Re: taking a "word" as input

arnuld <NoSpam@xxxxxxxxxx> writes:

On Wed, 30 Apr 2008 20:54:48 +0500, arnuld wrote:

by accident, it is actually exercise 6-4 of K&R2 :)

I don't have K&R2 so I don't know the end point of this exercise, so I
may have this wrong...

How about this code. It works fine:

/* A program that takes a single word as input. It will discard
* the whole input if it contains anything other than the 26 alphabets
* of English. If the input word contains more than 30 letters then only
* the extra letters will be discarded . For general purpose usage of
* English it does not make any sense to use a word larger than this size.
* Nearly every general purpose word can be expressed in a word with less
* than or equal to 30 letters.
*
* version 1.1
*
*/


#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>


enum MAXSIZE { WORDSIZE = 30 };

int getword( char *, int );


int main( void )
{
char ac[WORDSIZE];

if( getword( ac, WORDSIZE ) )
{
printf("%s\n", ac);
}

return EXIT_SUCCESS;

}


int getword( char *word, int max_length )
{
int c;
char *w = word;


while( isspace( c = getchar() ) )
{
;
}

I find { ; } a messy way of saying nothing, but that is a style
point. More important, if this will be used to read more than one
word (eventually) you need to skip anything that you don't count as a
word character, not just spaces.

while( --max_length )
{
if( isalpha( c ) )
{
*w++ = c;
}
else if( c == '\n' || c == EOF || isspace( c ) )
{
*w = '\0';
break;
}
else
{
return 0;

When the word ends because of this condition, why do you return 0
rather than the word you have read? You do have a word to return.

}

c = getchar();
}

/* I can simply ignore the if condition and directly write the '\0'
onto the last element because in worst case it will only rewrite
the '\n' that is put in there by else if clause.

I think the comment is confusing. Without the if below, you re-write
a 0 that is already there. A \n is never put into the buffer.

or in else if clause, I could replace break with return word[0].

I thought these 2 ideas will be either inefficient or
a bad programming practice, so I did not do it.
*/
if( *w != '\0' )
{
*w = '\0';
}

I'd just write *w = '\0';

return word[0];

That's a char. Given what you said about conversions and clarity, you
should really write return word[0] != '\0'; or maybe return !!word[0];

}

--
Ben.
.

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