How to printf vertically? - c

Question

This is my code. My point is to take string input and count how many of each letter occurs
#include <stdio.h>
#include <ctype.h>
int main()
{
int c1;
int a=0, b=0, c=0, d=0, e=0, f=0, g=0, h=0, i=0, j=0, k=0, l=0, m=0, n=0, o=0, p=0, q=0, r=0, s=0, t=0, u=0, v=0, w=0, x=0 , y=0, z=0;
while (( c1=getchar()) != EOF)
if (isalpha(tolower(c1)) != 0) {
if (tolower(c1) == 97) { // Character = 'a'
a += 1;
}
else if (tolower(c1) == 98) { // Character = 'b'
b += 1;
}
else if (tolower(c1) == 99) { // Character = 'c'
c += 1;
}
.
.
.
}
return 0;
}
Next I want to printf result in vertical. Could you give me some hints. For example,
input: ABC---Ddhhh
output:
*
* *
**** *
abcdefghijklmnopqrstuvwxyz

Answer 1

While it may be a struggle learning, it's important to spend time learning and not simply copy/pasting lines of code to have a big program. The best code is the briefest code that achieves the objective.
As others have said, the contiguous alphabet should cause you to use a contiguous array of counters, not individual variables. (One typo and your results will be wrong.) Nesting isalpha(tolower(c)) is not needed. And, you've shown how you might count, but not shown code that would output the desired results.
This is for you to study, learn and grow as a coder.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
// Try to write functions; blocks of code that do what you want in compartments
void countAndGraph( char *str ) {
int max = 0, cnt[26] = { 0 }; // initialises 'max' and 26 counters
for( char *p = str; *p; p++ ) // as long as there is a character
if( isalpha( *p ) ) { // if it is alphabetic
int i = tolower( *p ) - 'a'; // convert ASCII character to value 0-25
if( ++cnt[ i ] > max ) // increment the right counter and test against max
max = cnt[ i ]; // 'max' - the ceiling - keeps getting higher
}
for( ; max; max-- ) { // lower the ceiling
for( int i = 0; i < 26; i++ ) // for 26 alphabetic characters
putchar( " *"[cnt[i]>=max] ); // output ' ' or '*' if count at/above ceiling
putchar( '\n' ); // newline
}
puts( "abcdefghijklmnopqrstuvwxyz" ); // reference alphabet string
puts( str ); // the string that's been analysed
puts( "" );
}
int main() {
char *tst[] = { // three test strings to demonstrate
"The quick brown fox jumps over the lazy dogs",
"Sally sells seashells by the sea shore",
"I wish I was what I was when I wished I was what I am now",
};
const size_t nTst = sizeof tst/sizeof tst[0];
for( size_t i = 0; i < nTst; i++ ) // count and graph each of the strings
countAndGraph( tst[i] );
return 0;
}
*
* *
* * * ****
**************************
abcdefghijklmnopqrstuvwxyz
The quick brown fox jumps over the lazy dogs
*
*
* * *
* * *
* * *
* * * * *
* * * * * *
** * * * * *** *
abcdefghijklmnopqrstuvwxyz
Sally sells seashells by the sea shore
*
* *
* *
* * *
* ** * *
* ** * *
* ** * *
* * ** * ** *
* ** ** *** ** *
abcdefghijklmnopqrstuvwxyz
I wish I was what I was when I wished I was what I am now
Now that you have this code, an exercise would be to print the bar graph in ascending/descending order. Don't stop with this one version; push it to a new educational experience.
*
*
*
*
**
***
****
*********
*************
**************
****************
*****************
******************
fzcxqpjkuvywnbmdgrloshiate
'Twas brillig, and the slithy toves
Did gyre and gimble in the wabe:
All mimsy were the borogoves,
And the mome raths outgrabe.

Answer 2

Any time you have more than one or two variables that do the same thing, it is time to use a data structure; in this case, an array. You will want something like
int counts[26];
instead of twenty six different variables, and only have one statement instead of 26 different if clauses. Then you can say
counts[lower(c1) - 97]
instead of a, b... z (after checking that lower(c1) is one of the letters).
The way to print the output is based on these insights:
The number of lines above the alphabet is the highest number in counts (let's call it maxcount). It would be 3 in your example.
For each line, there is a * if that letter's count is at least maxcount - line (i.e. in 0th line, there is a * if there is at least 3 - 0 counts of that character), otherwise there's a space.

Answer 3

I think you can just print one at a time.
#include <stdio.h>
#include <ctype.h>
int main()
{
int c1;
int a=0, b=0, c=0, d=0, e=0, f=0, g=0, h=0, i=0, j=0, k=0, l=0, m=0, n=0, o=0, p=0, q=0, r=0, s=0, t=0, u=0, v=0, w=0, x=0 , y=0, z=0;
// I might do an array here, so you can loop through later.
int letters[] = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
while (( c1=getchar()) != EOF) {
if (isalpha(tolower(c1)) != 0) {
if (tolower(c1) == 97) { // Character = 'a'
a += 1;
letters[0]++;
}
else if (tolower(c1) == 98) { // Character = 'b'
b += 1;
letters[1]++;
}
else if (tolower(c1) == 99) { // Character = 'c'
c += 1;
letters[2]++;
}
.
.
.
}
}
int ii;
for (ii = 0; ii < 26; ii++) {
printf("%d\n", letters[ii]);
}
return 0;
}