计算机组成原理算法实现(四)的代码.doc

计算机组成原理课程设计报告源代码口令输入界面import java.awt.*;import java.awt.event.*;import javax.swing.*;public class Computer extends Frame implements ActionListener/*主窗口界面*/MenuBar bar;Menu Huanyuan,Bumajiajian,Yuanmacheng,Fudianjiajian,help;MenuItem t1,t2,t3,t4,t5;TextField password;Button ok;int count=0,m=1;Huanyuan biao; Bumajiajian bian;Yuanmacheng yuan; Fudianjiajian fu;Computer(String s)super(s);biao=new Huanyuan(this,机器数的真值还原（定点小数）,true);bian=new Bumajiajian(this,定点小数单符号位补码加减运算,true);yuan=new Yuanmacheng(this,定点小数原码乘法,true);fu=new Fudianjiajian(this,浮点数的加减法,true);bar=new MenuBar();t1=new MenuItem(定点小数真值还原);t2=new MenuItem(定点小数单符号位补码加减法);t3=new MenuItem(定点小数原码乘法);t4=new MenuItem(浮点加减法);t5=new MenuItem(帮助);t1.addActionListener(this);t2.addActionListener(this);t3.addActionListener(this);t4.addActionListener(this);t5.addActionListener(this);Huanyuan=new Menu(定点小数真值还原);Bumajiajian=new Menu(定点小数单符号位补码加减法);Yuanmacheng=new Menu(定点小数原码乘法);Fudianjiajian=new Menu(浮点加减法);help=new Menu(帮助);Huanyuan.add(t1);Bumajiajian.add(t2);Yuanmacheng.add(t3);Fudianjiajian.add(t4);help.add(t5);Huanyuan.setEnabled(false);Bumajiajian.setEnabled(false);Yuanmacheng.setEnabled(false);Fudianjiajian.setEnabled(false);help.setEnabled(false);bar.add(Huanyuan);bar.add(Bumajiajian);bar.add(Yuanmacheng);bar.add(Fudianjiajian);bar.add(help);setMenuBar(bar);Label label=new Label(计算机组成原理算法实现(四);label.setFont(new Font(TimesRoman,Font.BOLD,24);Panel panel0=new Panel();panel0.add(label);ok=new Button(确 认);ok.addActionListener(this);password=new TextField(20);password.setEchoChar(*); password.addActionListener(this);Box box=Box.createVerticalBox();box.add(panel0);Panel panel1=new Panel();panel1.add(new Label(输入口令:);panel1.add(password);box.add(panel1);Panel panel2=new Panel();panel2.add(ok);box.add(panel2);add(box);setBounds(100,100,510,350);setVisible(true);addWindowListener(new WindowAdapter() public void windowClosing(WindowEvent e) System.exit(0); );validate();public void actionPerformed(ActionEvent e)if(e.getSource()=ok&m真值); fan=new Button(反-真值); bu=new Button(补-真值); yi=new Button(移-真值); back=new Button(返回); input.addActionListener(this); yuan.addActionListener(this); fan.addActionListener(this); bu.addActionListener(this); yi.addActionListener(this); back.addActionListener(this); Panel p1,p2,p3,p4; p1=new Panel(); p2=new Panel(); p3=new Panel(); p4=new Panel(); Box box=Box.createVerticalBox(); Label label1=new Label(机器数的真值还原（定点小数）);label1.setFont(new Font(楷体,Font.BOLD,20); p1.add(label1); p2.add(new Label(请输入机器数：); p2.add(text1); p3.add(label); p3.add(text2); p4.add(input); p4.add(yuan); p4.add(fan); p4.add(bu); p4.add(yi); p4.add(back); box.add(p1); box.add(p2); box.add(p3); box.add(p4); add(box); setBounds(200,200,500,300); addWindowListener(new WindowAdapter() public void windowClosing(WindowEvent e) setVisible(false); ); validate(); public void actionPerformed(ActionEvent e) if(e.getSource()=input) text1.setText(null); text2.setText(null); text1.requestFocusInWindow(); if(e.getSource()=yuan)/*求原-真值*/ number=text1.getText(); if(number.startsWith(0) result=number.replaceFirst(0,+0); text2.setText(result); else