Integer overflow takes place even if I type cast to long int (C ) [closed]-CodePudding

I am trying to understand the phenomenon of overflow and decided to demonstrate the same by executing the piece of code provided below:

#include <bits/stdc  .h>

using namespace std;

int main(){
    int a = 100000;
    int b = 100000;
    cout << a * b << endl; 
    long int c = a * 1LL * b;
    cout << c;
    return 0;
}

The idea is that simply multiplying a and b (as done in line 6) would result in an integer overflow as 10^10 falls outside the range of the integer data type. To overcome this, the product (a*b) is multiplied with the literal '1LL' (as shown in line 7). The problem is that integer overflow still takes place as the output of the program is:

1410065408
1410065408

It's possible that I am making a super silly mistake somewhere, but I have spent a sufficient amount of time trying to understand where it's going wrong and am yet to find the reason. Hoping for some help/guidance here :)

CodePudding user response：

Type of 1LL is long long int. Its size is at least 8 bytes. Type of c is long int. Its size is at least 4 bytes. Thus you may still have overflow. To avoid this you may use auto:

#include <iostream>
#include <cstdint>

int main() {
  std::uint32_t a = 100'000;
  std::uint32_t b = 100'000;

  auto r1 = a * b;
  auto r2 = static_cast<std::uint64_t>(a) * b;

  std::cout << "r1 = " << r1 << '\n' 
    << "r2 = " << r2 << '\n';
}

Try Online

CodePudding user response：

Seems to be environment problem - works fine for me.

cat long_int_ex.cpp; g   -O2 long_int_ex.cpp ; a.out; g   --version
#include <bits/stdc  .h>

using namespace std;

int main(){
    int a = 100000;
    int b = 100000;
    cout << a * b << endl; 
    long int c = a * 1LL * b;
    cout << c << endl;
    return 0;
} 
1410065408
10000000000
g   (Ubuntu 9.3.0-17ubuntu1~20.04) 9.3.0
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