I need to write a program that prints its input one word per line. Here's what I got so far:
#include <stdio.h>
main(){
int c;
while ((c = getchar()) != EOF){
if (c != ' ' || c!='\n' || c!='\t')
printf("%c", c);
else
printf("\n");
}
}
The logic is pretty simple. I check to see if the input is not a newline, tab or space, and in that case it prints it, otherwise prints a newline.
When I run it, I get results like this:
input--> This is
output--> This is
It prints the whole thing. What goes wrong here?
if (c != ' ' || c!='\n' || c!='\t')
This will never be false.
Perhaps you meant:
if (c != ' ' && c!='\n' && c!='\t')
instead of using printf try putchar, also as per above comments, you should use && instead of ||.
here is my code-
#include<stdio.h>
main()
{
int c, nw; /* nw for word & c for character*/
while ( ( c = getchar() ) != EOF ){
if ( c != ' ' && c != '\n' && c != '\t')
nw = c;
else {
nw = '\n';
}
putchar (nw);
}
}
this code will give you the desired output
you can use if you want the strtok function in string.h library which can cut the input into many words by providing a delimiter.
Here is a perfect code commented which can fit to your needs
#include <stdio.h>
#include <string.h>
int main(int argc, char *argv[])
{
char line[1000]=""; // the line that you will enter in the input
printf("Input the line:\n>>");
scanf("%[^\n]",line); // read the line till the you hit enter button
char *p=strtok(line," !#$%&'()*+,-./'"); // cut the line into words
// delimiter here are punctuation characters (blank)!#$%&'()*+,-./'
printf("\nThese are the words written in the line :\n");
printf("----------------------------------------\n");
while (p!=NULL) // a loop to extract the words one by one
{
printf("%s\n",p); // print each word
p=strtok(NULL," !#$%&'()*+,-./'"); // repeat till p is null
}
return 0;
}
If we execute the code above we will get
Input the line:
>>hello every body how are you !
These are the words written in the line :
----------------------------------------
hello
every
body
how
are
you
suggest the code implement a state machine,
where there are two states, in-a-word and not-in-a-word.
Also, there are numerous other characters that could be read
(I.E. ',' '.' '?' etc) that need to be check for.
the general logic:
state = not-in-a-word
output '\n'
get first char
loop until eof
if char is in range a...z or in range A...Z
then
output char
state = in-a-word
else if state == in-a-word
then
output '\n'
state = not-in-a-word
else
do nothing
end if
get next char
end loop
output '\n'
I think the simple solution would be like
#include <stdio.h>
int main(void) {
// your code goes here
int c;
while((c=getchar())!=EOF)
{
if(c==' ' || c=='\t' || c=='\b')
{
printf("\n");
while(c==' ' || c=='\t' || c=='\b')
c=getchar();
}
if(c!=EOF)
putchar(c);
}
return 0;
}
Related
I'm writing a program which takes input and outputs it one word per line.
#include <stdio.h>
int main()
{
int c;
while( (c = getchar()) != EOF)
{
if( c != ' ')
putchar(c);
else
putchar('\n');
}
}
The program stops when it encounters the EOF character. When I input something like "Hello World^Z" (I am using Windows) I get the following output:
C:\Coding\C>a
Hello World^Z
Hello
World→
And it still waits for me to input something. What does the → symbol mean?
This is my C code:
#include <stdio.h>
int main()
{
int c = getchar();
while (c != EOF) {
if (c != '\n')
putchar(c);
else putchar(32);
c = getchar();
}
return 0;
}
I want to make a program that prints out a paragraph with newlines, by replacing the \n character with spaces. The problem is, it only prints out the last line, when I use the code provided above.
For, example, for the text:
This is
my
text
the result printed is text.
The paragraph is properly printed when I remove the if(), else conditions, and only leave the putchar(), without trying to replace anything.
What's the problem?
Your input file has CRLF newlines. You need to ignore the CR characters when you're replacing LF with space. Otherwise, printing the CR characters will go back to the beginning of the line and overwrite what was already printed.
#include <stdio.h>
int main()
{
int c;
while ((c = getchar()) != EOF) {
if (c == '\n') {
// replace newline with space
putchar(' ');
} else if (c == '\r') {
// ignore CR
} else {
putchar(c);
}
}
return 0;
}
#include <stdio.h>
int main() {
int c;
while(getchar() != EOF) {
if (getchar() == ' ') {
c++;
}
printf("%i", c);
}
}
I realized that typing in a sentence like the one you're reading right
I\nrealized\nthat\ntyping\nin\n\a\n ...
i believe that's how it's being read, getchar() does not reach the EOF to make the condition in the while parentheses false..
my goal here is to make a program that takes in input from me..
reads it
if there are any spaces
it counts on a counter
when EOF is reached
the condition to keep reading it becomes false
the counter value gets printed out on the screen
to show me how many spaces i had in my entire input..
is it impossible? is that why people just use scanf() ?
this is the output i get when trying something
user#user:/c# ./a.out
hello stackoverflow this does not do what i want it to
001111111222223344445666677
You need to put the result of getchar() into a variable:
int ch;
while ((ch = getchar()) != EOF)
You shouldn't call getchar() a second time to check if it's a space, since that will read a second character so you'll be testing every other character, just compare the variable:
if (ch == ' ')
And if you want to see the total number of spaces, put the printf() at the end of the loop, not inside it.
