1 using == to compare primitive wrappers objects such as Integer

Integer int1_1 = Integer.valueOf(“1”);Integer int1_2 = Integer.valueOf(1);System.out.println(“int1_1 == int1_2: ” + (int1_1 == int1_2)); // trueSystem.out.println(“int1_1 equals int1_2: ” + int1_1.equals(int1_2)); // trueInteger int2_3 = Integer.valueOf(“1000”);Integer int2_2 = Integer.valueOf(1000);System.out.println(“int2_3 == int2_4: ” + (int2_3 == int2_4)); // falseSystem.out.println(“int2_3 equals int2_4: ” + int2_3.equals(int2_4)); // true

The JVM maintains a cache of Integer objects for the range -128 to 127. For values in this range, the Integer. valueOf() will return the cached value rather than creating a new one. Thus, in the first example the Integer.valueOf(1) and Integer.valueOf(“1”) calls returned the same cached Integer instance. By contrast, in the second example the Integer. valueOf(1000) and Integer. valueOf(“1000”) both created and returned new Integer objects.

JVM維護(hù)范圍為128到127的整數(shù)對象緩存。對于此范圍內(nèi)的值，為整數(shù)。valueOf()將返回緩存的值，而不是創(chuàng)建新值。因此，在第一個(gè)示例中，Integer.valueOf(1) 和 Integer.valueOf(“1”)調(diào)用返回了相同的緩存整數(shù)實(shí)例。相比之下，在第二個(gè)示例中，valueOf(1000) 和 Integer. valueOf(“1000”)創(chuàng)建并返回了新的整數(shù)對象。

The == operator for reference types tests for reference equality (i. e. the same object). Therefore, in the first example int1_1 == int1_2 is true because the references are the same. In the second example int2_1 == int2_2 is false because the references are different.

引用類型的==運(yùn)算符測試引用相等性（即相同的對象）。因此，在第一個(gè)示例中，int1_1 == int1_2是真的，因?yàn)橐檬窍嗤?。在第二個(gè)示例中，int2_1 == int2_2為false，因?yàn)橐貌煌?/p>

2 using == to compare strings

public class Hello { public static void main(String[] args) { if(args.length > 0) if(args[0] == “hello”) { // always false, one point to heap , another point to iteral pool System.out.println(“Hello back to you”); else System.out.println(“Are you feeling grumpy today?”); }}

The correct way to test strings is to use the equals(Object) method.

public class Hello2 { public static void main(String[] args) { if(args.length > 0) { if(args[0].equals(“hello”)) System.out.println(“Hello back to you”); else System.out.println(“Are you feeling grumpy today?”); } }}

有字符串常量出現(xiàn)時(shí)，會(huì)先查找常量池，已定義的話不會(huì)再定義，引用相同空間的常量值即可：

public class Test1 { public static void main(String[] args) { String s1 = “hello”; // point to iteral pool String s2 = “hello”; // point to the same above if(s1 == s2) System.out.println(“same”); // same else System.out.println(“different”); }}

3 forgetting to free resources

private static void printFileJava6() throws IOException { FileInputStream input; try { input = new FileInputStream(“file.txt”); int data = input.read(); while(data != -1){ System.out.print((char) data); data = input.read(); } } finally { if(input != null) { input.close(); // don’t forget } }}

4 thinking of variables as objects

No Java variable represents an object.

String foo; // NOT AN OBJECT

Neither does any Java array contain objects.

String bar[] = new String[100]; // No member is an object.

The equality operator does NOT test that two objects are equal.

相等運(yùn)算符不會(huì)測試兩個(gè)對象是否相等。

Applying the equality (==) operator to reference values tests if the values refer to the same object. It does not test whether two (different) objects are “equal” in the intuitive sense.

將相等（==）運(yùn)算符應(yīng)用于引用值將測試這些值是否引用同一對象。它不測試兩個(gè)（不同）物體在直覺上是否“相等”。

Method calls do NOT pass objects at all.

方法調(diào)用根本不傳遞對象。

Java method calls use pass by value1 to pass arguments and return a result.

Java方法調(diào)用使用pass-by-value1傳遞參數(shù)并返回結(jié)果。

When you pass a reference value to a method, you’re actually passing a reference to an object by value, which means that it is creating a copy of the object reference.

向方法傳遞引用值時(shí)，實(shí)際上是按值傳遞對對象的引用，這意味著它正在創(chuàng)建對象引用的副本。

As long as both object references are still pointing to the same object, you can modify that object from either reference, and this is what causes confusion for some.

只要兩個(gè)對象引用仍然指向同一個(gè)對象，就可以從任意一個(gè)引用修改該對象，這就是導(dǎo)致某些人混淆的原因。

However, you are not passing an object by reference2. The distinction is that if the object reference copy is modified to point to another object, the original object reference will still point to the original object.

但是，您沒有通過引用2傳遞對象。區(qū)別在于，如果修改對象引用副本以指向另一個(gè)對象，則原始對象引用仍將指向原始對象。

void f(Point foo) { foo.x = 42; foo = new Point(3, 4); // Point local foo at a different object.}

5 Not understanding that String is an immutable class

New Java programmers often forget, or fail to fully comprehend, that the Java String class is immutable. This leads to problems like the one in the following example:

新的Java程序員經(jīng)常忘記或無法完全理解Java字符串類是不可變的。這會(huì)導(dǎo)致以下示例中的問題：

public class Shout { public static void main(String[] args) { for(String s : args) { s.toUpperCase(); System.out.print(s); System.out.print(” “); } System.out.println(); }}

The above code is supposed to print command line arguments in upper case. Unfortunately, it does not work, the case of the arguments is not changed.