if(number.startsWith(1) result=number.replaceFirst(1,-0); text2.setText(result); if(e.getSource()=fan)/*求反-真值*/ number=text1.getText(); if(number.startsWith(0) result=number.replaceFirst(0,+0); text2.setText(result); else if(number.startsWith(1) char c=number.toCharArray(); for(int i=0;ic.length;i+) if(ci=1) ci=0; else if(ci=0) ci=1; result=new String(c); String s=result.replaceFirst(0,-0); text2.setText(s); if(e.getSource()=bu) number=text1.getText(); if(number.startsWith(0) result=number.replaceFirst(0,+0); text2.setText(result); else if(number.startsWith(1) int j=number.lastIndexOf(1); char a=number.toCharArray(); for(int i=0;i真值*/ number=text1.getText(); if(number.startsWith(1) result=number.replaceFirst(1,+0); text2.setText(result); else if(number.startsWith(0) int j=number.lastIndexOf(1); char a=number.toCharArray(); for(int i=0;i=0;j-)/计算两个数的补码加法 if(aj=0 & bj=0) cj=0; dj=0; else if(aj=0& bj=1) cj=1; dj=0; else if(aj=1 & bj=0) cj=1; dj=0; else if(aj=1 & bj=1) cj=0; dj=1; else if(aj=. & bj =.) cj=.; dj=.; for(int k=d.length-1;k=1;k-) if(dk=1 & ck-1=0) ck-1=1; else if(dk=1 & ck-1=1) ck-1=0; dk-1=1; else if(dk=1 & ck-1=.& ck-2=0) ck-2=1; else if(dk=1 & ck-1=.& ck-2=1) ck-2=0; d0=1; if(d0=1& d2=1)|(d0=0& d2=0) if(c0=0) jia=new String(c); jia=jia.replaceFirst(0,+0); addResult.setText(jia); else if(c0=1) jia=new String(c); int j=jia.lastIndexOf(1); for(int i=0;ij;i+) if(ci=1) ci=0; else if(ci=0) ci=1; jia=new String(c); jia=jia.replaceFirst(0,-0); addResult.setText(jia); if(d0=0& d2=1)|(d0=1& d2=0) addResult.setText(结果溢出); JOptionPane.showMessageDialog(this,加法结果溢出,不正确！,错误,JOptionPane.ERROR_MESSAGE); if(e.getSource()=sub) shu1=text1.getText(); shu2=text2.getText(); char a=shu1.toCharArray(); char b=shu2.toCharArray(); int m=shu2.lastIndexOf(1); for(int i=0;i=0;j-)/计算两个数的补码减法 if(aj=0 & bj=0) cj=0; dj=0; else if(aj=0& bj=1) cj=1; dj=0; else if(aj=1 & bj=0) cj=1; dj=0; else if(aj=1 & bj=1) cj=0; dj=1; else if(aj=. & bj =.) cj=.; dj=.; for(int k=d.length-1;k=1;k-) if(dk=1 & ck-1=0) ck-1=1; else if(dk=1 & ck-1=1) ck-1=0; dk-1=1; else if(dk=1 & ck-1=.& ck-2=0) ck-2=1; else if(dk=1 & ck-1=.& ck-2=1) ck-2=0; dk-2=1; if(d0=1& d2=1)|(d0=0& d2=0) if(c0=0) jian=new String(c); jian=jian.replaceFirst(0,+0); subResult.setText(jian); else if(c0=1) jian=new String(c); int j=jian.lastIndexOf(1); for(int i=0;i0) if(a1r-1=0&a2r-1=0&cf=0)jgr-1=0;cf=0;else if(a1r-1=0&a2r-1=0&cf=1)|(a1r-1=0&a2r-1=1&cf=0)|(a1r-1=1&a2r-1=0&cf=0)jgr-1=1;cf=0;else if(a1r-1=0&a2r-1=1&cf=1)|(a1r-1=1&a2r-1=0&cf=1)|(a1r-1=1&a2r-1=1&cf=0) jgr-1=0;cf=1;else if(a1r-1=1&a2r-1=1&cf=1)jgr-1=1;cf=1;else jgr-1=0;r-; result=new String(jg);return result;/实现两数的相乘public String mul(String ss1,String ss2 )char a=ss1.toCharArray();char b=ss2.toCharArray(); int n1=ss1.length();int n2=ss2.length();char mul=new charn1+n2-2;int m=mul.length;char temp1=new charn1+n2-4;char temp2=new charn1+n2-4;/实现积的符号位和小数点if(a0=b0)mul0=0;else mul0=1;mul1=a1; for(int g=0;gn1+n2-4;g+)temp1g=0;for(int g=0;g=2;y-) if(by=1) ss1.getChars(2, n1, temp2, 0); else for(int g=0;gn1+n2-4;g+)temp2g=0; g1=add1(temp1,temp2);/得到两个数的和