helpfullpac

After upgrading from the 2.1.5 to 2.1.6 our code generation is failing with (we actually have it both failing on a Double and Integer conversion, example here with Double): Exception: java.lang.Double cannot be cast to java.lang.String a.

• Livy String Cannot Be Cast To Java.lang.integer Free

First, the return type of getTileCount should be an int.Then you want to declare and initialize a variable named count, which will be of type int.When you loop through the characters, your if statement should compare each of the tiles with the passed in tile.If you have a match then increment the count variable. Finally, return count. java.lang.Integer cannot be cast to java.lang.Long at scala.runtime.BoxesRunTime.unboxToLong(Unknown Source). Omitted stack is a code with asInstanceOfLong call for result I get with getObject from jdbc ResultSet. The stack is magic itself.

I have a String object and I need to convert to java.lang.Number.

If I try to cast the cellContents to Number directly,

it throws an exception:

I searched but could not get a complete answer as to how I can achieve this.Please help!!!

6 Answers

You can't just cast the reference type. You're getting an exception because the String object to which it points is not a Number object.

You can, however, cast the reference to a String, if you know it's a String. You can convert it to a real value with Double.valueOf( String ) or Float.valueOf( String ). Once you get a double, you can use auto-boxing to turn it into a Double, which isa Number.

Andy ThomasAndy Thomas

You can't cast a String to a Number, because they are not covariant (they don't fall in the same inheritance hierarchy).

You can convert the String to float or double using Float#parseFloat() and Double#parseDouble() respectively, which is what you would need in most cases.

Using Float#valueOf(String) and Double#valueOf(String) creates instances of wrapper classes.

So, depending upon what you need, you can use any of them. I'll show here the first parseXXX methods:

However, if you are dealing with monetory value, or some other value, where floating point precision is of concern, then you should use

Livy String Cannot Be Cast To Java.lang.integer Free

BigDecimal instead:Rohit JainRohit Jain

You can not simply convert a String into Number directly as they are not in same hierarchy.But good news is Java Wrapper API do most of the task for you automatically IE, convert a valid String into Wrapper you want and throws a NumberFormatException if the passed String is not a valid Number or other Wrapper.

You should use.

OR Office 365 keeps asking for activation.

they both return Number Integer and Double respectively and both ARE-A Number.

Sachin VermaSachin VermaThiloThilo

Firstly, you should cast your object to string.

You can try something like the following code:

You cannot cast '.475' to Integer. However you can cast it to Float or Double using

or

You should search more about Wrapper Classes in Java and AutoBoxing in java to get basic clear. Happy Coding :)

Abhishek SinghAbhishek Singh

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