Programming-Idioms

Implementation
Fortran

Be concise.

Be useful.

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Instead of changing the code of the snippet, consider creating another Fortran implementation.

Other implementations
def multiplyWillOverflow(x,y):
	return False
import core.checkedint;
bool multiplyWillOverflow(int x, int y)
{
    bool result;
    core.checkedint.muls(x, y, result);
    return result;
}
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; 
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 * $y) > PHP_INT_MAX;
}
def multiplyWillOverflow(x,y)
  false
end
fn multiply_will_overflow(x: i64, y: i64) -> bool {
    x.checked_mul(y).is_none()
}
sub multiply_will_overflow {
    my ($x, $y) = @_;
    return 'Inf' eq $x * $y;
}
public bool WillOverwflow(int x, int y) => int.MaxValue / x < y;