Files that do not end with newline?

Earlier I posted a program as a solution to a K&R exercise that reads a file/text stream and prints out a histogram of number of occurrences of word lengths. I think I've worked out nearly all the bugs and this is what I now have:

Code:

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/* Corresponding K&R section: 1.6 */

/* Prints vertical histogram of lengths of words in input*/

#include <stdio.h>

/* NOTE: given enough horizontal space, can scale to any two-digit maximum by
  modifying this alone... gets messy with three digits, but I could
  make the numbers on the x-axis display vertically down to make it infinitely
  scalable. */

#define MIN_WORD_LENGTH 1
#define MAX_WORD_LENGTH 20         

int main(void)
{
        //Holds characters
        int c = 0;                       
        int lastchar = 0;

        //current number of contiguous non-whitespace chars                                               
        int currentLength = 0;                       

        //holds number of occurrences of each length                               
        int wordLengthFrequencies[MAX_WORD_LENGTH] = {0};

        //highest number of occurrences encountered
        int maxFrequency = 0;

        /* ---------------- Collect length data --------------- */
       
        while((c = getchar()) != EOF)
        {
                //Are we currently inside a word?
                if(currentLength >= MIN_WORD_LENGTH)
                {
                        //Are we encountering whitespace?
                        if(c == ' ' || c == '\t' || c == '\n' || c == '\r' || c == '\v')
                        {
                                /* We've reached the end of a word. Update array */
                                wordLengthFrequencies[currentLength - 1]++;                               
                                currentLength = 0; //Reset
                        }
                       
                        /* No whitespace, so we're still inside a
                          word. Update currentLength if it hasn't
                          maxed out*/
                        else if(currentLength < MAX_WORD_LENGTH)
                                currentLength++;
                }
               
                /* Have we just encountered the start of a word? */
                else if(c != ' ' && c != '\t' && c != '\n' && c != '\r' && c != '\v')
                        currentLength = 1;

                lastchar = c;                       
        }

        /* Handle case where file does not end on newline */
        if(lastchar != '\n' && currentLength != 0)
                wordLengthFrequencies[currentLength - 1]++;       
       
        /* --------------- Print Histogram ------------- */
       
        printf("\n\n");

        int i = 0;

        /* Determine maximum frequency */
        for(i = MIN_WORD_LENGTH; i <= MAX_WORD_LENGTH; i++)
        {
                if(wordLengthFrequencies[i - 1] > maxFrequency)
                        maxFrequency = wordLengthFrequencies[i- 1];
        }


        /* Start printing the graph starting from the maximum frequency */
        for(c = maxFrequency; c >= 1; c--)
        {

                /* Make sure graph will still be aligned even for 7-digit frequencies
                        i.e. if we are operating on a large file */
                printf("%7d | ", c);       

                for(i = 0; i < MAX_WORD_LENGTH; i++)
                {
                        if(wordLengthFrequencies[i] >= c)
                                printf("*  ");        //fill in where appropriate
                        else
                                printf("  ");
                }
                printf("\n");
        }
       
        /* print the x-axis and legend */
        putchar('\t');
        for(i = MIN_WORD_LENGTH; i <= MAX_WORD_LENGTH; i++)
                printf("---");
       
        printf("\n\t");
        for(i = MIN_WORD_LENGTH; i <= MAX_WORD_LENGTH; i++)
                printf("%3d", i);
        putchar('+');
        printf("\n\nx-axis: word length\ny-axis: # of occurrences\n\n");
        return 0;
}

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I noticed that my program failed to count the last word in the file if the file did not end on a newline and had no trailing whitespace after the last word. So I added this particular snippet to take care of that:

Code:

---

/* Handle case where file does not end on newline */
if(lastchar != '\n' && currentLength != 0)
        wordLengthFrequencies[currentLength - 1]++;

---

By my logic, if the last character to be read before EOF was not '\n', this means
the file did not end on a newline. But, this does not necessarily mean I should
just go ahead and increment the array member corresponding to what's left
in currentLength. It may very well be 0 because I may have trailing whitespace after the last word which would cause currentLength to be reset when it is processed. What's worse, this would mean I'm incrementing the -1 index, which is out of bounds. No matter what whitespace is trailing, currentLength will be 0, so I just make sure it isn't. With that in place, I'm fairly certain my program is solid.

I might be being anal about this for an exercise out of a programming book, but I guess I see little point in trying to learn C with exercises if I don't make damned sure my solutions are airtight, given how easy it is to proverbially "shoot myself in the foot." I was wondering if anyone else could see any flaws in my logic above or if there's something I overlooked.

Keep track of the last character read and if you hit EOF, you can look at your last character. If it is a letter, it's part of a word, so you can just see the current length. If it's not, then you don't need to increment any word count, because the file ended in a space or something not-a-word.

Code:

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if( isalpha( lastchar ) )
    do something with this word
else
    not a word - don't do something with a word

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Alternately, you can decide if you want to check for numbers or letters or punctuation or something else. If all you care about is if it is white space, you can try flopping the if and else chunks, and use isspace, or some other similar function.


Quzah.

Those functions do look useful for what I'm doing. I think the point of the exercise, though, is to implement a solution using only the tools Kernighan and Ritchie have described so far. I feel like I'd be cheating if I used isspace() or isalpha() but I see your point and thanks for the info.

I was mainly wondering if, given my solution as it stands, there is any conceivable way someone could throw input at it that would screw it up. Should have made that clearer.

Code:

---

                if(currentLength >= MIN_WORD_LENGTH)
                {
                        //Are we encountering whitespace?
                        if(c == ' ' || c == '\t' || c == '\n' || c == '\r' || c == '\v')
                        {
                                /* We've reached the end of a word. Update array */
                                wordLengthFrequencies[currentLength - 1]++;                               
                                currentLength = 0; //Reset
                        }
                       
                        /* No whitespace, so we're still inside a
                          word. Update currentLength if it hasn't
                          maxed out*/
                        else if(currentLength < MAX_WORD_LENGTH)
                                currentLength++;
                }

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What happens if you hit your word length? I don't see you covering that scenario. Are you just ignoring the extra letters? So if the only thing in the file is a word of MAX_WORD_LENGTH+1 characters, you will just count it as a single word of MAX_WORD_LENGTH?

But I think you've pretty much got it covered.


Quzah.

Quote:

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Originally Posted by quzah

What happens if you hit your word length? I don't see you covering that scenario. Are you just ignoring the extra letters? So if the only thing in the file is a word of MAX_WORD_LENGTH+1 characters, you will just count it as a single word of MAX_WORD_LENGTH?

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Well, I realized I don't have infinite amounts of horizontal room over which to display every possible length, so I group 20 or more character-long words in the same category. When I display the histogram, I label that category as 20+.

Quote:

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Originally Posted by quzah

But I think you've pretty much got it covered.

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Sick. Onto the next exercise, then. Thanks for the help.

-Evan Williams