int_max , int_min, leetcode problem

[DonnaPin]

Hello, I am trying some leetcode problems but couldnt solve it and had to look at the solution. But I am a bit confused with it. Mostly the 2 lines with INT_MAX and INT_MIN. I understand that these are the max and min that an int can be, but I just cant figure out how it is working in this piece of code, especially with the >7 and <8.

If possible could someone please explain this to me

   int reverse(int x) 
    {
         int rev = 0;
        while (x != 0) {
            int pop = x % 10;
            cout <<"pop " << pop << "\n";
            x /= 10;
            cout << "x " << x << "\n";
            if (rev > INT_MAX/10 || (rev == INT_MAX / 10 && pop > 7)) return 0;
            if (rev < INT_MIN/10 || (rev == INT_MIN / 10 && pop < -8)) return 0;
            rev = rev * 10 + pop;
            cout << "rev " << rev << "\n";
        }
        return rev;      
    }

[jonnin]

the return zeros are error handling of sorts.
if its > int max / 10, then multiply by 10 on line 11 won't work right, so it quits.
if its < int min/10 you have a similar problem with a negative number.
the 7 and 8 are hard coded looking at the issue bit-wise. If you don't see it, look at an 8 bit number and it will pop out, its a lot of typing 1s and 0s to show...

[salem c]

It's basically to detect numeric overflow and underflow.

And because it's trying to reverse a number, the 7 and -8 come from the least significant digit of the number (which is about to become the most significant digit of the number).

$ cat baz.cpp
#include <iostream>
#include <climits>
using namespace std;

int main()
{
  cout << "INT_MIN=" << INT_MIN << endl;
  cout << "INT_MAX=" << INT_MAX << endl;
  return 0;
}
$ g++  baz.cpp
$ ./a.out 
INT_MIN=-2147483648
INT_MAX=2147483647