Order of operations not correct? (C++) [closed] - c++

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I was writing a small Least common multiple algorithm and encountered something I don't understand. This is the first and last part of the code:
long a = 14159572;
long b = 63967072;
int rest = 4;
long long ans;
.
. // Some other code here that is not very interesting.
.
else
{
//This appears correct, prints out correct answer.
ans = b/rest;
std::cout << a*ans;
}
But if I change the last "else" to this it gives an answer that is much smaller and incorrect:
else
{
std::cout << a*(b/rest);
}
Anyone know why this is? I don't think it's an overflow since it was no negative number that came out wrong, but rather just a much smaller integer (around 6*10^8) than the actual answer (around 2.2*10^14). As far as I understand it should calculate "b/rest" first in both cases, so the answers shouldn't differ?

Answer 1

Difference is not order of operations but data types:
ans = b/rest; // b/rest is long which upscaled to long long
std::cout << a*ans; // a converted to long long and result is long long
vs:
std::cout << a*(b/rest); // a*(b/rest) all calculations in long
so if you change your second variant to:
std::cout << a*static_cast<long long>(b/rest);
you should see the same result.
Update to why your cast did not work, note the difference:
long a,b;
// divide `long` by `long` and upscale result to `long long`
std::cout << static_cast<long long>( a / b );
// upscale both arguments to `long long` and divide `long long` by `long long`
std::cout << a / static_cast<long long>( b );

Answer 2

You're still encountering overflow. Just because you're not observing a negative number doesn't mean there's no overflow.
In your case specifically, long is almost certainly a 32-bit integer, as opposed to long long which is probably a 64-bit integer.
Since the maximum value of a 32-bit signed integer is roughly 2 billion, 14159572 * (63967072 / 4) is most definitely overflowing the range.
Make sure you perform your calculations using long long numbers, or else reconsider your code to avoid overflow in the first place.

Answer 3

The compiler assumes data types for each operand of your math equation and does the multiplication and division according to those assumed data types (refer to "integer division"). This also applies to intermediates of the computation. This also applies to the result passed to the stream since you don't pass a variable of an explicitly defined type.