create number 0 only using int in the C - c

Question

#include <stdio.h>
struct real_num
{
int int_num;
int frac_num;
};
void main()
{
struct real_num num1;
printf("input the number : ");
scanf("%d.%d",&num1.int_num,&num1.frac_num):
printf("%d.%d",num1.int_num,num1.frac_num);
}
i input 12.012 but buffer save 12.12 i want a 012 but this buffer save 12
what should i do? i want a save 012 (using only int)

Answer 1

Numbers are a matter of arithmetic. 1, 01, 1.0, 1.000, 0x01, 1e0 all describe the same number: whichever representation you use has the same mathematical properties, and behaves identically in calculation (ignoring the matter of computer storage of numbers as int or float or double... which is again another matter entirely).
The representation of a number is a matter of sequences of characters, or strings. Representations of numbers can be formatted differently, and can be in different bases, but can't be calculated with directly by a computer. To store leading zeroes, you need a string, not an int.
You typically convert from number representation to number at input, and from number to number representation at output. You would achieve your stated desire by not converting from number representation to number at input, but leaving it as a string.

Answer 2

You don't want to store 012, you want to store 0.012.
The value 0.012 in binary is (approximately):
0.00000011000100100110111010010111b
..and the value 12.012 is (approximately):
110.00000011000100100110111010010111b
Note that 0.012 is impossible to store precisely in binary because it would consume an infinite number of bits; in the same way that 1/3 can't be written precisely in decimal (0.333333333.....) because you'd need an infinite number of digits.
Let's look at 12.012. In hex it's this:
0x0000000C.03126E97
This makes it easier to see how the number would be stored in a pair of 32-bit integers. The integer part in one 32-bit integer, and the fractional part in another 32-bit integer.
The first problem is that you're using signed 32-bit integers, which means that one of the bits of the fraction is wasted for a sign bit. Essentially, you're using a "sign + 31 bit integer + wasted bit + 31 bit fraction" fixed point format. It'd be easier and better to use an unsigned integer for the fractional bits.
The second problem is that standard C functions don't support fixed point formats. This means that you either have to write your own "string to fixed point" and "fixed point to string" conversion routines, or you have use C's floating point conversion routines and write your own "floating point to fixed point" and "fixed point to floating point" conversion routines.
Note that the latter is harder (floating point is messy), slower, and less precise (double floating point format only supports 53 bits of precision while you can store 62 bits of precision).

Answer 3

A fraction does not consists of a single integer. A fraction consists of 2 integers: numerator/denominator.
Code needs to keep track of width of the fraction input. Could use "%n" to record offset in scan.
#include <stdio.h>
struct real_number {
int ipart;
int num;
int den_pow10;
};
void main(void) {
struct real_number num1;
printf("input the number : ");
fflush(stdout);
int n1 = 0;
int n2 = 0;
scanf("%d.%n%d%n",&num1.ipart, &n1, &num1.num , &n2):
if (n2 == 0) {
fprintf(stderr, "bad input\n");
return -1;
}
num1.den_pow10 = n2 - n1;
printf("%d.%*0d",num1.ipart,num1.den_pow10, num1.frac_num);
return 0;
}
Input/Output
input the number : 12.00056
Result 12.00056