How to count the amount of times a specific character appears in an array in C? -

so trying count amount of times specific character occurs in program. example, if entered, abcda, want program print, "there 2 a's." code follows:

int main(void) {   char array[10000];   printf("enter input: \n");   scanf("%s", array);   printf("array entered is: %s\n", array);    char a;          //variable want count   char *k          //used loop   int a_counter;   //number of times a, occurs     fgets(array, sizeof(array), stdin);   = fgetc(stdin);   a_counter = 0;    for(k = array; *k; k++)   {     if (*k == a)     {       a_counter++;     }    }    printf("number of a's: %d\n", a_counter);    return 0;   } 

the following loop found on forum attempted count specific character can not seem mine work. approach @ wrong? out of main confused on how so. appreciate given. thank you.

new attempt @ count loop not working. got rid of fgets because confusing me.

    int a_counter = 0;     if (array == 'a')     {        a_counter++;     }     printf("number of a's: %d\n", a_counter); 

attempt after @bjorn help.

#include <stdio.h> int main(void) {   char array[1000];   printf("enter input: \n);   scanf("%s", array);   printf("input is: %s\n", array);    int c,n =0;   while((c = getchar()) != eof)     if (c = 'a')       n++;   printf("amount of a's is: %d\n", n);   return 0;  } 

kism - keep simple, mate :) code way complicated simple task. here's alternative, illustrates mean:

#include <stdio.h>  int main(void) {     int c, n = 0;      while ((c = getchar()) != eof)         if (c == 'a')             n++;      printf("where %d characters\n", n);     return 0; }