Question

This question is a result of another SO question.

Example Code

#include <iostream>

int main()
{
    unsigned long b = 35000000;
    int i = 100;
    int j = 30000000;
    unsigned long n = ( i * j ) / b; // #1
    unsigned long m = ( 100 * 30000000 ) / b; // #2
    std::cout << n << std::endl;
    std::cout << m << std::endl;
}

Output

85
85

Compiling this code with g++ -std=c++11 -Wall -pedantic -O0 -Wextra gives the following warning:

9:28: warning: integer overflow in expression [-Woverflow]

Questions

1. Am I correct in thinking that #1 and #2 invoke undefined behavior because the intermediate result 100 * 30000000 does not fit into an int? Or is the output I am seeing well-defined?

2. Why do I only get a warning with #2?

Answer 1

Yes, it is undefined behaviour, and the result you get is usually¹ different if unsigned long is a 64-bit type.

¹ It’s UB, so there are no guarantees.