Help with K&R Book Exercise

This is a discussion on Help with K&R Book Exercise within the C Programming forums, part of the General Programming Boards category; Hi! I am learning C with the K&R book. I have a problem with exercise 1.13 from page 24 it ...

  1. #1
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    Help with K&R Book Exercise

    Hi! I am learning C with the K&R book.
    I have a problem with exercise 1.13 from page 24 it says
    "Write a program to print a histogram of the lengths of words in its input. It is easy to draw the histogram with the bars horizontal; a vertical orientation is more challenging"
    I 'm not sure if i understand what it wants (my native language is not English)
    For example if a i write "Hello to all!" the program should print something like
    1 **** (Hello)
    2 ** (to)
    3 **** (all!)
    is this right? if so any hints?


    Much thanks,
    Cheers!

  2. #2
    Kernel hacker
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    Yes, that looks about right.

    What portion of this are you struggling with. Whilst this may not count as homework per se, me giving the answer directly would not be the right solution.

    --
    Mats
    Compilers can produce warnings - make the compiler programmers happy: Use them!
    Please don't PM me for help - and no, I don't do help over instant messengers.

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    First all, thanks for the quick answer!

    The problem i have is how to store the number of letters every word have. I can make an array to hold int's with the number of letters every word have and then do a for ..i am doing something like this..which is far from good

    Code:
    int c, i, j, count, word_number = 0, word[SIZE];
    
        while ((c = getchar()) != '1') {
            if (c != ' ' && c != '\t' && c != '\n') {
                ++count;
            }
            else {
                word[word_number] = count;
                count = 0;
                ++word_number;
            }
    
        }
        for (i = 0; i < word_number; ++i) {
            printf("%d: ", i+1);
            for (j = 0; j < word[i]; ++j) {
                printf("*");
            }
            printf("\n");
        }
    PS_ Ignore that 1 in the while condition, is because right now i am in windows..and Crt+Z don't works like in Linux.

  4. #4
    CSharpener vart's Avatar
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    Quote Originally Posted by Alejandrito View Post
    PS_ Ignore that 1 in the while condition, is because right now i am in windows..and Crt+Z don't works like in Linux.
    And why do you think so?

    Code:
    #include <stdio.h> 
    int main(int argc, char *argv[]) 
    { 
    	int c = getchar();
    	if(c == EOF)
    		puts("EOF detected");
    	else
    		printf("Char found - %c", c);
    	return 0; 
    }
    I press Ctrl+z,Enter
    Output
    Code:
    ^Z
    EOF detected
    The first 90% of a project takes 90% of the time,
    the last 10% takes the other 90% of the time.

  5. #5
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    Quote Originally Posted by vart View Post
    And why do you think so?

    Code:
    #include <stdio.h> 
    int main(int argc, char *argv[]) 
    { 
    	int c = getchar();
    	if(c == EOF)
    		puts("EOF detected");
    	else
    		printf("Char found - %c", c);
    	return 0; 
    }
    I press Ctrl+z,Enter
    Output
    Code:
    ^Z
    EOF detected
    In linux, i don't need to press enter...
    Back to my question i did it.
    Code:
    #include <stdio.h>
    
    #define SIZE 100
    
    main()
    {
        int c, i, j, count = 0, word_number = 0, word[SIZE];
    
        while ((c = getchar()) != EOF) {
            if (c != ' ' && c != '\t' && c != '\n') {
                ++count;
            }
            else {
                word[word_number] = count;
                count = 0;
                ++word_number;
            }
    
        }
    
        for (i = 0; i < word_number; ++i) {
            printf("%d: ", i+1);
            for (j = 0; j < word[i]; ++j) {
                printf("*");
            }
            printf("\n");
        }
    }
    Cheers!

  6. #6
    CSharpener vart's Avatar
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    better make it
    int main(void)

    and add return 0; at the end

    to be more standard
    The first 90% of a project takes 90% of the time,
    the last 10% takes the other 90% of the time.

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