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1372 Creating HTML Documentation with javadoc (on CD) Appendix F 16 private int minute; // 0 - 59 17 private int second; // 0 - 59 18 19 20 /** 21 * Time3 constructor initializes each instance variable 22 * to zero. Ensures that Time object starts in a 23 * consistent state. 24 * @throws Exception in the case of an invalid time 25 */ 26 public Time3() throws Exception 27 { 28 setTime( 0, 0, 0 ); 29 } 30 31 32 /** 33 * Time3 constructor: hour supplied, minute and second 34 * defaulted to 0 35 * @param h the hour 36 * @throws Exception in the case of an invalid time 37 */ 38 public Time3( int h ) throws Exception 39 { 40 setTime( h, 0, 0 ); 41 } 42 43 /** 44 * Time3 constructor: hour and minute supplied, second 45 * defaulted to 0 46 * @param h the hour 47 * @param m the minute 48 * @throws Exception in the case of an invalid time 49 */ 50 public Time3( int h, int m ) throws Exception 51 { 52 setTime( h, m, 0 ); 53 } 54 55 /** 56 * Time3 constructor: hour, minute and second supplied 57 * @param h the hour 58 * @param m the minute 59 * @param s the second 60 * @throws Exception in the case of an invalid time 61 */ 62 public Time3( int h, int m, int s ) throws Exception 63 { 64 setTime( h, m, s ); 65 } 66 Fig. F.2Fig. F. A Java source code file containing documentation comments (part 2 of 5).
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Appendix F Creating HTML Documentation with javadoc (on CD) 1371 links to packages are the selected Web page is displayed in this frame displayed in this frame links to classes and interfaces are displayed in this frame. Interfaces are italicized Fig. F.1Java API documentation. Fig. 1 // Fig. F.2: Time3.java 2 // Time3 class definition with set and get methods 3 package com.deitel.jhtp4.appenF; 4 5 // Java core packages 6 import java.text.DecimalFormat; 7 8 /** 9 * This class maintains the time in 24-hour format. 10 * @see java.lang.Object 11 * @author Deitel & Associates, Inc. 12 */ 13 public class Time3 extends Object { 14 15 private int hour; // 0 - 23 Fig. F.2A Java source code file containing documentation comments (part 1 of 5). Fig. F.
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1370 Creating HTML Documentation with javadoc (on CD) Appendix F Outline F.1 Introduction F.2 Documentation Comments F.3 Documenting Java Source Code F.4 javadoc F.5 Files Produced by javadoc Summary Terminology F.1 Introduction In this appendix, we provide an introduction to the Java 2 Software Developent Kit s javadoc utility for creating HTML files that document Java code. This is the tool used by Sun Microsystems to create the Java API documentation (Fig. F.1). We discuss the special Java comments and tags required by javadocto create documentation based on your source code and how to execute the javadoc tool. For detailed information on javadoc, visit the javadoc home page at java.sun.com/j2se/1.3/docs/tooldocs/javadoc/index.html F.2 Documentation Comments Before HTML files can be generated with the javadoc tool, programmers must insert special comments called documentation comments into their source files. Documentation comments are the only comments recognized by javadoc. Documentation comments begin with /** and end with */. An example of a simple documentation comment is /** Sorts integer array using MySort algorithm */ Like other comments, documentation comments are not translated into bytecodes. Because javadoc is used to create HTML files, documentation comments can contain HTML tags. For example, the documentation comment /** Sorts integer array using MySort algorithm */ which contains the HTML emphasis tags and is valid. In the generated HTML files, MySort will appear as emphasized text (normally italic). As we will see, javadoctags can be inserted into the documentation comments to help javadoc document your source code. These tags which begin with an @ symbol are not HTML tags. F.3 Documenting Java Source Code Figure F.2 presents a modified version of the Time3class from Fig. 8.8 that contains documentation comments. In the text that follows the example, we thoroughly discuss each of the javadoc tags used in the documentation comments. We discuss how to use the javadoc tool to generate HTML documentation from this file in Section F.4.