So the whole thing should look like:
#include <stdio.h>
int main() {
int counter=0;
int ch;
while((ch = getchar()) != EOF) {
if (ch == ' ') {
counter++;
}
}
printf("%i\n", counter);
}
To send EOF from the terminal, type Control-d on Unix, Control-z on Windows.
I'm new to C, I have been asked to make a program in C asking to print each letter after a '.' after a user has entered an input.
For example if the user enters a..bcd..e.f..gh the output should be befg
which is the exact example I have been given in class.
I assume this would need to use pointers but I am unsure how to deal with this question, here is what I have tried to do so far. I know it is not correct, please help me understand how to use pointers to deal with this question.
#include <stdio.h>
int main() {
char *c, count =0;
printf("enter some characters");
scanf("%s", &c);
while( c != EOF ) {
if (c != '.') {
count ++;
}
else; {
printf("%s", c);
}
}
}
The program can look the following way
#include <stdio.h>
#define N 100
int main( void )
{
char s[N];
const char DOT = '.';
printf( "Enter some characters: " );
fgets( s, N, stdin );
for ( char *p = s; *p; ++p )
{
if ( p[0] == DOT && p[1] != DOT ) putchar( p[1] );
}
putchar( '\n' );
}
Its output might look like
Enter some characters: a..bcd..e.f..gh
befg
Take into account that here any symbol after a dot (except the dot itself) is printed. You can add a check that there is a letter after a dot.
You don't really need pointers for this, or even an array. Basically it's a simple state engine: read each character, if '.' is encountered, set a flag so the next character is printed.
#include <stdio.h>
int main() {
int c, flag = 0;
while ((c = getchar()) != EOF) {
if (c == '.')
flag = 1;
else if (flag) {
putchar(c);
flag = 0;
}
}
return 0;
}
There are some errors in your code:
- char* c means a pointer to one or more characters.
But where does it point to?
- scanf reads a string up to an "white space". White space characters are the space itself, a newline, a tab character or an EOF. scanf expects a format string and a pointer to a place in memory where it places what it reads. In your case c points to an undefined place and will overwrite whatever there is in memory.
- why do you place a ";" after the else? The else clause will end with the ";". So your program will do the print every time.
It helps you a lot if you format your code in a more readable way and give the variable names that give hint what they are used for.
Another very important thing is to initialize every variable that you declare. Errors with uninitialized variables are sometimes very hard to find.
I would do it this way:
#include <stdio.h>
int main(int argc, char* argv[])
{
// I read every single character. The getchar function returns an int!
int c = 0;
// This marks the program state whether we must print the next character or not
bool printNext = false;
printf("enter some characters");
// We read characters until the buffer is empty (EOF is an integer -1)
do
{
// Read a single character
c = getchar();
if ( c == '.')
{
// After a point we change our state flag, so we know we have to print the next character
printNext = true;
}
else if( c != EOF )
{
// When the character is neither a point nor the EOF we check the state
if( printNext )
{
// print the character
printf( "%c", c );
// reset the state flag
printNext = false;
}
}
// read until the EOF occurs.
}
while( c != EOF );
}
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int main()
{
char letter;
char *c;
c = malloc(256);
printf("enter the string : ");
scanf("%s", c);
while( (letter=*(c)) != '\0' )
{
if (letter == '.')
{
c++;
letter=*c;
if(letter!='.')
printf("%c",letter);
else
{
while(letter=='.')
{
c++;
letter=*c;
}
printf("%c",letter);
}
}
c++;
}
printf("\n");
}
I am new to C programming and I am currently trying to teach myself how to create a C program that can count words and lines in the input stream and print the two totals to the standard output.
What I am actually trying to do is to have the program count the number of lines and count the number of words depending on the definition of a word in which I feel that I am off.
I want the words to exclude blanks, tabs, newlines, hyphens, or colons. While having the program output the results (words and lines) as decimals.
#include<stdio.h>
int main()
{
int iochar;
int words;
int lines;
printf("Enter something here:\n\n");
while ((iochar = getchar ()) !=EOF)
{
if((iochar == ' ') || (iochar == '\t') || (iochar == '\n'))
putchar(iochar);
}
return 0;
}
Am I totally off on this program?
If your question is how to fix the compile error, that's simple. Add one more closing brace at the end.
But your program will still do only one pass through the loop and will print only one character if and only if the user types a space, tab or newline. No matter what the user types, the program will then terminate. I doubt that's what you wanted.
I suspect this is what you intended:
while ((iochar = getchar ()) !=EOF)
{
if((iochar == ' ') || (iochar == '\t') || (iochar == '\n'))
{
putchar(iochar);
}
}
return 0;
After your "I am trying to have thee numbers be right justified in an 8-column field ..." I cannot understand what you are trying to say :(
int words = 0;
int lines = 0;
char buffer[1024];
while(fgets(buffer, sizeof buffer, stdin))
{
lines++;
if(buffer[0] == '\n')
continue;
char *tmp = buffer-1;
while(tmp = strchr(tmp+1, ' '))
words++;
words++; /* count last word before \0*/
}
printf("lines: %d, words: %d\n", lines, words);
is that what you need/want?
The error message is:
Test.c:20:1: error: expected declaration or statement at end of input
It does not compile because you are missing a }.
Had you properly indented your code, like so, you would have found your mistake:
#include<stdio.h>
int main() {
int iochar;
int words;
int lines;
printf("Enter something here:\n\n");
while ((iochar = getchar ()) !=EOF)
{
if((iochar==' ')||(iochar=='\t')||(iochar=='\n'))
{
putchar(iochar);
iochar = getchar();
}
return 0;
}
Yet another example of the importance of readability :)