How to read a user input string and store it in an Array

Question

Trying to read user input string from the key board and assign it to an Array. It still confusing.

Also any Idea what char ch = 97 is in this program? Thanks.

#include<stdlib.h>

int main()
{
    int i = 0;
    int j = 0;
    int count[26]={0};
    char ch = 97;
    char string[100]="readmenow";

    for (i = 0; i < 100; i++)
    {
         for(j=0;j<26;j++)
         {
              if (tolower(string[i]) == (ch+j))
              {
                   count[j]++;
              }
         }
    }
    for(j=0;j<26;j++)
    {
        printf("\n%c -> %d",97+j,count[j]);
    }
}

Answer 1

to read user input do this:

  #include <stdio.h>  // for fgets
  #include <string.h> // for strlen

  fgets(string,sizeof(string),stdin);
  string[strlen(string)-1] = '\0'; // this removes the \n and replaces it with \0

make sure you include proper headers

Also ch= 97; is same as doing ch = 'a';

EDIT:

scanf is great for reading input as a string as long as the string doesn't have space. fgets is much better

EDIT 2

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main(){

    int i=0,j=0;

    char input[50]; // make the size bigger if you expect a bigger input

    printf("Enter string = ");
    fgets(input,sizeof(input),stdin);
    input[strlen(input)-1] = '\0';

    int count[26]={0};

    for (i = 0; i < strlen(input); i++)
    {
         for(j=0;j<26;j++)
         {
              if (tolower(input[i]) == ('a'+j))
              {
                   count[j]++;
              }
         }
    }
    for(j=0;j<26;j++)
    {
        printf("\n%c -> %d",'a'+j,count[j]);
    }


    return 0;
}

Output: $ ./test

Enter string = this is a test string

a -> 1
b -> 0
c -> 0
d -> 0
e -> 1
f -> 0
g -> 1
h -> 1
i -> 3
j -> 0
k -> 0
l -> 0
m -> 0
n -> 1
o -> 0
p -> 0
q -> 0
r -> 1
s -> 4
t -> 4
u -> 0
v -> 0
w -> 0
x -> 0
y -> 0
z -> 0

Answer 2

 char ch= 97

it means ch='a'
it use ASCII (American Standard Code for Information Interchange)