visual basic 2010 barcode generator The Bitwise NOT in Java

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The Bitwise NOT
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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 Bitwise AND
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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 42 &00001111 15 00001010 10
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The Bitwise OR - 63 -
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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 42 | 00001111 15 00101111 47
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The Bitwise XOR
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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 42 ^00001111 15 00100101 37
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Using the Bitwise Logical Operators
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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]);
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In this example, a and b have bit patterns which present all four possibilities for two binary digits: 0-0, 0-1, 1-0, and 1-1 You can see how the | and & operate on each bit by the results in c and d The values assigned to e and f are the same and illustrate how the ^ works The string array named binary holds the human-readable, binary representation of the numbers 0 through 15 In this example, the array is indexed to show the binary representation of each result The array is constructed such that the correct string representation of a binary value n is stored in binary[n] The value of ~a is ANDed with
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0x0f (0000 1111 in binary) in order to reduce its value to less than 16, so it can be printed by use of the binary array Here is the output from this program: a b a|b a&b a^b ~a&b|a&~b ~a = = = = = = = 0011 0110 0111 0010 0101 0101 1100
The Left Shift
The left shift operator, <<, shifts all of the bits in a value to the left a specified number of times It has this general form: value << num Here, num specifies the number of positions to left-shift the value in value That is, the << moves all of the bits in the specified value to the left by the number of bit positions specified by num For each shift left, the high-order bit is shifted out (and lost), and a zero is brought in on the right This means that when a left shift is applied to an int operand, bits are lost once they are shifted past bit position 31 If the operand is a long, then bits are lost after bit position 63 Java's automatic type promotions produce unexpected results when you are shifting byte and short values As you know, byte and short values are promoted to int when an expression is evaluated Furthermore, the result of such an expression is also an int This means that the outcome of a left shift on a byte or short value will be an int, and the bits shifted left will not be lost until they shift past bit position 31 Furthermore, a negative byte or short value will be sign-extended when it is promoted to int Thus, the high-order bits will be filled with 1's For these reasons, to perform a left shift on a byte or short implies that you must discard the high-order bytes of the int result For example, if you left-shift a byte value, that value will first be promoted to int and then shifted This means that you must discard the top three bytes of the result if what you want is the result of a shifted byte value The easiest way to do this is to simply cast the result back into a byte The following program demonstrates this concept: // Left shifting a byte value class ByteShift { public static void main(String args[]) { byte a = 64, b; int i; i = a << 2; b = (byte) (a << 2); Systemoutprintln("Original value of a: " + a); Systemoutprintln("i and b: " + i + " " + b);
The output generated by this program is shown here: Original value of a: 64 i and b: 256 0 Since a is promoted to int for the purposes of evaluation, left-shifting the value 64 (0100 0000) twice results in i containing the value 256 (1 0000 0000) However, the value in b contains 0 because after the shift, the low-order byte is now zero Its only 1 bit has been
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shifted out Since each left shift has the effect of doubling the original value, programmers frequently use this fact as an efficient alternative to multiplying by 2 But you need to watch out If you shift a 1 bit into the high-order position (bit 31 or 63), the value will become negative The following program illustrates this point: // Left shifting as a quick way to multiply by 2 class MultByTwo { public static void main(String args[]) { int i; int num = 0xFFFFFFE; for(i=0; i<4; i++) { num = num << 1; Systemoutprintln(num); }
The program generates the following output: 536870908 1073741816 2147483632 -32 The starting value was carefully chosen so that after being shifted left 4 bit positions, it would produce -32 As you can see, when a 1 bit is shifted into bit 31, the number is interpreted as negative
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