Justifying text with strings

Code:

#include<stdio.h>
#include<string.h>
char line[100];
int i, n, len, width, gaps, spacestoshare, equalshare, extras;
int main() {
	gaps=0;
	printf("Please enter the width of the column:");
	scanf("%i", &width);
	fflush (stdin);
	printf("\nPlease enter a line of text");
	gets (line);
	len = strlen (line);
	while (len>width) {
		printf("\nError, your line of text is too long, please re-enter a");
		printf(" shorter line of text:");
		gets (line);
		len = strlen (line);
	}
	for (i=0; i<=width; i++) {
		if (line[i]==' ') {
			gaps++;
		}
	}
	spacestoshare = width-len;
	equalshare = spacestoshare/gaps;
	extras = spacestoshare%gaps;
	for (i=0; i<=width; i++) {
		printf("%c", line[i]);
		if (line[i]==' ') {
			for (n=0;n<=equalshare;n++) {
				printf(" ");
			}
		 	if (extras!=0) {
			 	printf(" ");
			 	extras-=1;
			}
		}
	}
}

im trying to run this program, it works, but i need the cursor to be directly after the last letter of text, and when i run it it ends up being about 7 spaces after the last letter. any help is greatly appreciated

first use code tags. Dont' use gets, don't use fflush(stdin) see the FAQ for answers to why. If you would indent your code you would realize you are short on } braces. See if you can change those first. for gets try fgets, and for fflush(stdin); use while((c=getchar()!='\n') && (c=getchar()!=EOF)); Also dont' forget to indent

like i sed, the program runs and i just need to fix the spaces problem, we have been told to use fflush so i will thank you very much

try using these types of flags then

Code:

%4i
%3.2f
%3.0f
%5d

those move your variables around.

Originally Posted by linuxdude

use while((c=getchar()!='\n') && (c=getchar()!=EOF))

You do realise that this will actually read two characters here, and would then fail on if you had something like" \n" (space, enter) in your buffer, right?

Quzah.

> we have been told to use fflush
So now you know not to trust everything your teacher tells you.

>but i need the cursor to be directly after the last letter of text
Are you left justifying or right justifying? Your code says left, and that means that the spaces will be on the right side of the output. If you want to right justify then you'll have spaces on the left side of every word, which requires a slightly different algorithm:

Code:

/* Bad code. Use at your own peril */
#include<stdio.h>
#include<string.h>
char line[100];
char *t;
int i, n, len, width, gaps, spacestoshare, equalshare, extras;
int main() {
  gaps=0;
  printf("Please enter the width of the column:");
  scanf("%i", &width);
  fflush (stdin);
  printf("\nPlease enter a line of text");
  gets (line);
  len = strlen (line);
  while (len>width) {
    printf("\nError, your line of text is too long, please re-enter a");
    printf(" shorter line of text:");
    gets (line);
    len = strlen (line);
  }
  for (i=0; i<width; i++) {
    if (line[i]==' ') {
      gaps++;
    }
  }
  spacestoshare = width-len;
  if (++gaps == 0) {
    equalshare = spacestoshare;
    extras = 0;
  }
  else {
    equalshare = spacestoshare/gaps;
    extras = spacestoshare%gaps;
  }
  for (t=strtok (line, " "); t!=NULL; t=strtok (NULL, " ")) {
    for (n=0;n<equalshare;n++) {
      printf(" ");
    }
    if (extras!=0) {
      printf(" ");
      extras-=1;
    }
    printf ("%s", t);
  }
  printf("\n");

  return 0;
}

>we have been told to use fflush so i will thank you very much
You should be more selective in who you follow orders from. Your teacher is stupid, and we will continue to nag you about it when you post code like that.