Comments by ydlrr (Top 4 by date)

If I enter a number for x that is more than imax the result is always the imax number instead of cout << "overflow\n";

If cout << imax; no matter the input of the integer lets say 5 by the user the result will be 2147483647 so not sure what you mean by adding cout << imax;

I am new to C++ programming and have not gone over signed and unsigned but will research it. Thanks for your help.