上述代碼應(yīng)該以大寫形式打印命令行參數(shù)。不幸的是，它不起作用，參數(shù)的情況沒有改變。

In reality, the toUpperCase() method returns a String object which is an uppercase version of the String that you call it on. This will probably be a new String object, but if s was already all uppercase, the result could be the existing string.

實(shí)際上，toUpperCase()方法返回一個(gè)字符串對象，該對象是調(diào)用它的字符串的大寫版本。這可能是一個(gè)新的字符串對象，但如果s已經(jīng)全部大寫，則結(jié)果可能是現(xiàn)有字符串。

6 combining assignment and side-effects

i += a[i++] + b[i–]; // more sequence points and side-effects, not allowed

7 Using ‘==’ to test a boolean

Sometimes a new Java programmer will write code like this:

public void check(boolean ok) { if(ok == true) { // redundancy， Note ‘ok == true’ System.out.println(“It is OK”); }}

8 Using ‘assert’ for argument or user input validation

A question that occasionally on StackOverflow is whether it is appropriate to use assert to validate arguments supplied to a method, or even inputs provided by the user.

StackOverflow上偶爾出現(xiàn)的一個(gè)問題是，使用assert驗(yàn)證提供給方法的參數(shù)，甚至驗(yàn)證用戶提供的輸入是否合適。

The simple answer is that it is not appropriate.

簡單的答案是，這是不合適的。

Better alternatives include:

更好的替代方案包括：

Throwing an IllegalArgumentException using custom code.

使用自定義代碼引發(fā)IllegalArgumentException。

Using the Preconditions methods available in Google Guava library.

使用Google Guava庫中可用的前提條件方法。

Using the Validate methods available in Apache Commons Lang3 library.

使用Apache Commons Lang3庫中可用的驗(yàn)證方法。

9 Wildcard imports can make your code fragile

Consider the following partial example:

import com.example.somelib.*; // not recommendimport com.acme.otherlib.*; //public class Test { private Context x = new Context(); // from com.example.somelib …}

10 other syntax pitfall

10.1 Missing a ‘break’ in a ‘switch’ case

10.2 Declaring classes with the same names as standard classes

10.3 Leaving out braces: the “dangling if” and “dangling else” problems

10.4 Octal literals ( 010 is not 10)

10.5 Ignoring method visibility

10.6 Misplaced semicolons and missing braces

10.7 Overloading instead of overriding

10.8 Auto-Unboxing Null Objects into Primitives

11 Threads and Concurrency related

11.1 Too many threads makes an application slower

11.2 incorrect use of wait() / notify()

11.3 Shared variables require proper synchronization

11.4 Thread creation is relatively expensive

12 Nulls and NullPointerException

12.1 Using null to represent an empty array or collection

12.2 Not checking if an I/O stream isn’t even initialized when closing it

12.3 Returning null instead of throwing an exception

12.4 Unnecessary use of Primitive Wrappers can lead to NullPointerExceptions

13 Exception usage

13.1 Catching Throwable, Exception, Error or RuntimeException

13.2 Ignoring or squashing exceptions

13.3 Throwing Throwable, Exception, Error or RuntimeException

13.4 Using exceptions for normal flowcontrol

13.5 Directly subclassing `Throwable

13.6 Catching InterruptedException

13.7 Excessive or inappropriate stacktraces

14 memory leaks related

Java manages memory automatically. You are not required to free memory manually. An object’s memory on the heap may be freed by a garbage collector when the object is no longer reachable by a live thread.

Java自動(dòng)管理內(nèi)存。您無需手動(dòng)釋放內(nèi)存。當(dāng)活動(dòng)線程無法訪問對象時(shí)，垃圾收集器可能會(huì)釋放堆上的對象內(nèi)存。

However, you can prevent memory from being freed, by allowing objects to be reachable that are no longer needed. Whether you call this a memory leak or memory packratting, the result is the same — an unnecessary increase in allocated memory.

但是，您可以通過允許不再需要的對象可以訪問來防止內(nèi)存被釋放。無論您將其稱為內(nèi)存泄漏還是內(nèi)存打包，結(jié)果都是一樣的——分配的內(nèi)存會(huì)不必要地增加。

Memory leaks in Java can happen in various ways, but the most common reason is everlasting object references, because the garbage collector can’t remove objects from the heap while there are still references to them.

Java中的內(nèi)存泄漏可能以各種方式發(fā)生，但最常見的原因是永久的對象引用，因?yàn)槔占鳠o法在仍然存在對對象的引用時(shí)從堆中移除對象。

15 Performance Issues

15.1 String concatenation in a loop does not scale

15.2 Using size() to test if a collection is empty is inefficient

15.3 Interning strings so that you can use == is a bad idea

15.4 Using ‘new’ to create primitive wrapper instances is inefficient

15.5 Efficiency concerns with regular expressions

15.6 Small reads / writes on unbuffered streams are inefficient

15.7 Over-use of primitive wrapper types is inefficient

15.8 The overheads of creating log messages

15.9 Iterating a Map’s keys can be inefficient

15.10 Calling System.gc() is inefficient

15.11 Calling ‘new String(String)’ is inefficient

ref

《Java Notes For Professionals》

-End-