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F Creating HTML Documentation with javadoc (on CD) Objectives To introduce the javadoc J2SDK tool. To introduce documentation comments. To understand javadoc tags. To be able to generate HTML API documentation with javadoc. To understand javadoc generated documentation files. Oh I get by with a little help from my friends. John Lennon and Paul McCartney I feel The link of nature draw me. John Milton I think I shall never see A poem lovely as a tree. Joyce Kilmer There is only one religion, though there are a hundred versions of it. George Bernard Shaw What I like in a good author is not what he says, but what he whispers. Logan Pearsall Smith I shall return. Douglas MacArthur
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1368 Number Systems (on CD) Appendix E E.18 Binary: 512 256 128 64 32 16 8 4 2 1 256 128 64 32 16 8 4 2 1 (1*256)+(1*128)+(0*64)+(1*32)+(0*16)+(0*8)+(0*4)+(0*2)+ (1*1) 110100001 One s complement: 001011110 Two s complement: 001011111 Check: Original binary number + its two s complement 110100001 001011111 000000000 E.19 Zero. EXERCISES E.20 Some people argue that many of our calculations would be easier in the base 12 number system because 12 is divisible by so many more numbers than 10 (for base 10). What is the lowest digit in base 12? What might the highest symbol for the digit in base 12 be? What are the positional values of the rightmost four positions of any number in the base 12 number system? E.21 How is the highest symbol value in the number systems we discussed related to the positional value of the first digit to the left of the rightmost digit of any number in these number systems? E.22 Complete the following chart of positional values for the rightmost four positions in each of the indicated number systems: decimal1000100 10 1 base 6 …… 6… base 13 … 169… … base 3 27……… E.23 Convert binary 100101111010 to octal and to hexadecimal. E.24 Convert hexadecimal 3A7D to binary. E.25 Convert hexadecimal 765F to octal. (Hint: First convert 765F to binary, then convert that binary number to octal.) E.26 Convert binary 1011110 to decimal. E.27 Convert octal 426 to decimal. E.28 Convert hexadecimal FFFF to decimal. E.29 Convert decimal 299 to binary, to octal, and to hexadecimal. E.30 Show the binary representation of decimal 779. Then show the one s complement of 779, and the two s complement of 779. E.31 What is the result when the two s complement of a number is added to itself? E.32 Show the two s complement of integer value -1 on a machine with 32-bit integers.
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Appendix E Number Systems (on CD) 1367 E.19 What is the result when the one s complement of a number is added to itself? SELF-REVIEW ANSWERS E.1 10, 2, 8, 16. E.2 Fewer. E.3 False. E.4 Hexadecimal. E.5 False. The highest digit in any base is one less than the base. E.6 False. The lowest digit in any base is zero. E.7 1 (the base raised to the zero power). E.8 The base of the number system. E.9 Fill in the missing values in this chart of positional values for the rightmost four positions in each of the indicated number systems: decimal 1000 100 10 1 hexadecimal 4096 256 16 1 binary 8 4 2 1 octal 512 64 8 1 E.10 Octal 6530; Hexadecimal D58. E.11 Binary 1111 1010 1100 1110. E.12 Binary 111 011 001 110. E.13 Binary 0 100 111 111 101 100; Octal 47754. E.14 Decimal 2+4+8+32+64=110. E.15 Decimal 7+1*8+3*64=7+8+192=207. E.16 Decimal 4+13*16+15*256+14*4096=61396. E.17 Decimal 177 to binary: 256 128 64 32 16 8 4 2 1 128 64 32 16 8 4 2 1 (1*128)+(0*64)+(1*32)+(1*16)+(0*8)+(0*4)+(0*2)+(1*1) 10110001 to octal: 512 64 8 1 64 8 1 (2*64)+(6*8)+(1*1) 261 to hexadecimal: 256 16 1 16 1 (11*16)+(1*1) (B*16)+(1*1) B1
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1366 Number Systems (on CD) Appendix E TERMINOLOGY base digit base 2 number system hexadecimal number system base 8 number system negative value base 10 number system octal number system base 16 number system one s complement notation binary number system positional notation bitwise complement operator (~) positional value conversions symbol value decimal number system two s complement notation SELF-REVIEW EXERCISES E.1 The bases of the decimal, binary, octal, and hexadecimal number systems are , , and respectively. E.2 In general, the decimal, octal, and hexadecimal representations of a given binary number contain (more/fewer) digits