Im writing a simple program to count the number of character user is entered, and i wrote an if to check wether there is a newline but still printing it..
the code:
#include <stdio.h>
int main()
{
char ch;
int numberOfCharacters = 0;
printf("please enter a word, and ctrl + d to see the resault\n");
while ((ch = getchar()) != EOF)
{
if (numberOfCharacters != '\n')
{
numberOfCharacters++;
}
}
printf("The number of characters is %d", numberOfCharacters);
return 0;
}
what am i doing wrong?
Think about this line:
if (numberOfCharacters != '\n')
how can it make sense? You are comparing the number of characters read so far with a newline, it's like comparing apples to oranges and surely won't work. It's another variable that you should check...
Change your loop to this.
while ((ch = getchar()) != EOF)
{
if(ch != '\n')
numberOfCharacters++;
}
Related
I don't understand why the printf() call after the while loop does not get executed?
int main(){
while((getchar()) != EOF){
characters ++;
if (getchar() == '\n'){
lines++;
}
}
printf("lines:%8d\n",lines);
printf("Chars:%8d",characters);
return 0;
}
I think you are trying to do that
#include<stdio.h>
int main()
{
int characters=0,lines=0;
char ch;
while((ch=getchar())!= EOF)
{
if (ch == '\n')
lines++;
else
{
characters++;
while((ch=getchar())!='\n'&&ch!=EOF); //is to remove \n after a character
}
}
printf("lines:%8d\n",lines);
printf("Chars:%8d",characters);
return 0;
}
Output:
a
s
d
f
^Z
lines: 1
Chars: 4
Process returned 0 (0x0) execution time : 8.654 s
Press any key to continue.
Note: ^Z(ctrl+z) is to send EOF to stdin (in windows)
You have to be careful of you're treatment in the while loop. Indeed, you are missing every caracter read in your while statement. You have to save this input, in order to use it afterwards.
The proper syntax would be while(( c = getchar()) != EOF)
You are probably looking for something like this:
#include <stdio.h>
int main()
{
int characters = 0;
int lines = 0;
int c;
while ((c = getchar()) != EOF) {
characters++;
if (c == '\n') {
lines++;
characters--; // ignore \n
}
}
printf("lines: %8d\n", lines);
printf("Chars: %8d", characters);
return 0;
}
while ((c = getchar()) != EOF) might look a bit confusing.
Basically it calls getchar, puts the returned valuee into c ands then checks if c equals EOF.
So i have this code but when i run it it always says that the amount of words are 1 no matter how many i put in and hopefully it is in easy fix. I tried changing the scanf to just %s but that didn't work because it only printed out the first word but it got the number of words right.
#include <stdio.h>
int main()
{
int words = 0;
char ch,sen[100]="", i;
printf("Enter a sentence ended by a '.', a '?', or a '!':");
scanf("%[^\n]", sen);
while ((ch = getchar()) != '\n') {
if (ch == ' ')
words++;
}
words++;
for(i=0;sen[i];i++)
{
if( (sen[i]>=97) && (sen[i]<=122) )
sen[i]-=32;
}
printf("Capitalized sentence: %s\n", sen);
printf("Total number of words:%d\n", words);
return 0;
}
Your program has a major bug. scanf() won't read/store the newline. Then the newline is read by getchar(). This loop will only execute once.
while ((ch = getchar()) != '\n') {
if (ch == ' ')
words++;
}
Hence you are getting only 1 word. Why you are using 2 methods to take input.
Either use scan() and manipulate variable "sen" or use getchar() and store character 1 by 1 in sen.
// don't use scanf() in this case
i=0;
while ((ch = getchar()) != '\n') {
if (ch == ' ')
words++;
sen[i++] = ch;
}
Recommended will be to use fgets() to get such inputs. Learn about it.
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.
I have just started learning C, been reading a C textbook by Keringhan and Ritchie. There was this example in the textbook, counting characters from user input. Here's the code:
#include <stdio.h>
main()
{
long nc;
nc = 0;
while(getchar() != EOF) {
if (getchar() != 'q')
++nc;
else
break;
}
printf("%ld\n", nc);
}
The problem is, when I execute the code, if I input only one character per line, when I input "q" to break, it doesn't do so. I have to type some word per line, only after that it will break the loop. Also, it only counts the half of the characters of the word. I.e. if I input
a
b
russia
it will only print '5' as final result.
Could you please explain to me why is this happening?
This works, but only when you finish off with an Enter. So, this will count the characters until the first "q" appears. That is just how getchar() and getc(stdin) work.
#include <stdio.h>
int main() {
char c = 0;
long count = 0;
short int count_linebreak = 1; // or 0
while((c = getchar()) != EOF) {
if(c != 'q' && (count_linebreak || (!count_linebreak && c != '\n'))) {
++count;
}else if(c == 'q') {
printf("Quit\n");
break;
}
}
printf("Count: %ld\n",count);
return 0;
}
A StackOverflow question about reading stdin before enter
C read stdin buffer before it is submit
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/