[java] A long bigger than Long.MAX_VALUE

How can I get a long number bigger than Long.MAX_VALUE?

I want this method to return true:

boolean isBiggerThanMaxLong(long val) {
    return (val > Long.MAX_VALUE);
}

If triangle.lborderA is indeed a long then the test in the original code is trivially true, and there is no way to test it. It is also useless.

However, if triangle.lborderA is a double, the comparison is useful and can be tested. isBiggerThanMaxLong(1e300) does return true.

  public static boolean isBiggerThanMaxLong(double in){
    return in > Long.MAX_VALUE;
  }

You can't. If you have a method called isBiggerThanMaxLong(long) it should always return false.

If you were to increment the bits of Long.MAX_VALUE, the next value should be Long.MIN_VALUE. Read up on twos-complement and that should tell you why.

Firstly, the below method doesn't compile as it is missing the return type and it should be Long.MAX_VALUE in place of Long.Max_value.

public static boolean isBiggerThanMaxLong(long value) {
      return value > Long.Max_value;
}

The above method can never return true as you are comparing a long value with Long.MAX_VALUE , see the method signature you can pass only long there.Any long can be as big as the Long.MAX_VALUE, it can't be bigger than that.

You can try something like this with BigInteger class :

public static boolean isBiggerThanMaxLong(BigInteger l){
    return l.compareTo(BigInteger.valueOf(Long.MAX_VALUE))==1?true:false;
}

The below code will return true :

BigInteger big3 = BigInteger.valueOf(Long.MAX_VALUE).
                  add(BigInteger.valueOf(Long.MAX_VALUE));
System.out.println(isBiggerThanMaxLong(big3)); // prints true