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The reason Java (and most other computer languages) uses two s complement is easy to see when you consider the issue of zero crossing Assuming a byte value, zero is represented by 00000000 In one s complement, simply inverting all of the bits creates 11111111, which creates negative zero The trouble is that negative zero is invalid in integer math This problem is solved by using two s complement to represent negative values When using two s complement, 1 is added to the complement, producing 100000000 This produces a 1 bit too far to the left to fit back into the byte value, resulting in the desired behavior, where 0 is the same as 0, and 11111111 is the encoding for 1 Although we used a byte value in the preceding example, the same basic principle applies to all of Java s integer types Because Java uses two s complement to store negative numbers and because all integers are signed values in Java applying the bitwise operators can easily produce unexpected results For example, turning on the high-order bit will cause the resulting value to be interpreted as a negative number, whether this is what you intended or not To avoid unpleasant surprises, just remember that the high-order bit determines the sign of an integer no matter how that high-order bit gets set

Floppy disks have been produced in a variety of sizes and capacities over the years These include the following:

The bitwise logical operators are &, |, ^, and ~ The following table shows the outcome of each operation In the discussion that follows, keep in mind that the bitwise operators are applied to each individual bit within each operand

525" Single Sided, Double Density: 180 KB 525" Double Sided, Double Density: 360 KB 525" Double Sided, High Density: 12 MB 35" Double Sided, Double Density: 720 KB 35" Double Sided, High Density: 144 MB (This is the type most commonly

Also called the bitwise complement, the unary NOT operator, ~, inverts all of the bits of its operand For example, the number 42, which has the following bit pattern: 00101010 becomes 11010101 after the NOT operator is applied

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The AND operator, &, produces a 1 bit if both operands are also 1 A zero is produced in all other cases Here is an example: 00101010 & 00001111 00001010 42 15 10

used today) Floppy disks have three key disadvantages:

Part I:

The OR operator, |, combines bits such that if either of the bits in the operands is a 1, then the resultant bit is a 1, as shown here: 00101010 | 00001111 00101111 42 15 47

They are very slow They don t hold very much Diskettes degrade and lose data after a short amount of time

The XOR operator, ^, combines bits such that if exactly one operand is 1, then the result is 1 Otherwise, the result is zero The following example shows the effect of the ^ This example also demonstrates a useful attribute of the XOR operation Notice how the bit pattern of 42 is inverted wherever the second operand has a 1 bit Wherever the second operand has a 0 bit, the first operand is unchanged You will find this property useful when performing some types of bit manipulations 00101010 ^ 00001111 00100101 42 15 37

Before you do anything from the client NFS system, you first need to configure its host-based firewall to allow NFS traffic on IP port 2049 If you don t, you ll spend hours banging your head on the wall wondering why it doesn t work (Trust me, I know this from experience) You can mount an NFS export in two different ways First, you can mount it from the shell prompt using the mount command To do this, you first need to verify that

This parallel beam will let me compare value A to value B. Remember, I want to check the statement A greater than B and determine if it is True or False. The first thing I need to do is drop the COMPARE block onto the new beam, as shown in Figure 15-6.

The following program demonstrates the bitwise logical operators:

// Demonstrate the bitwise logical operators class BitLogic { public static void main(String args[]) { String binary[] = { "0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111" }; int a = 3; // 0 + 2 + 1 or 0011 in binary int b = 6; // 4 + 2 + 0 or 0110 in binary int c = a | b; int d = a & b; int e = a ^ b; int f = (~a & b) | (a & ~b); int g = ~a & 0x0f; Systemoutprintln(" a Systemoutprintln(" b Systemoutprintln(" a|b Systemoutprintln(" a&b Systemoutprintln(" a^b Systemoutprintln("~a&b|a&~b Systemoutprintln(" ~a } } = = = = = = = " " " " " " " + + + + + + + binary[a]); binary[b]); binary[c]); binary[d]); binary[e]); binary[f]); binary[g]);

FIGURE 14-54

4:

FIGURE 14-55

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