A LIST ANAGRAM WORD PAIRS USING 2d ARRAY

Question

I'm trying to do an exercise which is an anagram.

Sample input: GARDEN DANGER WHO HOW DO ODD

SAMPLE OUTOUT: A LIST OF ANAGRAM WORD PAIRS.

WHO : HOW

DANGER : GARDEN

this the error:

main.c:24:6: warning: incompatible pointer to integer conversion assigning to 'char' from 'char [10]' [-Wint-conversion] ch = word[i];

main.c:38:15: warning: comparison between pointer and integer ('char *' and 'int') if( word[i ]== 1){

main.c:39:19: warning: more '%' conversions than data arguments [-Wformat] printf("%s : %s", word[i]);

Here is my code:

#include <stdio.h>

int main() {
  char word[5][10];
  char ch;
  int counts1[26];
  int counts2[26];
  int i = 0 , n = 5;

  printf("Input a word: ");
  
  //Read the user input
  while(i <= n){
  scanf("%s[^\n]", word[i]);
  i++;
  }
  //Count the characters
  for(i = 0; i < 26; ++i) {
  counts1[i] = 0;
  counts2[i] = 0;
  }
  i = 0;
  while(word[i] != 0) {
  ch = word[i];
  if(ch >= 'A' && ch <= 'Z') {
  ch += 32;
    }
  counts1[ch-'a']++;
  i++;
  }
  i = 0;
  for(i = 0; i < 26; ++i) {
  if(counts1[i] != counts2[i]) {
  return 0;
    }
  }
  //Check if Anagram or not
  if( word[i]== 1){
    printf("%s : %s", word[i]);
  }
  else{
    printf("%s is not an anagram", word[i]);
  }
}

Answer 1

In :

while(word[i] != 0) {
  ch = word[i];

word[i] is an array of chars (not an int) and ch is a char.

And :

if( word[i]== 1){

means nothing : you compare an array of char (word[i]) with an integer