Preventing user mistakes

[joexoxox]

Here is the deal , I'm writing a function that makes the user guess for a number already set by me , the user will enter a single digit number , and if it exists , it's printed in its right position on the screen, like hangman but for numbers :P
I am reading the input of the user using scanf and i dont want to use gets, so plz let ur answers differ from that ....
Im just checking if the user_number is between 0 and 9 to make sure that it's a single digit number, the code is simple :
if(user_number >= 0 || user_number <= 9) // if correct, enter the function ...
{

}
else printf("the number does not exist");

Note that the user has a certain amount of tries to guess ...
But the deal is that how can i prevent the user to type 'a' for example, if he types 'a' or any other character, the program will print the message of "the number does not exist " but will also instantly go through all the number of tries loop and finish it.

Can anyone tell me how can i prevent that?? is there a way to flush the screen and preventing the loop to happen??

PS: I dont want to use the gets function to read the input, i must stick to an array of integers to store my number in.

Thx

[Ganon11]

Well, I know you can use the return value of scanf to determine if your input was correct (i.e. an integer). scanf returns an integer, which holds the number of values properly read. For example:

int try1 = scanf("%d", &mynum); //User enters 23, try1 holds 1
int try2 = scanf("%f %d %f", &myfloat1, &myint, &myfloat2); //User enters 3.14 10 MYBAD, try2 holds 2
int try3 = scanf("%d", &user_number); //User enters 'a', try3 holds 0

See if this will work for you.

[joexoxox]

Thx for that , but it didn't solve the problem ...

it's knowing that the value is not between 0 and 9 and it's printing the right message.
The problem is that it's in a loop, and when it's supposed to prompt the user for a number each time, it's just assuming that the user is enterin 'a' each time , so it's printing the message 10 times then exists...this doesnt occur if the user types '33' in that case, its working just fine.

[oler1s]

The problem is that it's in a loop, and when it's supposed to prompt the user for a number each time, it's just assuming that the user is enterin 'a' each time , so it's printing the message 10 times then exists...this doesnt occur if the user types '33' in that case, its working just fine.

Not surprising. By the way, we almost never, ever, ever recommend using scanf. If you want to convert numbers, the recommended way is fgets then strtol.

Anyway, what you need to realize about scanf is how it deals with failed conversions. When scanf fails to convert something into a number, it leaves whatever that is on the input stream.

So if you had in the input:
a123

Scanf would fail on a. But it would leave a in the input stream. So the next time you run scanf, you pick up the a again. And fail. And again. And fail. See the problem.

You need to flush the input stream. See cprogramming.co m on how to do this.