检测 C 中的有符号 integer 乘法溢出

Question

I'm writing this code for more than 3 hours already..

I gave up about the overflow thing and tried to google and look it up on stackoverflow.

I did not find any solution besides the one that I wrote in my code as you can see in lines 27-28 (where it returns 0). But this condition also does not work.

#include <stdio.h>

int reverse(int x) {
  int pos = 0;
  int reversed = 0;
  int numOfDigits = 0;
  int tenPower = 1;
  if (x < 0) {
    pos = -x;
  } else
    pos = x;
  while (pos > 0) {
    pos = (pos - (pos % 10)) / 10;
    numOfDigits++;
  }
  while (numOfDigits > 0) {
    for (int i = numOfDigits - 1; i > 0; i--) {
      if (numOfDigits == 1)
        tenPower = 1;
      else
        tenPower *= 10;
    }
//overflow check - does not work
    if (x % 10 != 0 && ((x % 10) * tenPower) / (x % 10) != tenPower)
      return 0;
    reversed += (x % 10) * tenPower;
    numOfDigits--;
    x = (x - (x % 10)) / 10;
    tenPower = 1;
  }
  if (x < 0)
    return -reversed;
  else
    return reversed;
}

int main() {
  int arr[5] = {-30, 120, 1501, 321, 0};
  for (int i = 0; i < 5; i++) {
    printf("Original number is: %d \n", arr[i]);
    printf("Reversed number is: %d \n", reverse(arr[i]));
  }
}

The input that is not working due to overflow is:

1534236469

The error code on leetcode is

Line 25: Char 28: runtime error: signed integer overflow:

1000000000 * 9 cannot be represented in type 'int' [solution.c]

Line: if (x%10 != 0 &&((x%10)*tenPower) / (x%10) != tenPower)

Other than that, the code is working and every number (positive & negative numbers) is being successfully reversed.

I'll be glad to hear you out about a possible solution and also let me know what do you think about my code and the way I decided to complete this task, I know that's the most basic and naïve way to do it, but I'll be glad to know how could I improve it.

The task is:

Given a signed 32-bit integer x, return x with its digits reversed. If reversing x causes the value to go outside the signed 32-bit integer range [-231, 231 - 1], then return 0.

Assume the environment does not allow you to store 64-bit integers (signed or unsigned).

Examples:

Input: x = 123 Output: 321, Input: x=-120 Output = -21

Answer 1

The main thing you want to check for overflow is this:

reversed += (x % 10) * tenPower;

So what you want to know is if this is true:

((x % 10) * tenPower) > INT_MAX

Or this is true:

(reversed + (x % 10) * tenPower) > INT_MAX

Of course these can never be true in code due to overflow, but we can rearrange the terms:

if ((x % 10) != 0 && (tenPower > INT_MAX / (x % 10)) || 
                      (((x % 10) * tenPower) > INT_MAX - reversed))
    return 0;

Answer 2

You can do something like this.

int reverse(int n) {
    // 1st overflow checking...
    // Check if the absolute value is greater than INT32_MAX.
    // I converted "int" to "long" to get correct overflow value.
    if (n < 0 && (long) n * -1l > INT32_MAX) {
        return 0;
    }
    
    int res = 0;
    // Convert to absolute value.
    int tmp = n < 0 ? n * -1 : n;
    
    while (tmp > 0) {
        // Get the right most digit and add it to the current result.
        res += tmp % 10;
        // Remove the right most digit.
        tmp /= 10;
        
        // "tmp" still has remaining numbers.
        if (tmp > 0) {
            // 2nd overflow checking...
            // Check if reversed value will be greater than INT32_MAX when appending 0 to right most.
            // I converted "int" to "long" to get correct overflow value.
            if ((long) res * 10l > INT32_MAX) {
                return 0;
            }
            
            // Append 0 to right most value of result.
            // If result is equal to 0, do not append 0.
            res *= res == 0 ? 1 : 10;
        }
    }
    
    // Return result.
    // If original value is negative, return negative result value..
    return n < 0 ? res * -1 : res;
}

Answer 3

OP's code is subject to int overflow in various places

  if (x < 0) { pos = -x; }
  ...
  (x % 10) * tenPower
  ....
  reversed += (x % 10) * tenPower;
  ...
  if (x < 0) return -reversed;

A simple overflow pre-test involves INT_MAX/INT_MIN

  if (x < 0) { 
    if (x < -INT_MAX) { puts("Overflow"); return 0; }
    pos = -x;
  }

  // x is >= 0 here, tenPower >= 1
  int digit = x % 10;
  if (digit > INT_MAX/tenPower)     { puts("Overflow"); return 0; }
  int digit10 = digit * tenPower;
  if (reversed > INT_MAX - digit10) { puts("Overflow"); return 0; }
  reversed += digit10;

---

Or use stand-alone full range tests .

// Return 1 on overflow
int is_undefined_mult1(int a, int b) {
  if (a > 0) {
    if (b > 0) {
      return a > INT_MAX / b;       // a positive, b positive
    }
    return b < INT_MIN / a;         // a positive, b not positive
  }
  if (b > 0) {
    return a < INT_MIN / b;         // a not positive, b positive
  }
  return a != 0 && b < INT_MAX / a; // a not positive, b not positive
}

  int digit = x % 10;
  if (is_undefined_mult1(digit, tenPower))  { puts("Overflow"); return 0; }
  int digit10 = digit * tenPower;
  if (is_undefined_add1(reversed, digit10)) { puts("Overflow"); return 0; }
  reversed += digit10;

Answer 4

You can use bit_length() to check integer size

if digit.bit_length() >= 32:
        return 0
    else:
        return reversed