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一丶运算的前提是数据类型必须一致(掌握)
(1) 观点一:(掌握)
// A: byte short char-int-long-float-double 默认大小转化时向最大的数据类型靠近, (除了 下面的观点的数据类型外)
// B: byte short char 在运算时默认转化成int 类型进行运算
// A: byte short char-int-long-float-double
// B: byte short char 在运算时默认转化成int 类型进行运算
14
14
1
@Test
2
public void test1()
3
{
4
byte b = 4;
5
int i = 3;
6
7
// int sum = b + i;
8
// byte sum = b + i;
9
short s = 2;
10
byte b1 = 1;
11
// (1) 验证成功, byte short char 在运算时默认转化成int 类型进行运算
12
// short sum = s + b1;
13
System.out.println("sum = " + sum);
14
}
注意: double和float精度不正确问题,二进制问题,使用大数据类型BigDecimal
(2) 面试题:(掌握)
/*
面试题:
byte b1=3,b2=4,b;
b=b1+b2;
b=3+4;
哪句是编译失败的呢?为什么呢?
b = b1 + b2;是有问题的。
因为变量相加,会首先看类型问题,最终把结果赋值的也会考虑类型问题。
常量相加,首先做加法,然后看结果是否在赋值的数据类型范围内,如果不是,才报错。
*/
11
11
1
class DataTypeDemo6 {
2
public static void main(String[] args) {
3
//定义了三个byte类型的变量,b1,b2,b3
4
//b1的值是3,b2的值是4,b没有值
5
byte b1 = 3,b2 = 4,b;
6
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//b = b1 + b2; //这个是类型提升,所有有问题
8
9
b = 3 + 4; //常量,先把结果计算出来,然后看是否在byte的范围内,如果在就不报错。
10
}
11
}
注意:(掌握)
常量之间的运算结果是在编译成字节码文件时便被计算完毕, 成为一个新的常量例如: int a = 3 + 7 编译后便成为 int a = 10, 即使是 int a = 0; a = 3 + 7;, 在编译时也会变成 int a = 3 + 7; 这就是编译器的优化效果, 但是常量是存在数据类型, 从 c++ 方面便可简单的看出来 const int a = 10; 这个便是常量, 在java中也是存在常量 的 例如 : private final String str = "aaaa"还存在常量对象, private final String str1 = new String("aaa");还有 从 float 和 double 类型的常量可以看出来常量存在数据类型. 0.01f这些都是常量
(3)案例思考题:(掌握)
17
17
1
// 思考题5
2
// 'a' = 97
3
@Test
4
public void test6()
5
{
6
System.out.println('a');
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// 这个将被虚拟机设置成 98
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System.out.println('a' + 1);
9
// 这个在编译期间就被编译成 "helloa1"
10
System.out.println("hello" + 'a' + 1);
11
// "98hello"
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System.out.println('a' + 1 + "hello");
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// 5 + 5 = 55
14
System.out.println("5 + 5 = " + 5 + 5);
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// 10= 5 + 5
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System.out.println(5 + 5 + "= 5 + 5");
17
}
9
9
1
// 思考题4
2
// byte b = (byte)120; -127 ~ 126;
3
@Test
4
public void test5()
5
{
6
// byte b = -126;
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byte b = (byte)130;
8
System.out.println(b);
9
}
17
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1
// 思考题3:
2
// byte b1 = 3, b2 = 4, b;
3
// b = b1 + b2;
4
// b = 3 + 4;
5
@Test
6
public void test4()
7
{
8
byte b1 = 3, b2 = 4, b;
9
// b1 b2 在运行时, 类型提升了, 所以有问题
10
// b = b1 + b2;
11
// 常量, 先把结果计算出来, 然后看是否在 byte 的范围内,
12
// 如果不在就报错
13
b = 3 + 4;
14
System.out.println(b);
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b = 10;
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System.out.println(b);
17
}
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1
// 思考题2:
2
// 下面这两种方法有没有区别??
3
// float f = (float)12.345;
4
// float f = 12.345f;
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// 一个是强制转化, 一个是原先就是 float 类型
7
@Test
8
public void test3()
9
{
10
// 但是我在反编译之后发现 这下面两句都一样的处理方式
11
float f = (float)12.345;
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float f1 = 12.345f;
13
}
10
10
1
// 思考题1:
2
// double d = 12.345;
3
// float f = d; ???
4
@Test
5
public void test2()
6
{
7
double d = 12.345;
8
float f = (float) d;
9
System.out.println(f);
10
}
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16
1
// A: byte short char-int-long-float-double
2
// B: byte short char 在运算时默认转化成int 类型进行运算
3
@Test
4
public void test1()
5
{
6
byte b = 4;
7
int i = 3;
8
9
// int sum = b + i;
10
// byte sum = b + i;
11
short s = 2;
12
byte b1 = 1;
13
// (1) 验证成功, byte short char 在运算时默认转化成int 类型进行运算
14
// short sum = s + b1;
15
// System.out.println("sum = " + sum);
16
}
(4) 交换两个变量的方式, 面试题(掌握)
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32
1
// 思考题1
2
// 交换两个变量值的话题, 解决方案, 平时使用和面试使用
3
@Test
4
public void test1()
5
{
6
int a = 10;
7
int b = 20;
8
9
// 方式一: 使用第三发变量存储 工作中使用...
