c - Need some suggestions on how to print a histogram more neatly -

i'm writing program read input , give histogram of character count k & r - ex. 1.13

any suggestions on how can improve code? matter whether or not if test status in condition or out first? have noticed in examples people test see if c blank or tab first.

i think need revisit histogram. doesn't scale results. draws hyphen based on length.

revised make little bit more readable think.

// print histogram of length of words in it's input.  #include <stdio.h> #define in 1 #define out 2 #define max 99  int main(){      int c;  // character     int countofletters = 0;     int insideword = out;     int frequencyoflengths[max];     int longestwordcount = 0;     int i, j; // counters      (i = 0; < max; i++){         frequencyoflengths[i] = 0;     }      while ((c = getchar()) != eof){         if (c == ' ' || c == '\n' || c == '\t'){             if (insideword == in){                 if (countofletters > max){                     return 1;                 }                 ++frequencyoflengths[countofletters];                 if (countofletters >= longestwordcount) longestwordcount = countofletters;             }             countofletters = 0;         }         else {             countofletters++;             insideword = in;         }     }      (i = 1; <= longestwordcount; i++){         printf("%3i : %3i     ", i, frequencyoflengths[i]);         (j = 0; j < frequencyoflengths[i]; j++){             printf("*");         }         printf("\n");     }      return 0; } 

definitely scale results, check out character histogram horizontal scaling histogram.

also, benefit y-axis label. it's hard tell bar kind of word length. have no idea bar word length.

i added code right before display histogram, halves every value, throw off bar number labels. can figure out!

// iterates , tells frequent word length int mostfrequent = 0; (i = 1; < maxword; i++)     if (charcount[i] > mostfrequent)         mostfrequent = charcount[i];  // if bar big, cut every value in half while (mostfrequent > 60) {     (i = 1; < maxword; i++)         if (charcount[i] > 0) {             charcount[i] /= 2;             charcount[i] |= 1;         }      // check again find frequent word length category     mostfrequent = 0;     (i = 1; < maxword; i++)         if (charcount[i] > mostfrequent)             mostfrequent = charcount[i]; } 

honestly bars hard read, maybe use single row of characters such █ !

great book far, we're practically reading , on same page!

cheers