Dark Horse programmer-java Basic-java Basic Syntax 1

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02.01 Overview and use of keywords

Keywords Overview: Words that are given a specific meaning by the Java language

Keywords feature: The letters that make up the keywords are all lowercase

Keyword considerations: Goto and Const exist as reserved words, not currently used, advanced Notepad like notepad++, with special color tags for keywords, very intuitive

Reserved words: May be promoted to keywords in new versions of JDK

02.02 Overview and composition rules for identifiers

Identifier Overview: A sequence of characters to use when naming classes, interfaces, methods, variables, etc.

Composition rules: English uppercase and lowercase letters, numeric characters, $ and _

Precautions:

1. Cannot start with a number

2. Cannot be a keyword in Java

3. Strictly case-sensitive

02.03 common naming conventions in identifiers

Common naming rules requirements: see Names and meanings

Package: All lowercase, multi-level package when used. Separate

Class or interface: The first letter of a word is capitalized, and the first letter of each word is capitalized when multiple words

Methods and variables: The first letter of a word is lowercase, and multiple words start with a second word with each first letter capitalized

Constants: All uppercase, multiple words separated by _

02.04 Overview of annotations and their classifications

Comment Overview: Text used to interpret the description program

Comment Classification format in Java

Single-line Comment: Format://Comment text

Multiline Comment: Format:/* Comment Text */

Document comments: Format:/** Note text */

1. For single-line and multiline comments, the annotated text is not interpreted by the JVM (Java Virtual machine).

2. For a document comment, a Java-specific comment, where the comment content can be resolved by the tool Javadoc provided by the JDK, generating a set of documentation for the program as a Web page file.

02.05 HelloWorld Case Add Comment version

Cases:

1/* 2 Requirements: Write a program in console output HelloWorld 3 analysis: 4     1. Write a Java program that defines the class first. 5     2. The main method must be defined for the program to be able to be called by the JVM. 6     3. If the program wants to have output results, it must use the output statement. 7  8 Implementation: 9     1. The class keyword is used to define classes, followed by the class name     2.main method basic format: public static void Main (string[] args)     3. Output Statement basic format: System.out.println ("HelloWorld"), */14//helloworld case, class Demo16 {    //main method, public    static void Main (string[] args) +      //Output statement      System.out.println ("HelloWorld");    }23}

02.06 function of the annotations

1. Explain procedures to improve the reading of the program

2. Can help us to debug the program

02.07 Overview and use of constants

Constants Overview: Their values cannot be changed during program execution

Constants in Java categories: Literal constants and custom constants

Literal constants:

1. String constants: Examples of contents enclosed in double quotation marks: "HelloWorld"

2. Integer constants: All integers example: 12,23

3. Decimal constants: All decimal Examples: 12.34,56.78

4. Character constants: Examples of content enclosed in single quotes: ' A ', ' a ', ' 0 '

5. Boolean constants: Only True and false two values

6. NULL constant: null

02.08 binary Overview and binary, octal, hex

Java provides 4 representations of integer constants: binary, octal, decimal, hexadecimal

Introduction: The system is a carry system, is a kind of carry method.

For any kind of binary--x, it means that the number operation at a certain position is every x in a bit.

Binary is every two in one, eight into the system is every eight into a, decimal is every ten into one, Hex is 16 into one.

Conversion:

1byte (bytes) = 8 bit (bits)

1 k = 1024byte

Law, the larger the system, the shorter the form of expression.

02.09 representations of the different binary data

Binary: 0, 1, full 2 in 1, beginning with 0b

Octal: 0~7, full 8 in 1, with 0 beginning to indicate

Decimal: 0~9, full 10 in 1, integer by default is decimal

Hex: 0~9, A~f, full 16 in 1, starting with 0x

Any data exists in the computer as a binary form

02.10 conversion of any binary to decimal

Binary: 110 to decimal: 1*22 + 1*21 + 1*20 = 6

Octal: 110 to decimal: 1*82 + 1*81 + 1*80 = 72

Hex: 110 to decimal: 1*162 + 1*161 + 1*160 = 272

02.11 any decimal-based exercise

0b10101 to decimal: 1*24 + 0*23 + 1*22 + 0*21 + 1*20 = 21

0123 decimal: 1*82 + 2*81 + 3*80 = 83

0x3C to decimal: 3*161 + 12*160 = 60

02.12 decimal to any binary conversion

Decimal: 56 is converted into binary:

56/2 = 28...0

28/2 = 14...0

14/2 = 7 ... 0

7/2 = 3 ... 1

3/2 = 1 ... 1

1/2 = 0 ... 1

Combine the remainder from bottom to top, i.e. 0b 111000

Decimal: 56 into octal:

56/8 = 7...0

7/8 = 0...7

Combine the remainder from bottom to top, i.e. 0 70

Decimal: 56 to Hex:

56/16 = 3...8

3/16 = 0...3

Combine the remainder from bottom to top, i.e. 0x 38

02.13 decimal to arbitrary binary exercises

Convert decimal 52 to binary, octal, hexadecimal, respectively

Decimal: 52 into binary: 0b 110100

52/2 = 26...0

26/2 = 13...0

13/2 = 6 ... 1

6/2 = 3 ... 0

3/2 = 1 ... 1

1/2 = 0 ... 1

Decimal: 52 to octal: 0 64

52/8 = 6...4

6/8 = 0...6

Decimal: 52 to Hex: 0x 34

52/16 = 3...4

3/16 = 0...3

02.14 Fast Conversion method of the binary

Fast conversion of decimal and binary binary

8421 yards:

