将数字的数字打印为字符的程序（C语言）

Question

I have to do a program to take a number and print all the digits one by one as characters.

The code I have written is here:

   #include <stdio.h>
   #include <stdlib.h>
   char digito(int, int);
   int numberofdigits(int);

   int main(void)
{
int num=0, test=0, kdigits=0, k=0;
char ch='\0';
printf("Enter a positive number:\n");
test=scanf("%d", &num);
if (teste!=1 || num<=0)
{
    printf("Error: not a valid number.\n");
    exit(EXIT_FAILURE);
}

kdigits = numberofdigits(num);


for(k=0; k<kdigits; k++)
{
    ch= digito(num, k);
    printf("The digit %d of the number is %c\n", k, ch);
}

return EXIT_SUCCESS;
}


int numberofdigits(int _n)
{
int count=0;
while (_n!=0)
{
    _n/=10;
    count++;
}

return count;
}


char digito(int _number, int _kdigit)
{
int flag=0, digit=0;
char ch='\0';
for (flag=0; flag<=_kdigit; flag++)
{
    digit=_number%10;
    _number/=10;
}

ch= digit + '0';
return ch;
}

Now the code is working pretty fine for relatively small numbers (up to 8 or 9 digits I would say).

But then something odd is happening: I tried to print the digits of the number 11111111111122 and obtained

The digit 0 of the number is 3;

The digit 1 of the number is 7;

The digit 2 of the number is 3;

The digit 3 of the number is 6;

The digit 4 of the number is 1;

The digit 5 of the number is 7;

The digit 6 of the number is 0;

The digit 7 of the number is 3;

I wonder why? Is it because it's a very large number? Because I tried even larger numbers and what happens is that the program enters the initial if clause that verifies the scanf reading. And that's ok. But the problem is that the program should also do the same with this number since it's bigger that the largest int. Can someone help me fix this please?

Thanks!

Answer 1

There are two answers to two questions:
Q1) Why do I get wrong digits for the example?
A1) As others have commented: Because the example number is too high for the data type used, it matches the number which you find in the lower 32 bits: 30716370.

Q2) Why is the initial check "<0" not triggering for the example but does trigger for other, higher examples?
A2) Because the 30716370 is smaller than the biggest positive number which can be represented by a 32 bit signed int, which is 2147483647 == 0x7FFFFFFF. That one however is smaller (even in the number of digits, but that is not the point) than 11111111111122.
30716370 <
2147483647 <
11111111111122
The even higher numbers will by chance have the bit31 set, which makes the lower 32bit look negative. You could probably find other numbers which are too high but do not seem negative.