Why on while(getchar()!=EOF) it iterates extra time? - c

Question

I'm trying to count chars from input, and I noticed that while(getchar()!=EOF) produces an extra count, Is it because it counts the null-terminated from input?
This is my code:
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
#define LINE 5
#define MEM_SIZE 10
char *getInput(int *counter);
void printInput(char *input);
int main() {
char *mainInput;
int counter = 0;
printf("Please enter your input:\n");
mainInput = getInput(&counter);
if (mainInput == NULL) {
return 1;
}
printf("\n\n\nOutput:\n%s\n", mainInput);
printf("number of chars: %d\n", counter);
printInput(mainInput);
free(mainInput);
return 0;
}
char *getInput(int *counter) {
char *p = (char *)malloc(MEM_SIZE * sizeof(char));
char *q = p; /* backup pointer, if realloc fails to allocate, function will return last address values stored */
int c;
int temp_counter = 0;
long current = 0, last = MEM_SIZE - 1;
while ((c = getchar()) != EOF) {
if (current >= last) {
q = p;
p = (char *)realloc(q, last + (MEM_SIZE * sizeof(char)));
if (p == NULL) {
printf("Memory allocation failed, printing only stored values \n");
return q;
}
last += MEM_SIZE;
}
p[current] = c;
temp_counter++;
printf("number of chars: %d\n", temp_counter);
++current;
}
p[current] = '\0';
(*counter) = temp_counter - 1;
return p;
}
void printInput(char *input) {
int i, j = 0;
while (input[j] != '\0') {
for (i = 0; i < LINE; i++) {
if (input[j] == '\0')
break;
putchar(input[j]);
++j;
}
if (input[j] != '\0')
putchar('\n');
}
}