error : The literal 1111111111111000 of type int is out of range

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the literal of type long is out of range
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This is the method that I am testing

public static char[] convertDecimalTo2sComp(int decimal) {
    char [] bits = new char [16];
    double temp;
    int tempInt;
    String bitStr = " ";


    for(int i = bits.length-1; i>=0; i--) {
        //negative input
        if(decimal <1) {
            temp = Math.pow(2,16); //add 2^number of bits, then convert to binary
            decimal += (int)temp;
        }
        //non-negative input
        tempInt = decimal%2;
        bitStr = tempInt + bitStr;
        decimal/=2;

    }
    bits = bitStr.toCharArray();

    return bits;
}

And this is what I'm getting the error: The literal 1111111111111000 of type int is out of range, in JUnit from

    @Test
public void testNegativeDecimal() { 
    int data = -8; 
    assertEquals(1111111111111000, Convert.convertDecimalTo2sComp(data)); }

How do I fix this?

Your problem is that the number you've given is outside of the range of an int (which has a max value of 2^31-1 https://docs.oracle.com/javase/tutorial/java/nutsandbolts/datatypes.html)

Since you've declared it as a literal java assumes that you number is an int.

The easiest way to declare that the number should be a Long instead (which has a max value of 2^63-1) assertEquals(1111111111111000L, Convert.convertDecimalTo2sComp(data));

Of course, you could also choose to make it explicit for the sake of readability

long expected = 1111111111111000;
assertEquals(expected, Convert.convertDecimalTo2sComp(data));

The literal xyz of type int is out of range, Add a capital L to the end: long value = 9223372036854775807L;. Otherwise, the compiler will try to parse the literal as an int , hence the error� The literal xyz of type int is out of range (4) I am working with data types at the moment in Java, and if I have understood correctly the type long accepts a value between the ranges of: -9,223,372,036,854 to +9,223,372,036,854,775,807.

1111111111111000 as int is indeed outside the int range. 10^16 = ca. 2^48 Indeed you intended to have a char array:

public void testNegativeDecimal() { 
    int data = -8; 
    char[] expected = "1111111111111000".toCharArray();
    assertEquals(expected, Convert.convertDecimalTo2sComp(data));
}

or a more lazy, less error prone solution:

public void testNegativeDecimal() { 
    int data = -8; 
    String expected = "0000000000000000" + Integer.toBinaryString(data);
    expected = expect.substring(expected.length() - 16);
    assertEquals(expected, new String(Convert.convertDecimalTo2sComp(data)));
}

The latter having the advantage that on error you see the (human readable) Strings.

The literal -1 of type int is out of range � Issue #285 � groovy/groovy , As soon as you have any other compile errors in your code, any constant (static final fields) of type int and long produce an additional compiler� I think you need to put an "L" on the end of your literal: long number=300425737572L; Literals are (again, I think) ints by default, and this is too big for an int.

Let me start by addressing your underlying confusion.

Judging from your method-name convertDecimalTo2sComp and your parameter-name decimal, you're expecting the argument to be "decimal" (base-10) in some sense. But in fact, that's not true: the argument is just an integer in the range [−231, 231−1], not specifically a base-10 representation of that integer. Internally, in fact, it's actually represented as a two's-complement binary integer — for example, if the integer is −8, then it's represented internally as 1111 1111 1111 1111 1111 1111 1111 1000 — but that only matters when using bit-level operations such as & (bitwise "and") and >> (right-shift). It may seem like integers are represented in base 10, because conversions to and from strings use base 10 by default (e.g., Integer.toString(32) is equivalent to Integer.toString(32, 10) and Integer.valueOf("32") is equivalent to Integer.valueOf("32", 10)), and because we usually write them in decimal form in the source code (e.g., we usually write the integer 32 as 32 rather than 0x20 or 040 or 0b10_0000); but that's all just for our convenience, and has nothing to do with the internal representation.

So, for example, these statements are all exactly equivalent:

System.out.println(32);
System.out.println(0x20);
System.out.println(040);
System.out.println(0b100000);
System.out.println(0b_0010_0000);

because in all cases, the integer 32 is represented internally as 0000 0000 0000 0000 0000 0000 0010 0000.


With that cleared up . . .

To assert that something is a character array containing thirteen '1'-s followed by three '0'-s, you can either use Assert.assertArrayEquals [link], or you can convert the char[] to a String and use Assert.assertEquals. So, any of these:

assertArrayEquals(
    new char[] {
        '1', '1', '1', '1',  '1', '1', '1', '1',
        '1', '1', '1', '1',  '1', '0', '0', '0'  },
    Convert.convertDecimalTo2sComp(data);

assertArrayEquals(
    "1111111111111000".toCharArray(),
    Convert.convertDecimalTo2sComp(data));

assertEquals(
    "1111111111111000",
    new String(Convert.convertDecimalTo2sComp(data)));

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The literal of type int is out of range?, Hello, I've just started a new piece of code and am getting an error already: The literal 600851475143 of type int is out of range Java Code:� // Error! literal 42332200000 of type int is out of range int myVariable1 = 42332200000;// 42332200000L is of type long, and it's not out of range long myVariable2 = 42332200000L; Integer literals can be expressed in decimal, hexadecimal and binary number systems. The numbers starting with prefix 0x represents hexadecimal.