Integer overflow and std::stoi - c++

Question

if x > INT_MAX or if x > INT_MIN the function will return 0... or that's what i'm trying to do :)
in my test case i pass in a value that is INT_MAX + 1... 2147483648 ... to introduce integer overflow to see how the program handles it.
i step through... my IDE debugger says that the value immediately goes to -2147483648 upon overflow and for some reason the program executes beyond both of these statements:
if (x > INT_MAX)
if (x < INT_MIN)
and keeps crashes at int revInt = std::stoi(strNum);
saying out of range
Must be something simple, but it's got me stumped. Why isn't the program returning before it ever gets to that std::stoi() given x > INT_MAX? Any help appreciated. Thanks! Full listing of function and test bed below: (sorry having trouble with the code insertion formatting..)
#include <iostream>
#include <algorithm>
#include <string> //using namespace std;
class Solution {
public: int reverse(int x)
{
// check special cases for int and set flags:
// is x > max int, need to return 0 now
if(x > INT_MAX)
return 0;
// is x < min int, need to return 0 now
if(x < INT_MIN)
return 0;
// is x < 0, need negative sign handled at end
// does x end with 0, need to not start new int with 0 if it's ploy numeric and the functions used handle that for us
// do conversion, reversal, output:
// convert int to string
std::string strNum = std::to_string(x);
// reverse string
std::reverse(strNum.begin(), strNum.end());
// convert reversed string to int
int revInt = std::stoi(strNum);
// multiply by -1 if x was negative
if (x < 0)
revInt = revInt * -1;
// output reversed integer
return revInt;
}
};
Main:
#include <iostream>
int main(int argc, const char * argv[]) {
// test cases
// instance Solution and call it's method
Solution sol;
int answer = sol.reverse(0); // 0
std::cout << "in " << 0 << ", out " << answer << "\n";
answer = sol.reverse(-1); // -1
std::cout << "in " << -1 << ", out " << answer << "\n";
answer = sol.reverse(10); // 1
std::cout << "in " << 10 << ", out " << answer << "\n";
answer = sol.reverse(12); // 21
std::cout << "in " << 12 << ", out " << answer << "\n";
answer = sol.reverse(100); // 1
std::cout << "in " << 100 << ", out " << answer << "\n";
answer = sol.reverse(123); // 321
std::cout << "in " << 123 << ", out " << answer << "\n";
answer = sol.reverse(-123); // -321
std::cout << "in " << -123 << ", out " << answer << "\n";
answer = sol.reverse(1024); // 4201
std::cout << "in " << 1024 << ", out " << answer << "\n";
answer = sol.reverse(-1024); // -4201
std::cout << "in " << -1024 << ", out " << answer << "\n";
answer = sol.reverse(2147483648); // 0
std::cout << "in " << 2147483648 << ", out " << answer << "\n";
answer = sol.reverse(-2147483648); // 0
std::cout << "in " << -2147483648 << ", out " << answer << "\n";
return 0;
}

Answer 1

Any test like (x > INT_MAX) with x being of type int will never evaluate to true, since the value of x cannot exceed INT_MAX.
Anyway, even if 2147483647 would be a valid range, its reverse 7463847412 is not.
So I think its better to let stoi "try" to convert the values and "catch" any out_of_range-exception`. The following code illustrates this approach:
int convert() {
const char* num = "12345678890123424542";
try {
int x = std::stoi(num);
return x;
} catch (std::out_of_range &e) {
cout << "invalid." << endl;
return 0;
}
}