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// int tmp = a;
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// a = b;
13
// b = tmp;
14
// System.out.println("a = " + a + " b = " + b);
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// 方式二: 面试专用, 使用 a ^ b ^ b ==> a 的方法
17
// 效率最高
18
// a = a ^ b;
19
// b = a ^ b; // b = a == (a ^ b) ^ b;
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// a = a ^ b; // a = b == (a ^ b) ^ a
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// System.out.println("a = " + a + " b = " + b);
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// 方式三: 用变量相加减的方法
24
// a = a + b;
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// b = a - b;
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// a = a - b;
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// System.out.println("a = " + a + " b = " + b);
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// 方式四: 一句话搞定
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b = (a + b) - (a = b);
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System.out.println("a = " + a + " b = " + b);
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}
(5) 运算符与和或(掌握)
28
28
1
// 逻辑 && || 和 位 的 & 和 | 区分
2
// && 和 || 的短路效果
3
// && 出现左边是 false 右边就不执行了
4
// || 出现左边是 true 右边就不执行了
5
@Test
6
public void test2()
7
{
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boolean a = false;
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boolean b = true;
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System.out.println("(1)");
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System.out.println("a = " + a);
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System.out.println("b = " + b);
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if (a != true || (b = false))
14
{
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// 输出的值 是 true
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System.out.println("b = " + b);
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}
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a = false;
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b = true;
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System.out.println("(2)");
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System.out.println("a = " + a);
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System.out.println("b = " + b);
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if (b == false && (a = true))
25
{
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System.out.println("a = " + a);
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}
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}
(6) 面试题
20
20
1
/*
2
* 面试题1
3
* (1)
4
* short s = 1;
5
* s = s + 1;
6
* (2)
7
* short s = 1;
8
* s += 1;
9
* */
10
@Test
11
public void test1()
12
{
13
short s = 1;
14
// 还是同一个问题
15
// 变量(short char byte )在运算的时候会被上升为 int
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// 类型, 这个时候计算完毕, 但是赋值失败, 类型不符合
17
// s = s + 1;
18
s += 1;
19
System.out.println(s);
20
}
(7) 位移运算符(要会)
// << 左移: 左边最高位置丢弃, 右边补齐0
// >> 右移: 最高位是0 左边补齐 0; 最高位是1 左边补齐1
// >>> 无符号右移: 无论最高位是 0 还是 1, 左边补齐 0
44
44
1
@Test
2
public void test1()
3
{
4
//<< 把<<左边的数据乘以2的移动次幂
5
System.out.println(3 << 2);// 12
6
7
//>> 把>>左边的数据除以2的移动次幂
8
System.out.println(24 >> 2);// 6
9
System.out.println(24 >>> 2);// 6
10
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System.out.println(-24 >> 2);// 6
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System.out.println(-24 >>> 2);// 乱码
13
}
14
/*
15
计算出3的二进制:11
16
00000000 00000000 00000000 00000011
17
(00)000000 00000000 00000000 0000001100
18
19
>>的移动:
20
计算出24的二进制:11000
21
原码:10000000 00000000 00000000 00011000
22
反码:11111111 11111111 11111111 11100111
23
补码:11111111 11111111 11111111 11101000
24
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11111111 11111111 11111111 11101000
26
1111111111 11111111 11111111 111010(00) 补码
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补码:1111111111 11111111 11111111 111010
29
反码:1111111111 11111111 11111111 111001
30
原码:1000000000 00000000 00000000 000110
31
32
结果:-6
33
34
>>>的移动:
35
计算出24的二进制:11000
36
原码:10000000 00000000 00000000 00011000
37
反码:11111111 11111111 11111111 11100111
38
补码:11111111 11111111 11111111 11101000
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40
11111111 11111111 11111111 11101000
41
0011111111 11111111 11111111 111010(00)
42
43
结果:
44
*/
(8) 面试题
/*
面试题:
请用最有效率的方式写出计算2乘以8的结果?