than the binary number contains. E.3 (True/False) A popular reason for using the decimal number system is that it forms a convenient notation for abbreviating binary numbers simply by substituting one decimal digit per group of four binary bits. E.4 The (octal / hexadecimal / decimal) representation of a large binary value is the most concise (of the given alternatives). E.5 (True/False) The highest digit in any base is one more than the base. E.6 (True/False) The lowest digit in any base is one less than the base. E.7 The positional value of the rightmost digit of any number in either binary, octal, decimal, or hexadecimal is always . E.8 The positional value of the digit to the left of the rightmost digit of any number in binary, octal, decimal, or hexadecimal is always equal to . E.9 Fill in the missing values in this chart of positional values for the rightmost four positions in each of the indicated number systems: decimal 1000 100 10 1 hexadecimal … 256 … … binary … … … … octal 512 … 8 … E.10 Convert binary 110101011000 to octal and to hexadecimal. E.11 Convert hexadecimal FACE to binary. E.12 Convert octal 7316 to binary. E.13 Convert hexadecimal 4FEC to octal. (Hint: First convert 4FEC to binary then convert that binary number to octal.) E.14 Convert binary 1101110 to decimal. E.15 Convert octal 317 to decimal. E.16 Convert hexadecimal EFD4 to decimal. E.17 Convert decimal 177 to binary, to octal, and to hexadecimal. E.18 Show the binary representation of decimal 417. Then show the one s complement of 417, and the two s complement of 417.
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Appendix E Number Systems (on CD) 1365 by adding the two s complement of valueto a as follows: x = a + (~value + 1); Suppose a is 27 and value is 13 as before. If the two s complement of value is actually the negative of value, then adding the two s complement of value to a should produce the result 14. Let us try this: a (i.e., 27) 00000000 00000000 00000000 00011011 +(~value + 1) +11111111 11111111 11111111 11110011 00000000 00000000 00000000 00001110 which is indeed equal to 14. SUMMARY When we write an integer such as 19 or 227 or -63 in a Java program, the number is automatically assumed to be in the decimal (base 10) number system. The digits in the decimal number system are 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. The lowest digit is 0 and the highest digit is 9 one less than the base of 10. Internally, computers use the binary (base 2) number system. The binary number system has only two digits, namely 0 and 1. Its lowest digit is 0 and its highest digit is 1 one less than the base of 2. The octal number system (base and the hexadecimal number system (base 16) are popular primarily because they make it convenient to abbreviate binary numbers. The digits of the octal number system range from 0 to 7. The hexadecimal number system poses a problem because it requires sixteen digits a lowest digit of 0 and a highest digit with a value equivalent to decimal 15 (one less than the base of 16). By convention, we use the letters A through F to represent the hexadecimal digits corresponding to decimal values 10 through 15. Each number system uses positional notation each position in which a digit is written has a different positional value. A particularly important relationship that both the octal number system and the hexadecimal number system have to the binary system is that the bases of octal and hexadecimal (8 and 16 respectively) are powers of the base of the binary number system (base 2). To convert an octal number to a binary number, simply replace each octal digit with its three-digit binary equivalent. To convert a hexadecimal number to a binary number, simply replace each hexadecimal digit with its four-digit binary equivalent. Because we are accustomed to working in decimal, it is convenient to convert a binary, octal or hexadecimal number to decimal to get a sense of the number s real worth. To convert a number to decimal from another base, multiply the decimal equivalent of each digit by its positional value, and sum these products. Computers represent negative numbers using two s complement notation. To form the negative of a value in binary, first form its one s complement by applying Java s bit- wise complement operator (~). This reverses the bits of the value. To form the two s complement of a value, simply add one to the value s one s complement.
