conversion in c - c

Question

I'm new to C and I need to write a function in c which converts an integer to a string in the specified base and print it.
If I were given an input value of 1234 (base 10) it should return 2322 (base 8).
Here is the code structure I'm currently working on:
void int2ascii(int value, int base){
int a=0;
if (value > base) {
a = a + int2char(value); //recursive case
int2ascii(value/base, base); //base case
}
printf("%s\n",a);
}
The program won't run, can someone enlighten me?
Thanks

Answer 1

Analysis of your program
I noted first off that a is declared int, but it was being passed to printf with %s as the format specifier. %s designates the associated argument will be a string, which a is not. The result will be undefined behavior, and is a possible cause for your crash.
You do not specify what int2char() does, but let's assume that it converts a numeric "digit" into a corresponding char value. With that in mind, let us suppose its implementation is similar to:
int int2char(int d) {
return "0123456789abcdefghijklmnopqrstuvwxyz"[d];
}
In your code, you pass value to int2char(). In my hypothetical implementation, this would cause out-of-bounds access of the array, and thus undefined behavior. This is indicates a logic error, and another possible cause for your crash.
I note that if value is less than base, a remains 0. Probably, you really mean to compute a value for a even when value is less than base. This indicates another logic error.
Base conversion
The number dKdK-1 .. d0, where each di is in (0 .. 9), is a short form for ∑di×10i. To discover the base 10 digits of a number N, the process is:
di = ⌊N / 10i⌋ mod 10
But, as it turns out, you can replace 10 with some other base number to compute the digits for a number in that base:
di = ⌊N / Bi⌋ mod B
In your code, the recursive call that divides by the base represents the first part of the calculation. However, in your calculation of a, you were missing the "mod" part of the calculation.
Solution 1
Assuming base is from 2 to 36, and that your int2char(d) does something more or less as illustrated earlier:
void int2ascii(int value, int base){
int a=0;
a = a + int2char(value%base);
if (value > base) {
int2ascii(value/base, base);
}
printf("%c",a);
}
Because you are not passing a into your recursive call, you can only use it to store the current digit to be printed. So, move the code to store the value in a out of the if check, since you always want a sensible value to print at the end. The digit in a is in the radix base, so you need the modulus result to get the right digit value. Since a represents a character, change the format string to match, and remove the \n so that all the digits end up on the same line.
Solution 2
The first solution is an attempt to leave most of your original code intact. Since a is initialized to 0, the extra addition does not make it incorrect. But, since the print doesn't happen until the very end, the value doesn't really need to be stored at all, and can be calculated at the point you actually want to print it. With that in mind, the program can be simplified to:
void int2ascii(int value, int base){
if (value > base) {
int2ascii(value/base, base);
}
printf("%c",int2char(value%base));
}

Answer 2

#include <stdio.h>
#include <limits.h>
char *int2ascii_aux(int n, int base, char *s){
//base {x| 2 <= x <= 36 }
static const char *table = "0123456789abcdefghijklmnopqrstuvwxyz";
if(n){
*--s = table[n % base];
return int2ascii_aux(n/base, base, s);
} else {
return s;
}
}
char *int2ascii(int n, int base){
//base {x| 2<= x <= 36 }
static char s[sizeof(int)*CHAR_BIT+1];
if(n <= 0){
*s = '0';
return s;
}
if(base < 2 || base > 36){
*s = '\0';
return s;
}
return int2ascii_aux(n, base, s + sizeof(s) -1);
}
int main(){
printf("%s\n", int2ascii(1234, 8));
return 0;
}