Thread: Exercise 1-13 in K&M

Hey guys. Have been self teaching myself programming and got stuck on another problem. For problem 1-13 in the K&M, it asks for a program that makes a histogram of lengths of words. I figured that was a too too difficult and decided instead to just make a program that counts the numbers of words of different lengths, and to make it even easier I decided to just limit it to 10. Yet I can't even get that to work =\

Code:

#include <stdio.h>

main(){

#define IN 1
#define OUT 0

int c, i, j, state;
int word[10];

i = 0;
state = OUT;

for (j=0;j<=9;++j)
word[j]=0;


while ((c = getchar()) != EOF)
{
    if ((c == ' '|| c == '\b'|| c == '\n') && state == IN )
    {
    state = OUT;
    ++word[i - '1'];
    i = 0;
    }
    else if ((c==' '|| c=='\b'|| c=='\n') && state == OUT);

    else
    {
    ++i;
    state == IN;
}
}

printf("The number of words length 1 2 3 4 5 6 7 8 9 10 is\n\t\t\t\t");

for (j=0;j<=9;++j)
printf("%d ", word[j]);
}

when I print it just prints out all 0's. Not quite sure what I'm doing wrong. Thanks in advance.

It's mostly okay -- you just have a couple of mistakes. The logic of the code, which is the difficult bit, looks to be fine.

Code:

else
    {
        ++i;
        state == IN;
    }

Looks like a typo I think you wanted '=' (assignment) not '==' (equality).

Code:

    ++word[i - '1'];

'1' in single quotes is the character "1", not the number 1. This line will subtract the character encoding for 1 from i!

Finally,

Code:

    else if ((c==' '|| c=='\b'|| c=='\n') && state == OUT);

Nothing wrong with this, it's legal C and will work, but I don't really like the style. I know it's exactly the same but I'd write:

Code:

    else if ((c==' '|| c=='\b'|| c=='\n') && state == OUT)
        ; // do nothing

Ah. Yeah that got it. Thanks a bunch smokeyangel