Thread: help needed to know how the code is running

In Java for an arithmetic expression involving two char variables, the operands are promoted to ints and the result is an int.

Then, when you use println("whatever" + (c1+c2)) it prints the integer value of the sum of integer 64 and integer 187

On the other hand...

When you write char c3 = 64+187, it automatically converts the integer sum of 64 and 187 to a char and stores it in c3. Then when you use println("whatever" + c3) you get the Java char corresponding to the integer value 251 (shows up on my system as a lower-case Latin u with circumflex).

The topic Numeric Promotions and its implications is spelled out in section 5.6 of the Java Language Specification. Maybe there is some information in whatever reference material you are learning Java from. (Class notes? Textbook? What?)

Maybe you can play around with something like

public class Z
{
    public static void main(String [] args)
    {
        char c1, c2, c3, c4;
        c1 = 64;
        c2 = 187;
        c3 = 64 + 187;
        c4 = (char)(c1 + c2);// Cast needed to prevent compiler warning
 
        System.out.println("          c1  = " + c1);
        System.out.println("          c2  = " + c2);
        System.out.println("          c3  = " + c3);
        System.out.println("          c4  = " + c4);
        System.out.println();
 
        System.out.println("     (int)c1  = " + (int)c1);
        System.out.println("     (int)c2  = " + (int)c2);
        System.out.println("     (int)c3  = " + (int)c3);
        System.out.println("     (int)c4  = " + (int)c4);
        System.out.println();
 
        System.out.println("       c1+c2  = " + (c1+c2));
        System.out.println("(char)(c1+c2) = " + (char)(c1+c2));
    }
}

See if the output makes sense. If not, tell us what it printed and what you expected it to print and what you don't understand about the difference(s).

Cheers!

Z

Originally Posted by pbrockway2

What has this got to do with the subject of this thread?

Actually it is related because it concerns the effects of Numeric Promotions, and this question is a continuation of investigation of the concepts illustrated by the original post in this thread and by my previous response. The problem (from my point of view) is that the title of the thread is singularly uninformative.

The other recent thread from this Poster has a similar title, but is about a completely different topic, namely, about using an int condition in an if() statement.

To the Original Poster: Try to choose appropriate titles for your threads.

To the Community: Sometimes when people start threads they don't know the exact terminology to make the titles something that is "textbook-precise." If I think I can help, I'll give it a shot. If others think that it's not worth their efforts to try to figure out where it's coming from and don't want to try, well, as my dear old Earthling-adoptive grannie used to say, "Chacun à son goût!" Actually, she used the cajun equivalent "A chacun son goût," but her snobbish son (my adoptive Uncle Whose Name Must Not Be Spoken) tried to refine it by using textbook French. But that's another story...


Anyhow...

Originally Posted by Dark knight

why this code is generating -1

In the expression (byte)(b >>> 10), b is promoted to an int. The int is shifted right (with zeros shifted into the upper bits), and then you cast it to a byte (which merely chops off the upper 24 bits of the int. Since the eight lower bits are '1' the result is decimal -1.

Here's an example, where I have separated out the conversion to byte. I lined up the bits of the byte variable below the place in the int variable where the byte got its value. I put everything in a loop so that we can see the results as it is shifted one bit at a time:

public class z
{
    public static void main(String [] args)
    {
        byte b =-2;
 
        for (int i = 0; i < 11; i++)
        {
            int intc = (b >>> i);
            byte bytec = (byte)intc; // Cast is necessary to make the compiler tolerate the loss of precision
            System.out.printf("i = %2d: Decimal value of intc  = %d\n", i, intc);// output -1
            System.out.printf("        Binary  value of intc  = %32s\n", binaryInt(intc));
            System.out.printf("        Binary  value of bytec = %32s\n", binaryByte(bytec));
            System.out.printf("        Decimal value of bytec = %d\n\n", bytec);// output -1
        }
    }
 
    // Convert a byte to a binary string with leading zeros
    static String binaryByte(byte x)
    {
 
        String str = "";
        for (int mask = 0x80; mask != 0; mask >>>= 1)
        {
            if ((mask & x) == 0) 
            {
                str += '0';
            }
            else
            {
                str += '1';
            }
        }
        return str;
    }
 
    // Convert an int to a binary string with leading zeros
    static String binaryInt(int x)
    {
 
        String str = "";
        for (int mask = 0x80000000; mask != 0; mask >>>= 1)
        {
            if ((mask & x) == 0) 
            {
                str += '0';
            }
            else
            {
                str += '1';
            }
        }
        return str;
    }
}

Output:

i =  0: Decimal value of intc  = -2
        Binary  value of intc  = 11111111111111111111111111111110
        Binary  value of bytec =                         11111110
        Decimal value of bytec = -2
 
i =  1: Decimal value of intc  = 2147483647
        Binary  value of intc  = 01111111111111111111111111111111
        Binary  value of bytec =                         11111111
        Decimal value of bytec = -1
 
i =  2: Decimal value of intc  = 1073741823
        Binary  value of intc  = 00111111111111111111111111111111
        Binary  value of bytec =                         11111111
        Decimal value of bytec = -1
 
i =  3: Decimal value of intc  = 536870911
        Binary  value of intc  = 00011111111111111111111111111111
        Binary  value of bytec =                         11111111
        Decimal value of bytec = -1
 
i =  4: Decimal value of intc  = 268435455
        Binary  value of intc  = 00001111111111111111111111111111
        Binary  value of bytec =                         11111111
        Decimal value of bytec = -1
.
.
.
i = 10: Decimal value of intc  = 4194303
        Binary  value of intc  = 00000000001111111111111111111111
        Binary  value of bytec =                         11111111
        Decimal value of bytec = -1

The bottom line for me is: If you have to use a cast, somewhere ("just to keep the compiler happy"), make absolutely sure that you understand the ramifications.


Cheers!

Z