How to store multiple value of printf in a variable? - c

Question

The program must accept an integer N with even number of digits as the input.
The program must reverse every two digits in N and print the modified N as the output.
Boundary Condition(s): 10 <= N < 10^16
Input Format: The first line contains N.
Output Format: The first line contains the modified N.
Example Input/Output 1:
Input: 214587
Output: 125478
Explanation: The first two digits are 2 and 1 which are reversed as 1 and 2. The second two digits are 4 and 5 which are reversed as 5 and 4. The third two digits are 8 and 7 which are reversed as 7 and 8.
Example Input/Output 2:
Input: 504786
Output: 57468
I have proceeded in a logic of printing the values inside a while loop by printing b/10 and b%10; where the output should be reversed and I am stuck in it, please help in my logic how to proceed further thank you.
scanf("%d",&a);
while(a>0) {
b=a%100;
printf("%d%d",b/10,b%10);
a=a/100;
}

Answer 1

As I noted in three comments (1, 2, 3), I think you need to use recursion, and some care. Here is code that works, and a test harness (which is bigger than the code being tested) demonstrating that it works:
#include <inttypes.h>
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
static bool reverse_digit_pairs(uint64_t value)
{
uint64_t b = value % 100;
uint64_t v = value / 100;
if ((v > 0 && v < 10) || (v == 0 && b < 10))
return false;
int t = b / 10; // tens
int u = b % 10; // units
bool ok = true;
if (v >= 10)
ok = reverse_digit_pairs(v);
if (ok)
{
if (u == 0 && v == 0)
printf("%d", t);
else
printf("%d%d", u, t);
}
return ok;
}
static void test_reverse_digit_pairs(uint64_t v)
{
printf("V: %20" PRIu64 " = ", v);
if (!reverse_digit_pairs(v))
{
fprintf(stderr, "Odd number of digits in %" PRIu64 "\n", v);
printf("bogus input");
}
putchar('\n');
fflush(0); // Ensure standard output (and standard error) are flushed.
}
int main(void)
{
test_reverse_digit_pairs(UINT64_C(214587)); // 125478 per question
test_reverse_digit_pairs(UINT64_C(504786)); // 57468 per question
test_reverse_digit_pairs(UINT64_C(5047000086)); // Pairs of zeros
test_reverse_digit_pairs(UINT64_C(5047900086)); // Pair of zeros
test_reverse_digit_pairs(UINT64_C(1234567890123456)); // 16 digits
test_reverse_digit_pairs(UINT64_MAX); // 20 digits
test_reverse_digit_pairs(UINT64_C(123456789012345)); // Odd digits - failure
test_reverse_digit_pairs(UINT64_C(0)); // Odd digits - failure
test_reverse_digit_pairs(UINT64_C(1)); // Odd digits - failure
test_reverse_digit_pairs(UINT64_C(9)); // Odd digits - failure
test_reverse_digit_pairs(UINT64_C(10)); // 2 digits
test_reverse_digit_pairs(UINT64_C(99)); // 2 digits
test_reverse_digit_pairs(UINT64_C(100)); // Odd digits - failure
test_reverse_digit_pairs(UINT64_C(999)); // Odd digits - failure
test_reverse_digit_pairs(UINT64_C(1000)); // 4 digits
test_reverse_digit_pairs(UINT64_C(1098)); // 4 digits
test_reverse_digit_pairs(UINT64_C(1234)); // 4 digits
return 0;
}
The output I get from it is:
V: 214587 = 125478
V: 504786 = 57468
V: 5047000086 = 574000068
V: 5047900086 = 574090068
V: 1234567890123456 = 2143658709214365
V: 18446744073709551615 = 81447644707390556151
Odd number of digits in 123456789012345
V: 123456789012345 = bogus input
Odd number of digits in 0
V: 0 = bogus input
Odd number of digits in 1
V: 1 = bogus input
Odd number of digits in 9
V: 9 = bogus input
V: 10 = 1
V: 99 = 99
Odd number of digits in 100
V: 100 = bogus input
Odd number of digits in 999
V: 999 = bogus input
V: 1000 = 100
V: 1098 = 189
V: 1234 = 2143
The test cases for the numbers with 1 to 4 digits were necessary to flush out some issues. I'm not completely happy with the first condition in reverse_digit_pairs(), but I didn't get the output I think is required when I tried some simplifications.