reverse an unsigned int number causing overflow - c

Question

Enter a unsigned int, reverse it and see if its' still in range, if it is, print the reverse number, if not, print your number is out of range
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
int main() {
unsigned int a = UINT_MAX; // 0xffff
printf("max unsigned int = %u\n\n", a);
unsigned int x = 0;
printf("please enter any unsigned int,\nit will show the reverse number if it's in range\n(enter 10 digit only)\n\n");
while (scanf("%u", &x) != EOF) {
//printf("x = %u\n", x);
// above %d will print each UINT's binary oder, ex: enter 4294967292, x = -1
// if change %u, it will print entered number, ex: enter 4294967292, x = 4294967292
unsigned int temp = x, result = 0;
int m = 0;
while (temp > 0) {
unsigned int digit = temp % 10;
if (result > 429496729) {
m++; //if reversed 9digits is already bigger 429496729,
//then 'result = result * 10 + digit', it will have over flow problem.
}
result = result * 10 + digit;
temp /= 10;
if (temp == 0) break;
}
printf("m=%d\n", m);
if (m >= 1)
printf("out of range\n");
else
printf("result is %u\n", result);
printf("\n");
}
return 0;
}

Answer 1

You never return the reversed number, all you return is the final (rightmost) digit.
And your function doesn't "return itself", it returns the result of calling itself, which is not the same.

Answer 2

Your test for overflow is incorrect because you compare the unsigned int result with a value that is larger than the range of this type, so the test always fails.
There are 2 ways to check for overflow:
taking advantage of unsigned arithmetics, you can compare is the result is smaller than the previous result.
alternately, you can compare if result is larger than UINT_MAX / 10 or if it is equal, if digit is larger than UINT_MAX % 10. This method works for all integer types, unsigned and signed.
Here is the modified code:
#include <stdio.h>
#include <limits.h>
int main(void) {
unsigned int x;
printf("max unsigned int = %u\n\n", UINT_MAX);
printf("please enter any unsigned int,\n"
"it will show the reverse number if it is in range\n"
"(enter 10 digit only)\n\n");
while (scanf("%u", &x) == 1) {
unsigned int temp = x, result = 0;
int overflow = 0;
while (temp > 0) {
unsigned int digit = temp % 10;
if (result > UINT_MAX / 10
|| (result == UINT_MAX / 10 && digit > UINT_MAX % 10)) {
overflow++; // the result will not fit
break;
}
result = result * 10 + digit;
temp /= 10;
}
if (overflow)
printf("out of range for unsigned int type\n\n");
else
printf("result is %u\n\n", result);
}
return 0;
}

Answer 3

Thanks for your answer, much appreciated!
1. Regarding to "compare if result is larger than UINT_MAX / 10 or if it is equal, if digit is larger than UINT_MAX % 10. This method works for all integer types, unsigned and signed."
Is (result == UINT_MAX / 10 && digit > UINT_MAX % 10)) necessary?
I assume input below digits
(reminder: only enter digit within UINT_MAX)
unsigned int 4294967295 (10digit), reverse will be 592769492(4) /out
range
unsigned int 4294967294 (10digit), reverse will be 492769492(4)
/out range
unsigned int 4227694924 (10digit), reverse will be 429496722(4) / in
range
unsigned int 927694924(9digit), reverse will be 429496729 / in
range
the closest input is 3927694924 i can think of, and it will fit (result == UINT_MAX / 10 && digit > UINT_MAX % 10)) / in range
However, I've done the same for signed int, please see the code at below, but im not sure how to apply your method of signed int:
#include <stdio.h>
#include <limits.h>
int main(void) {
int x;
printf("max int = %d\n", INT_MAX);
printf("min int = %d\n\n", INT_MIN);
printf("please enter any integer within -2,147,483,648 ~ 2,147,483,647,\n"
"it will show the reverse number if it is in range\n"
"(enter 10 digit or less)\n\n");
while (scanf("%d", &x) == 1) {
int temp = x, result = 0;
int overflow = 0;
while (temp >= INT_MIN || temp <= INT_MAX) {
int digit = temp % 10;
printf("temp is %u\n", temp);
printf("result is %u\n", result);
if (result > INT_MAX / 10 || (result == INT_MAX / 10 && digit > INT_MAX % 10)) {
overflow++; // the result will not fit
break;
}
if (result < INT_MIN / 10 || (result == INT_MIN / 10 && digit > INT_MIN % 10)) {
overflow++; // the result will not fit
break;
}
result = result * 10 + digit;
temp /= 10;
if(temp == 0) break;
}
if (overflow)
printf("out of range for int type\n\n");
else
printf("result is %d\n\n", result);
}
return 0;
}