Detailed explanation of Chapter 3 of Java programming thought summary notes

There is not much to summarize in this chapter, but you need to pay attention to the details, some of which are easy to forget.

Chapter 3

Table of Contents:

• 3.1 Simpler print statements

• 3.2 Using Java operators

• 3.3 Priority

• 3.4 Assignment

• 3.5 Arithmetic operators

• 3.6 Automatic increment and decrement

• 3.7 Relational operators

• 3.8 Logical operators

• 3.9 Direct constants

• 3.10 Bitwise operators

• ##3.11 Shift operators

• 3.12 Ternary operator if-else

• ##3.13 String operators + and +=

• 3.14 Common mistakes when using operators

##3.15 Type conversion operator
3.16 Java does not have sizeof
3.17 Operator summary
• 3.18 Summary

  
  
  

##3.2 Using Java operators

The operators "=", "==" and "==" can operate on all objects.

3.4 Assignment

If the object uses c=d, then both c and d point to the object that only d originally pointed to.

#When t1 = t2 (alias phenomenon), then modifying t1 will also change t2, because t1 and t2 contain the same reference and point to the same object.

If you want to keep the two objects independent of each other, you can do this t1.level = t2.level; .

3.6 Automatic increment and decrement

a = 15，++a、--a： System.out.printf("~output:" + a++);   //~ output:15a = 15，a++、a--： System.out.printf("~output:" + ++a);   // ~ output:16

3.7 Relational Operator

The contents of the above two Integer objects are the same, but the references are different , and == and != compare object references (basic types directly compare values, without references). If you want to compare object contents, use the equals() method.

Note: The default behavior of the equals() method is to compare references, so you must override the e quals() method in your new class, otherwise it will not appear the effect you want. Most Java class libraries override the equals() method so that the contents of objects can be compared. Examples are as follows:

##3.9 Direct constants

Hexadecimal: Prefix 0x, Octal: Prefix 0, Binary has no direct representation of constants.

Exponent counting method: 1.39e-43^{f means 1.39 * 10}-43 ^{e means "the power of 10".}

3.10 Bitwise operators

Bitwise operations The operand of the operator is a binary "bit" (bit). Java was originally designed to be embedded in a TV set-top box, so this bottom-level operation is still retained. But the operator is rarely used.

##&And |or Avoid confusion with logical NOT. Bitwise operators are similar to logical operators, but without the short-circuiting effect.

3.11 Shift operator

The operand is a binary "bit" and can only be used to process integer types (Boolean types do not work).

Negative numbers need to be converted to two’s complement first and then operated on. I won’t introduce them here.

Shift left (n high bits of

0

are discarded, and n bits of the lowest bit are filled with n bits of 0)<<： The binary form of 11 is 1011. The binary form after 11<<2 is 101100. So 11<<2

= 44 is equivalent to the integer 11*2

n ^{##Shift right (n numbers in the low bits are moved out, and n zeros are added in the high bits)>>：}

The binary form of ##11 is 1011 11>>2 The binary form after 2 is 0010 So 11>>2 = 2 is equivalent to the integer 11/2

n

## Unsigned right shift operator>>>: Regardless of positive or negative, n numbers are shifted to the low bits and n zeros are added to the high bitsNote:

Perform unsigned right shift on byte or short value (

>>>), the result may not be correct result. They are first converted to int type, then right-shifted, and then truncated and assigned to the original type. In this case, a result of -1 may be obtained. The example is as follows:

The assembly-level execution speed of bit operations is very fast, so during the interview you may ask: What is the most commonly used method in Java? What is the efficient way to calculate the value of 2 times 8? Answer: 2 ##, The code is not intuitive.

#3.16 Java does not have sizeof

Java does not require the sizeof() operator, because all data types have the same size in all machines, and there is no need to consider "transplantation" issues.

Summary: Knowledge points that are too easy and often used do not appear in the notes. All appear. Might as well read the book again.

I saw such a complaint on a certain software today

When I saw the knowledge points in the second chapter of Java programming ideas, I immediately thought of them:

When a variable is used as a member of a class, Java ensures that its default value is given to ensure that the basic type member variable is initialized (the initial value may not be your If you want, it's better to initialize it yourself). Note that the default initialization method does not apply to field variables that are not of a certain class. If you forget to initialize, Java will return an error to you during compilation.

Stack: Located in general-purpose RAM (random access register), the Java compiler must know the size of all data stored in the stack ## and life cycle, the "stack pointer" moves downward to allocate new memory, and moves upward to release memory. The speed is second only to the register, where basic data types and references are stored.

Local variables are allocated on the stack during runtime. They are large in amount and have a short life cycle. If the virtual machine initializes each local variable, it will be a big deal. A lot of overhead, but it's unsafe to use variables without initializing them to their default values.

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