Back to scanf and dont run the loops

Question

Ok why my code always back me to scanf? I don't know what i made wrong in this code but the code only backup me at scanf only reading numbers

int main()
{
    int m,n,i,tmp,min=9999999;
    scanf("%d%d",&m,&n); // scanf m and n right ?
    for(i=m;i<=n;i++) // for loop
    {
        tmp=i; // we give tmp a i value 
        while(tmp%10%2 == 0) 
        {
            tmp/=10;// in this while loop we check if the number is made only of even numbers
        }
        if(tmp!=0) // when the while loop break  we go down here right ? and if the number isn't made only of even digits we printf No
        printf("NO");
        else{ // else if its made of we check that that number is smaller then min number made of even digits
            if(i<min)
            min=i;
        }

    }
printf("%d",min); // on the end of the for loop we printf the min number made of even digits 
}

So why my compiler don't wont to run this program?

Answer 1

Your program enters infinite loop on the first number made of every even number. You cut digit and guide tmp till zero which falls to infinite loop: 0 % 10 % 2 is zero, and result of 0 / 2 is zero. You may fix it this way:

while (tmp > 0 && (tmp % 10) % 2 == 0) { /* ... */ }

Answer 2

I'm guessing you aren't having trouble compiling, but, instead, you run the code and nothing happens.

I think you're imagining that with "scanf("%d%d",&m,&n)", if you type in "1 30", that m will be 1 and n will be 30 and scanf will return 2. But actually scanf will return 1 because it stops parsing when it sees the space that isn't reflected in your format string, so m will be 1, n will be technically random, but probably 0 in your case, and your loop will "run" from 1 to 0, ie, zero times. Try "%d %d". And check the return code from scanf to make sure 2 numbers are entered.