1 1 1 1 1 1 1 1

128 64 32 16 8 4 2 1

Cases:

100 turn into binary: 0b 01100100

101101 turn into decimal: 32+8+4+1 = 45

Binary and octal, 16 binary conversion

1. Use decimal as a bridge

2. Binary to octal: binary to octet from low to high for each three-bit group, insufficient high 0

Example: 100100 = (100) (100) =0 44

3. Binary to 16 binary: binary to hexadecimal from low to high every four bits of a group, insufficient high 0

Example: 100100= (0010) (0100) =0x 24

02.15 Original code anti-code complement

Symbolic data representation: In a computer, there are 3 notation numbers: The original code, the inverse code, and the complement.

All data is performed in the complement.

The original code: is the binary fixed-point notation, that is, the highest bit is the sign bit, "0" is positive, "1" is negative, the remaining bits represent the size of the value.

Anti-code: The inverse code of positive number is the same as its original code, and the inverse code of negative number is the inverse of its original code, except the sign bit.

Complement: The complement of a positive number is the same as its original code; the complement of a negative number is the minus 1 of its inverse code.

Cases:

7 The original code: 0 0000111

7 Anti-code: 0 0000111

Complement of 7:0 0000111

-7 of the original code: 1 0000111

-7 of the inverse code: 1 1111000 (on the basis of the original code in addition to the symbol bit by bit reversed)

-7 Complement: 1 1111001 (reverse code and 1)

02.16 Original code Anti-code complement practice

It is known that a number of the original code is 10110100, to seek its complement and anti-code

Original code: 1 0110100

Anti-code: 1 1001011

Complement: 1 1001100

The complement of a certain number is known to be 11101110, to seek its original code

Complement: 1 1101110

Anti-code: 1 1101101 (complement minus 1)

Original code: 1 0010010 (bitwise reversed)

02.17 Overview and format of variables

Variable Overview: The amount of value that can change within a range during program execution

Variable definition Format: Data type variable name = initialization value;

02.18 Overview and classification of data types

The Java language is a strongly typed language that defines specific data types for each data and allocates different sizes of memory space in memory

02.19 defining variables for different data types

Cases:

1//define a byte variable 2 byte b = 10; 3  4//define a short integer variable 5 shorter s = 7; 6      8//define an integer variable 9 int i = n; 10     100000000//define a long integer variable with a suffix plus l or l mark one-time 0000L;12     13//define a floating-point variable with a single-precision floating-point number suffix plus f or f mark float F = 12.34f;15  

02.20 Considerations for using variables

Considerations for Using Variables:

Scope: The variable is defined in the curly braces, which range is the scope of the variable. Two variables of the same name cannot be defined in the same scope.

Initialization value: No initialization value can be used directly

It is recommended that you define only one variable on a line, but you can define multiple, but do not recommend

02.21 default conversion of data type conversions

The Boolean type cannot be converted to another data type

Default conversion: Byte,short,char→int→long→float→double

Byte,short,char do not convert to each other, they participate in the operation first converted to int type

02.22 different data type variables participate in operation interpretation

Cases:

1 byte a = 3;2 int b = 4;3 byte c = a + b;//error 4 int d = a + B; That's right

Explanation: The data-participation operation of byte type is promoted by default to int, the result of a + B is int, and cannot be assigned to a byte type C

02.23 Casting of data type conversions

Forced conversions

Target type variable name = (target type) (data to be converted);

Cases:

1 byte a = 3;2 int b = 4;3 byte c = (byte) (A + B);

The preceding statement discards the first three bytes of the result int of a + B, leaving the last byte assigned to C

02.24 study questions of the cast

1. The following assignment has no problem

Double d = 12.345;

float F = d;

A: There is a problem, you cannot assign a double type D to the float type F.

Required cast: float F = (float) D;

2. Are there any differences between the definitions below?

Float F1 = (float) 12.345;

float F2 = 12.345F;

A: There are differences, F1 is strong by double type to float type. The F2 itself is a type of float.

02.25 the distinguishing questions of the addition of variables and the addition of constants

BYTE B1 = 3,b2 = 4,b;

B = B1 + b2;//compilation failed, variable, type promoted first, b1+b2 result int type

b = 3 + 4; Correct, constant, calculate the result first, see if it is within the byte range, do not error

02.26 result calculation after casting data overflow

byte B = 130; Is there a problem? What do you do if you want to assign the correct value? How much is the result?

A: There is a problem, 130 is out of byte range, you can use cast byte B = (byte) 130;

Results:

int type 130 Original code: 00000000 00000000 00000000 10000010

int type 130 complement: 00000000 00000000 00000000 10000010

Cast drop high three-bit result is: 10000010

Complement: 1 0000010

Anti-code: 1 0000001

Original code: 1 1111110→ result is decimal-126

02.27 character data participation operations

System.out.println (' a '); A

System.out.println (' a ' + 1); 98, the character a participates in the operation using the corresponding 97 in the ASCII code table

02.28 String Participation Operations

System.out.println (' A ' +1+ "Hello"); 98hello

System.out.println ("Hello" + ' a ' + 1); Helloa1

System.out.println ("5+5=" +5+5); 5+5=55

System.out.println (5+5+ "=5+5"); 10=5+5

Dark Horse programmer-java Basic-java Basic Syntax 1