why getch() function works twice?

Question

I write small program in C.
The code of the program is below.

#include<stdio.h>
#include<conio.h>
int main()
{
  char ch;
  int count = 0;
  while(1){
   ch = getch();
   count++;
   printf("%d\n",count);
  }
  return 0;
}

when i run this application when i press any key count increases one by one,
but when i press ARROW KEYS count increases two by two.
What is the problem. And how to fix it?

OS:Windows 7
IDE: Dev-Cpp with MINGW

EDIT #1:
when i print ch on the screen like

printf("%d",ch);

it shows two digits: for example -32 and 77 for left arrow key.
so how can i fix it.

Answer 1

check by pressing arrow and then backspace.

In GCC when I press Up arrow

this below string appears. this is the combination of three characters.

^[[A ==> ^[  [  A  

because of this count is incremented totally by 4 adding Enter also.

to fix this you can read input into string and then assign first char into c but after entering character you need to press Enter. and here you have choice when ever if you press arrow you have chance to escape count increment and assignment by checking string length.

  char c,ch[2];
  int count = 0;
  while(1){
   scanf("%s",ch);
   c =ch[0];
   count++;
   printf("%ld\n",strlen(ch));
   printf("%d\n",count);
  }

Answer 2

The docs for Windows says

"The _getch and_getwch functions read a single character from the console without echoing the character. None of these functions can be used to read CTRL+C. When reading a function key or an arrow key, each function must be called twice; the first call returns 0 or 0xE0, and the second call returns the actual key code."

Answer 3

The reason is most probably because arrow keys are represented by a scan code. Which is normally 2 hex values. In your case two characters. What you can do is to print ch and see what gets printed when you press the arrow key.

printf("%c\n",ch);