Thread: printing long text

I am trying to print a sting. I want 50 char a line and i don't want the word at the end of the line to break....

I can't figure it out!!!


Here is what I have so far

int main(void)
{
const char text[]="This is a text I am trying to print this long\
text without having to break a word.";

int n;

while (text!=EOF)
{
n=strlen(text);
if (n>50)
{
n--;
while (text[n]!=' ')
{
continue;
}
text[n]='\n';
text[n+1]='\0';
}
printf("%s",text);
}
return 0;
}

I already posted this in the "word wrap" thread

here...

if you're looking for wordwrap in console,
look here.....

Code:

void mywordwrap(FILE *fp, char string[], short maxLineWidth) {

   short startIndex;
   short endIndex;
   short actualLineWidth;

   if (fp == NULL || string == NULL || maxLineWidth <= 0) {
      return;
   }

   
   startIndex = 0;
   endIndex = strlen(string);


    while (startIndex < endIndex) {

      
     while (startIndex < endIndex && isspace( string[startIndex] )) {
	 ++startIndex;
      }

      
      actualLineWidth = maxLineWidth;
      if (endIndex - startIndex > maxLineWidth) {
         while (actualLineWidth > 0 &&
            !isspace( string[ startIndex+actualLineWidth-1 ]) &&
            !isspace( string[ startIndex+actualLineWidth   ]) ) {
            --actualLineWidth;
         }
	}


       if (startIndex < endIndex && actualLineWidth > 0) {
	 fprintf(fp,"%.*s\n",actualLineWidth, &string[ startIndex ]);
      }

      
      startIndex += actualLineWidth;
   }
}

//syntax mywordwrap(stdout, chararray, 70);

put the above function above your main function,

in main(), you have

char text[]="This is a text I am trying to print this long
text without having to break a word.";

you said you wanted a 50 char line....
so do this

mywordwrap(stdout, text, 50)

got it now ?


GOOD LUCK

thanks that is nice of you but I am a beginner C This is a homework problem....I need something much simpler

I tried to use printf("%0.50s",text); it works but it prints only one line

quzah I hope you are still online

I am having trouble writting the codes for your instructions...I need help

This is what I have so far:

int main(void)
{
const char text[]="This is a text I am trying to print this long\
text without having to break a word.";

//count the caracters in a word
//if the total characters + the word length + 1 is >= X
// print a new line
// total characters = 0
//print the word
//total character += word length + 1
//print a space


int k;

for (k=0;k<strlen(text);k++)
{

if (((strlen(text)) + (word length + 1))>=60)
printf("\n");


printf(" ");
}
return 0;
}

ok dude...
here's a clue...

1) Take a list of all the places in the sentence where
there are spaces

2) find the nearest space to 50 (if you want to allow 50 characters
per line)

3) use a for loop to print till that place

4) loop it again for the whole sentence

Code:

	
                int array[50]; //holds places where space character is in the given string

	for (i = 0; i<strlen(text); i++)
	{
		if (text[i] == ' ')
		{
			array[k] = i+1;
                                                printf("%d\n", array[k]);
                                                k++;
			
		}
	}

The above code will give you wherever there are spaces in
a given string.

Try working the rest of it for yourself...

hope this worked as a clue for you..

something like this will probably work fine:

Code:

int spaceplace,counter=0,printcounter=0;
char sentence;


....
....
while(couner<strlen(sentence))
{
  while(counter<25)
  {
    if(sentence[counter]=' ')
      spaceplace = counter;
    counter++;
  }
  for(printcounter=0;printcounter<spaceplace;printcounter++)
    printf("%c",sentence[printcounter]);
  printf("\n");

  counter=printcounter+1;
}

I think somethin like that will work, or atleast give you an idea