Cannot convert from type object to long

2019-02-11 20:50发布

问题:

I have a hashtable named table. The type value is long. I am getting values using .values(). Now I want to access these values.

Collection val = table.values();

Iterator itr = val.iterator();
long a  =   (long)itr.next();

But when I try to get it, it gives me error because I can't convert from type object to long. How can I go around it?

回答1:

Try this:

  Long a = (Long)itr.next();

You end up with a Long object but with autoboxing you may use it almost like a primitive long.

Another option is to use Generics:

  Iterator<Long> itr = val.iterator();
  Long a = itr.next();


回答2:

Number class can be used for to overcome numeric data-type casting.

In this case the following code might be used:

long a = ((Number)itr.next()).longValue();



I've prepared the examples below:

Object to long example - 1

// preparing the example variables
Long l = new Long("1416313200307");
Object o = l;

// Long casting from an object by using `Number` class
System.out.print(((Number) o).longValue() );

Console output would be:

1416313200307



Object to double example - 2

// preparing the example variables
double d = 0.11;
Object o = d;

// Double casting from an Object -that's a float number- by using `Number` class
System.out.print(((Number) o).doubleValue() + "\n");

Console output would be:

0.11



Object to double example - 3

Be careful about this simple mistake! If a float value is converted by using doubleValue() function, the first value might not be equal to final value.
As shown below 0.11 != 0.10999999940395355.

// preparing the example variables
float f = 0.11f;
Object o = f;

// Double casting from an Object -that's a float number- by using `Number` class
System.out.print(((Number) o).doubleValue() + "\n");

Console output would be:

0.10999999940395355



Object to float example - 4

// preparing the example variables
double f = 0.11;
Object o = f;

// Double casting from an Object -that's a float number- by using `Number` class
System.out.print(((Number) o).floatValue() + "\n");

Console output would be:

0.11


回答3:

Try : long a = ((Long) itr.next()).longValue();



回答4:

You should use the new Generics features from Java 5.

When you take an element out of a Collection, you must cast it to the type of element that is stored in the collection. Besides being inconvenient, this is unsafe. The compiler does not check that your cast is the same as the collection's type, so the cast can fail at run time.

Generics provides a way for you to communicate the type of a collection to the compiler, so that it can be checked. Once the compiler knows the element type of the collection, the compiler can check that you have used the collection consistently and can insert the correct casts on values being taken out of the collection.

You can read this quick howto or this more complete tutorial.



回答5:

in my case I have an array of Objects that I get from a flex client,

sometimes the numbers can be interpreted by java as int and sometimes as long,

so to resolve the issue i use the 'toString()' function as follows:

public Object run(Object... args) {

  final long uid = Long.valueOf(args[0].toString());


回答6:

long value = Long.parseLong((String)request.getAttribute(""));


回答7:

I faced the same problem but while doing JSP coding. The above mentioned suggestions regarding Long and generics either did not work or did not fit into the code fragment.

I had to solve it like this(in JSP):

<%Object y=itr.next(); %>

and afterwards use my Object y like <%=y%> as we would use any other Java variable in scriptlet.