Find a two digit number with recursion - c

Question

I'm trying to code something in C by using recursion.
The user writes two positive numbers of same length and the program gives him a new number, which is composed like this :
new number unity digit = the smallest digit in the second positive number that the user wrote.
new number ten digit = the biggest digit in the first positive number that the user wrote.
Very simple in fact, here is an example :
5642 and 2371
will give us : 61.
I tried something like this :
#include <stdio.h>
int calcPair(int a, int b){
int number = calcPair(a/10, b/10);
int digit1 = (number/10);
int digit2 = number%10;
if(digit1 < a%10){
digit1 = a%10;
}
if(digit2 > b%10){
digit2 = b%10;
}
return(number);
}
int main()
{
int a, b, number=0;
printf("Please enter two positive number of same length:\n");
scanf("%d", &a);
scanf("%d", &b);
calcPair(a, b);
printf("The two-digit number composed from %d, %d is: %d", a, b, number);
return 0;
}
BUT the program doesn't run at all.. and closes.
Maybe someone can correct me ? Or helping me finding the mistake.
Thanks by advance.

Answer 1

Your recursion can never end. Consider the following line in calcPair:
int number = calcPair(a/10, b/10);
This statement will always be executed unless you make it conditional, such as:
int number;
if((a != 0) || (b != 0))
number = calcPair(a/10, b/10);
Eventually, because you're dividing both numbers by 10, this condition will prove FALSE.

Answer 2

Something like this:
int calcPair(int a, int b){
int number;
if (a < 10 && b < 10) {
number = a*10 + b;
} else {
int digita = a%10;
int digitb = b%10;
number = calcPair(a/10, b/10);
if(digita > number/10){
number = digita*10 + number%10;
}
if(digitb < number%10){
number = (number/10)*10 + digitb;
}
}
return number;
}
Also, a small fix to the main:
int main()
{
int a, b, number=0;
printf("Please enter two positive number of same length:\n");
scanf("%d", &a);
scanf("%d", &b);
number = calcPair(a, b);
printf("The two-digit number composed from %d, %d is: %d", a, b, number);
return 0;
}

Answer 3

I think you can refactor your code to be more expressive of your requirement, with a few helper functions.
int greater(int a, int b)
{
return (a>b);
}
int less(int a, int b)
{
return (a<b);
}
int pickDigit(int n, int (*func)(int, int))
{
int ret = n%10;
n /= 10;
while ( n > 0 )
{
if ( fun(n%10, ret) )
{
ret = n%10;
}
n /= 10;
}
return ret;
}
int getBiggestDigit(int n)
{
return pickDigit(n, greater);
}
int gteSmallestDigit(int n)
{
return pickDigit(n, less);
}
int numDigits(int n)
{
int ret = 0;
while (n > 0 )
{
++ret;
n /= 10;
}
return ret;
}
int calcPair(int a, int b)
{
if ( numDigits(a) != numDigits(b) )
{
// Deal with error.
}
return betBiggestDigit(a)*10+getSmallestDigit(b);
}

Answer 4

Whether or not you are allowed to (you did not specify in OP),
here is a recursive search method using strings :
Strings are just an array of char. because you are interested in distinguishing the individual digits within a larger integer, the char data type will be a sufficient size container to facilitate the comparison.
Using arrays of char (strings) within a recursive function with exit criteria of strlen() > 0 will allow you to walk through each integer, and select the appropriate value (min or max).
This approach uses two recursive functions: getMinDigit() and getMaxDigit(), both returning a char representing the maximum value digit, or minimum value digit of their respective original multi-digit integer. These results are then concatenated, and converted back into a two digit integer.
Here is the example code that given:
5642 and 2371
will give us : 61.
char getMinDigit(char *digit)
{
static char val='9';//largest single digit base 10
int len=0;
if(strlen(digit) > 0)
{
len = strlen(digit);
if(digit[len-1] < val) //test for smallest char in string
{
val = digit[len-1];
digit[len-1] = 0;
getMinDigit(digit);
}
else
{
digit[len-1] = 0;
getMinDigit(digit);
}
}
return val;
}
char getMaxDigit(char *digit)
{
static char val='0'; //smallest single digit base 10
int len=0;
if(strlen(digit) > 0)
{
len = strlen(digit);
if(digit[len-1] > val) //search for largest char in string
{
val = digit[len-1];
digit[len-1] = 0;
getMaxDigit(digit);
}
else
{
digit[len-1] = 0;
getMaxDigit(digit);
}
}
return val;
}
int calcPair(int a, int b)
{
char big[10]={""}, small[10]={""};
char Big, Small;
char result[3]={""};
sprintf(big, "%d", a);
sprintf(small, "%d", b);
Big = getMaxDigit(big); //recursive function
Small = getMinDigit(small); //recursive function
sprintf(result, "%c%c", Big, Small);
return atoi(result);
}
int main(void)
{
int result = calcPair(5642, 2371);
printf("%d", result);
return 0;
//for illustration, hard coded to OP values
//int a, b, number=0;
//printf("Please enter two positive number of same length:\n");
//scanf("%d", &a);
//scanf("%d", &b);
//calcPair(a, b);
//printf("The two-digit number composed from %d, %d is: %d", a, b, number);
//return 0;
}