2 * 8
2 << 3
*/
(9) 三目运算符(要会)
--------------------------------------------------------------------------------
// 1. 获得两个数的最大值
// 2. 获得三个数的最大值
// 3. 比较两个整数是否相等
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1
@Test
2
public void test3()
3
{
4
// 3. 比较两个整数是否相等
5
int m = 20;
6
int n = 10;
7
boolean flag = (m == n) ? true : false;
8
System.out.println(flag);
9
}
10
@Test
11
public void test2()
12
{
13
// 2. 获得三个数的最大值
14
int x = 10;
15
int y = 30;
16
int z = 20;
17
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// (1)
19
int tmp = (x > y) ? x : y;
20
int max = (tmp > y) ? tmp : y;
21
System.out.println("max = " + max);
22
// (2) 一步到位
23
int max1 = (x > y) ? (x > z ? x : z) : (y > z ? y : z);
24
System.out.println("max1 = " + max1);
25
}
26
@Test
27
public void test1()
28
{
29
// 1. 获得两个数的最大值
30
int x = 10;
31
int y = 20;
32
int max = (x > y) ? x : y;
33
System.out.println("max = " + max);
34
}
--------------------------------------------------------------------------------
注意:
4
4
1
// 写一个程序需要考虑:
2
// (1) 成功的数据
3
// (2) 错误的错误
4
// (3) 边界问题
二、switch的使用方法
1. switch(表达式)中表达式的返回值必须是下述几种类型之一:byte,short,char,int,枚举,String;
2. case子句中的值必须是常量,且所有case子句中的值应是不同的;
3. default子句是可任选的,当没有匹配的case时,执行default
4. break语句用来在执行完一个case分支后使程序跳出switch语句块;如果没有break,程序会顺序执行到switch结尾
5. 附加跳出多层循环的方法, goto tab;
三、方法的重载(overload)
1、重载的概念
在同一个类中,允许存在一个以上的同名方法,只要它们的参数个数或者参数类型不同即可。
2、重载的特点
与返回值类型无关,只看参数列表,且参数列表必须不同。(参数个数或参数类型)。调用时,根据方法参数列表的不同来区别。------ 这个和函数签名有关, 函数签名只要函数名和函数参数, 并没有函数的返回值
3. 重载示例
//返回两个整数的和
1
1
1
int add(int x,int y){return x+y;}
//返回三个整数的和
1
1
1
int add(int x,int y,int z){return x+y+z;}
//返回两个小数的和
1
1
1
double add(double x,double y){return x+y;}
9
9
1
public class PrintStream{
2
public static void print(int i) {……}
3
public static void print(float f) {……}
4
private static void print(String s) {……}
5
public static void main(String[] args){
6
print(3);
7
print(1.2f);
8
print(“hello!”);
9
}
4. 判断方式:
与void show(int a,char b,double c){}构成重载的有:
7
7
1
a)void show(int x,char y,double z){} //no
2
b)int show(int a,double c,char b){} //yes
3
c) void show(int a,double c,char b){} //yes
4
d) boolean show(int c,char b){} //yes
5
e) void show(double c){} //yes
6
f) double show(int x,char y,double z){} //no
7
g) void shows(){double c} //no
函数重载只看参数个数和类型,不看返回值。
例子:
int biggest(int a,int b,int c);
{
......
}
float biggest(float a,float b)
{
......
}
float biggest(float a,float b,float c)
{
......
}
以上三个函数都可以。
但不能出现只是返回值不同的重载。
如
int biggest(int a,int b);
{
......
}
float biggest(int a,int b)
{
......
}
四、体会可变个数的形参
//下面采用数组形参来定义方法
1
1
1
public static void test(int a ,String[] books);
//以可变个数形参来定义方法
1
1
1
public static void test(int a ,String…books);
说明:
1.可变参数:方法参数部分指定类型的参数个数是可变多个
2.声明方式:方法名(参数的类型名...参数名)
3.可变参数方法的使用与方法参数部分使用数组是一致的
4.方法的参数部分有可变形参,需要放在形参声明的最后
17
17
1
public void test(String[] msg){
2
System.out.println(“含字符串数组参数的test方法 ");
3
}
4
public void test1(String book){
5
System.out.println(“****与可变形参方法构成重载的test1方法****");
6
}
7
public void test1(String ... books){
8
System.out.println("****形参长度可变的test1方法****");
9
}
10
public static void main(String[] args){
11
TestOverload to = new TestOverload();
12
//下面两次调用将执行第二个test方法
13
to.test1();
14
to.test1("aa" , "bb");
15
//下面将执行第一个test方法
16
to.test(new String[]{"aa"});
17
}
五、方法的参数传递
1. 方法,必须有其所在类或对象调用才有意义。若方法含有参数:
Ø形参:方法声明时的参数
Ø实参:方法调用时实际传给形参的参数值
2. Java的实参值如何传入方法呢?
Java里方法的参数传递方式只有一种:值传递。 即将实际参数值的副本(复制品)传入方法内,而参数本身不受影响。
六、字段名和属性名的概念
Bean层的属性名和变量名
属性名便是在方法 set 和 get 之后的名字便是属性名
属性名不再是变量的名字
3
1
Bean层的属性名和变量名
2
属性名便是在方法 set 和 get 之后的名字便是属性名
3
属性名不再是变量的名字
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