How does this code convert between bases? - c

Question

#include<stdio.h>
#include<string.h>
void baseconversion(char s[20], int, int);
main()
{
char s[20];
int base1, base2;
printf("Enter the number and base:");
scanf("%s%d", s, &base1);
printf("Enter the base to be converted:");
scanf("%d", &base2);
baseconversion(s, base1, base2);
}
void baseconversion(char s[20], int b1, int b2)
{
int count = 0, r, digit, i, n = 0, b = 1;
for(i = strlen(s) - 1; i >= 0; i--)
{
if(s[i] >= 'A' && s[i] <= 'Z')
{
digit = s[i] - '0' - 7;
}
else
{
digit = s[i] - '0';
}
n = digit * b + n;
b = b * b1;
}
while(n != 0)
{
r = n % b2;
digit = '0' + r;
if(digit > '9')
{
digit += 7;
}
s[count] = digit;
count++;
n = n / b2;
}
for(i = count - 1; i >= 0; i--)
{
printf("%c", s[i]);
}
printf("\n");
}
I know this code converts chars to integers, but I've never seen it before, never used C.
If someone could explain a bit of what's going on with the conversions I'd appreciate it, thank you.
I understand that at some point the digits get reversed.

Answer 1

It does it through two steps, the first one is converting the number into its decimal form, in this part:
for(i = strlen(s) - 1; i >= 0; i--) //Start from right to left
{
if(s[i] >= 'A' && s[i] <= 'Z')
digit = s[i] - '0' - 7; //Get the integer equivalent to the letter
else
digit = s[i] - '0'; //Get the integer equivalent to the numerical character
n = digit * b + n; //Add the value of this character at this position
b = b * b1; //The value of the next character will be higher b times
}
Then it transforms the result to the desired base, in this part:
while(n != 0)
{
r = n % b2; //The remaining will be the rightmost value for the new base
digit = '0' + r; //Get the integer for the new digit
if(digit > '9')
digit += 7; //Here the digit will be a letter
s[count] = digit;
count++;
n = n / b2; //Remove the rightmost digit to get the next one
}