替换键盘中断（Interrupt 9）进行两次调用而不是一次调用

Question

Each keyboard press will enter to the function, which replaced the interrupt, twice. Why is that happing?

Maybe the "enter" after scanf is interfering? But it goes to the function twice each press and not only after first try

What am I doing wrong? How can I make it enter only once to the function for one press each time? In the pic you can see I press 5 only twice instead of 5 times

#include<stdio.h>
#include<dos.h>

volatile int ctrl_break_flag;   //counter
void interrupt(*Int9Save)(void);   


void interrupt my_func8(void)
{
  ctrl_break_flag++;
  printf("%d\n",ctrl_break_flag);
  Int9Save();
  }
void main()
{
 int N=0,i;
 Int9Save=getvect(9); //Save pointer to original interrupt.
 printf("Please enter number: ");
 scanf("%d",&N);
 setvect(9,my_func8);//Set interrupt pointer to our function.
 ctrl_break_flag=0;

 while(ctrl_break_flag<N);
   printf("End");
 setvect(9,Int9Save);//Return to original interrupt.
 return;
}

Answer 1

In order to avoid it you can simply set integer to count the times there was an interrupt and then ignore all the odd for example.

code example:

void interrupt Keyboard(void){
    check++;
    Keyboard_Flag=1;
    if(check%2==0) //print only button pressed(will print only 
                                               odd interrupts)
    printf(" KeyPressed");
    Int9Save(); // return state
    
}