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1364 Number Systems (on CD) Appendix E Positional values: 25616 1 Symbol values: 177 and thus decimal 375 is equivalent to hexadecimal 177. E.6 Negative Binary Numbers: Two s Complement Notation The discussion in this appendix has been focussed on positive numbers. In this section, we explain how computers represent negative numbers using two s complement notation. First we explain how the two s complement of a binary number is formed, and then we show why it represents the negative value of the given binary number. Consider a machine with 32-bit integers. Suppose int value = 13; The 32-bit representation of valueis 00000000 00000000 00000000 00001101 To form the negative of value we first form its one s complement by applying Java s bit- wise complement operator (~): onesComplementOfValue = ~value; Internally, ~value is now value with each of its bits reversed ones become zeros and zeros become ones as follows: value: 00000000 00000000 00000000 00001101 ~value (i.e., value s ones complement): 11111111 11111111 11111111 11110010 To form the two s complement of value we simply add one to value s one s complement. Thus Two s complement of value: 11111111 11111111 11111111 11110011 Now if this is in fact equal to -13, we should be able to add it to binary 13 and obtain a result of 0. Let us try this: 00000000 00000000 00000000 00001101 +11111111 11111111 11111111 11110011 00000000 00000000 00000000 00000000 The carry bit coming out of the leftmost column is discarded and we indeed get zero as a result. If we add the one s complement of a number to the number, the result would be all 1s. The key to getting a result of all zeros is that the twos complement is 1 more than the one s complement. The addition of 1 causes each column to add to 0 with a carry of 1. The carry keeps moving leftward until it is discarded from the leftmost bit, and hence the resulting number is all zeros. Computers actually perform a subtraction such as x = a - value;
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Appendix E Number Systems (on CD) 1363 Positional values: 64 32 16 8 4 2 1 Then we discard the column with positional value 64 leaving: Positional values: 32 16 8 4 2 1 Next we work from the leftmost column to the right. We divide 32 into 57 and observe that there is one 32 in 57 with a remainder of 25, so we write 1 in the 32 column. We divide 16 into 25 and observe that there is one 16 in 25 with a remainder of 9 and write 1 in the 16 column. We divide 8 into 9 and observe that there is one 8 in 9 with a remainder of 1. The next two columns each produce quotients of zero when their positional values are divided into 1 so we write 0s in the 4 and 2 columns. Finally, 1 into 1 is 1 so we write 1 in the 1 column. This yields: Positional values: 32 16 8 4 Symbol values: 1 1 1 0 2 0 1 1 and thus decimal 57 is equivalent to binary 111001. To convert decimal 103 to octal, we begin by writing the positional values of the columns until we reach a column whose positional value is greater than the decimal number. We do not need that column, so we discard it. Thus, we first write: Positional values: 51264 8 1 Then we discard the column with positional value 512, yielding: Positional values: 648 1 Next we work from the leftmost column to the right. We divide 64 into 103 and observe that there is one 64 in 103 with a remainder of 39, so we write 1 in the 64 column. We divide 8 into 39 and observe that there are four 8s in 39 with a remainder of 7 and write 4 in the 8 column. Finally, we divide 1 into 7 and observe that there are seven 1s in 7 with no remainder so we write 7 in the 1 column. This yields: Positional values: 648 1 Symbol values: 147 and thus decimal 103 is equivalent to octal 147. To convert decimal 375 to hexadecimal, we begin by writing the positional values of the columns until we reach a column whose positional value is greater than the decimal number. We do not need that column, so we discard it. Thus, we first write Positional values: 4096 256 16 1 Then we discard the column with positional value 4096, yielding: Positional values: 25616 1 Next we work from the leftmost column to the right. We divide 256 into 375 and observe that there is one 256 in 375 with a remainder of 119, so we write 1 in the 256 column. We divide 16 into 119 and observe that there are seven 16s in 119 with a remainder of 7 and write 7 in the 16 column. Finally, we divide 1 into 7 and observe that there are seven 1s in 7 with no remainder so we write 7 in the 1 column. This yields:
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