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Write boolean function multiplyWillOverflow which takes two integers x, y and return true if (x*y) overflows.

An overflow may be above the max positive value, or below the min negative value.
New implementation

Be concise.

Be useful.

All contributions dictatorially edited by webmasters to match personal tastes.

Please do not paste any copyright violating material.

Please try to avoid dependencies to third-party libraries and frameworks.

Other implementations
public bool WillOverwflow(int x, int y) => int.MaxValue / x < y;
public bool multiplyWillOverflow(int x, int y) 
{
	if (x == 0)
		return false;

	if (y > int.MaxValue / x)
		return true;

	if (y < int.MinValue / x)
		return true;

	return false;
}
import core.checkedint;
bool multiplyWillOverflow(int x, int y)
{
    bool result;
    core.checkedint.muls(x, y, result);
    return result;
}
  logical function multiply_will_overflow (x, y) result(res)
    integer, intent(in) :: x, y
    integer, parameter :: ik = selected_int_kind (int(digits(y)*log10(2.)*2))

    res = int(x,kind=ik) * int(y,kind=ik) > huge(x)
  end function multiply_will_overflow
func multiplyWillOverflow(x, y uint64) bool {
   if x <= 1 || y <= 1 {
     return false
   }
   d := x * y
   return d/y != x
}
static boolean multiplyWillOverflow(int x, int y) {
	return Integer.MAX_VALUE/x < y;
}
function multiplyWillOverflow($x, $y)
{
      return ($x * $y) > PHP_INT_MAX;
}
function MultiplyWillOverflow(x, y: Integer): Boolean;
begin
  if ((x and y) = 0) then
    Result := False
  else
  begin
    if ((x > 0) and (y > 0)) or ((x < 0) and (y < 0)) then
      Result := ((High(Integer) div Abs(x)) > Abs(y))
    else
      Result := Abs(Low(Integer) div Abs(x)) > Abs(y);
  end;
end; 
sub multiply_will_overflow {
    my ($x, $y) = @_;
    return 'Inf' eq $x * $y;
}
def multiplyWillOverflow(x,y):
	return False
def multiplyWillOverflow(x,y)
  false
end
fn multiply_will_overflow(x: i64, y: i64) -> bool {
    x.checked_mul(y).is_none()
}