Thread: Hey guys, I need a little help here~

By checking the input.

If you posted your current code, I might be more inclined to be less vague

Your compiler will have one or more functions to transform a number, into a string. So, you need a char array set up for it, and then see if your compiler has itoa. If not, peek around your help topics, because it will have something like it.

Code:

#include <stdlib.h>
#include <stdio.h>

int main(void)
{
   int number = 12345;
   char string[25];

   itoa(number, string, 10);
   printf("integer = %d string = %s\n", number, string);
   return 0;
}

You can also "roll your own...battery", <little inside joke>, but it's much faster to just use a function like itoa(), if you have it.

<LOL> If you posted your current code, I might be more inclined to be less vague

Here's my code:

Code:

#include <stdio.h>


int main(){

    
    int b2, power=1;


...//Deleted //...



 
    printf("_%d", b2);



    scanf("\n");

    
    
    return 0;



}

Yes. Directly check every number by taking one digit at a time, off the right hand side (or some side, but right hand, for now), and comparing it with the base.

We can treat the 21012 number as if it was any suitably sized base, so let's use base 10, just for convenience.

Code:

int num = 21012;
int badnum, remNum, baseNum=2 

badnum = 0;
while(num > 0) {
  remNum = (num % 10);
  //remNum now equals 2 on the first loop, 1 on the second, etc.;

  if(remNum >= baseNum) {
    badNum = 1;
    break;
  }
  num /= 10; (num = num / 10)
}

which "peels off the most right hand digit, completely, with each loop. Continue looping until num < 1, and if you haven't got a badNumber==1, it's OK.

I haven't run this code, but it's very close. Any errors, let me know.