#include <stdio.h>
#include <conio.h>
void main(){
int turn,i=1,num;
char answer;
for(i>0;i++;){
printf("Please enter a number in the range 1-5:");
scanf("%d",&num);
if (num == 1){
printf("Disconnecting\n");
continue;
}
else if(num == 2){
continue;
}
else if(num == 3){
printf("Are you sure you would like to finish your order??? \nplease enter one char:");
scanf("%c", &answer);
if(answer == 'y'){
printf("Canceled");
break;
}
else{
continue;
}
}
else if(num == 4){
turn=i-1;
printf("your position in queue is:%d\n",turn);
continue;
}
else if(num == 5){
break;
}
else {
printf("Wrong input\n");
continue;
}
}
getch();
}
I'm using c language for this, if You look at the
else if(num == 3)
it should function in a way that when I enter the letter y,Y it will say canceled and end the program and if not it will just reset the loop. now when I use the number 3 and give the variable c the letter 'y' it just says nothing and acts like I gave it the command "continue" + pressed the number 3 again although all I have done is press y or Y and enter. everything else is good. I would be glad if anyone can tell me how to fix that.