case '.':
sentences++; // any dot is considered end of a sentence and a word
break;
case ' ': // a space is the end of a word
break;
default:
in_word = 1; // any non-space non-dot char is considered part of a word
end_word = 0; // cancel word ending
}
// handle word termination
if (in_word and end_word)
{
in_word = 0;
words++;
}
} while (ch != '\n');
A general approach to these parsing problems is to write a finite-state machine that will read one character at a time and react to all the possible transitions this character can trigger.
In this example, the machine has to remember if it is currently parsing a word, so that one new word is counted only the first time a terminating space or dot is encountered.
This piece of code uses a switch for concision. You can replace it with an if...else if sequence to please your teacher :).
If your teacher forced you to use only while loops, then your teacher has done a stupid thing. The equivalent code without other conditional expressions will be heavier, less understandable and redundant.
Since some people seem to think it's important, here is one possible solution:
int sentences = 0;
int words = 0;
int characters = 0;
int in_word = 0; // state of our parser
int ch;
// read initial character
ch = getch();
// do it with only while loops
while (ch != '\n')
{
// count characters
characters++;
// count words
while (in_word)
{
in_word = 0;
words++;
}
// skip spaces
while (ch == ' ')
{
ch = -1;
}
// detect sentences
while (ch == '.')
{
sentences++;
ch = -1;
}
// detect words
while ((ch != '\n')
{
word_detected = 1;
ch = -1;
}
// read next character
ch = getch();
}
Basically you can replace if (c== xxx) ... with while (c== xxx) { c = -1; ... }, which is an artifical, contrieved way of programming.
An exercise should not promote stupid ways of doing things, IMHO.
That's why I suspect you misunderstood what the teacher asked.
Obviously if you can use while loops you can also use if statements.
Trying to do this exercise with only while loops is futile and results in something that as little or nothing to do with real parser code.
All these solutions are incorrect. The only way you can solve this is by creating an AI program that uses Natural Language Processing which is not very easy to do.
Input:
"This is a paragraph about the Turing machine. Dr. Allan Turing invented the Turing Machine. It solved a problem that has a .1% change of being solved."
Checkout OpenNLP
https://sourceforge.net/projects/opennlp/
http://opennlp.apache.org/

Resources