编译原理课程中,编了一个简单的语法分析预测程序,这个程序时根据固定的文法得到预测分析表,然后编写程序来判断表达式是否会正确推到出来。
前提是程序没有左递归符合LL(1)文法:
文法如下:
E→TE'
E’ →+TE'|ε
T→FT'
T’ →*FT'|ε
F→(E)|i
为了程序便于编写将E'替换为e,T'替换为t
(2)FIRST集
FIRST(E)={(,i};
FIRST(E’)={+, ε};
FIRST(T)={(,i};
FIRST(T’)={ *, ε};
FIRST(F)={(,i};
(3)FALLOW集
FOLLOW(E)={),#};
FOLLOW(E’)={),#};
FOLLOW(T)={+,),#};
FOLLOW(T’)={+,),#};
FOLLOW(F)={*,+,),#};
(4)预测分析表
i + * ( ) # E E->TE’ E->TE’ E’ E’->+TE’ E’->ε E’->ε T T->FT’ T->FT’ T’ T’->ε T’->*FT’ T’->ε T’->ε F F->i F->(E)
一、Stack.java
public class Stack {
private char s[];
private int top;
private int base;
private final int MAX = 200;
/**
*
* <p>Title: </p>
* <p>Description: 初始化栈</p>
*/
public Stack() {
initStack();
}
private void initStack() {
s = new char[MAX];
base = 0;
top = 0;
}
public boolean isEmpty() {
if(top == base) {
return true;
}else {
return false;
}
}
/**
*
* <p>Title: getTop</p>
* <p>Description: 获取栈顶元素</p>
* @return 返回栈顶元素
*/
public char getTop() {
return s[top-1];
}
/**
*
* <p>Title: push</p>
* <p>Description: 进栈方法</p>
* @param str 进栈的字符
*/
public void push(String str) {
for (int i = str.length() - 1; i >= 0; i--) {
s[top++] = str.charAt(i);
}
}
/**
*
* <p>Title: clear</p>
* <p>Description: 清空栈</p>
*/
public void clear() {
top = base;
}
/**
*
* <p>Title: pop</p>
* <p>Description: 出栈</p>
* @return 栈顶元素出栈并返回出栈的元素
*/
public char pop() {
return s[--top];
}
/**
*
* <p>Title: size</p>
* <p>Description: 返回栈中元素个数</p>
* @return 栈中元素个数
*/
public int size() {
return top;
}
/**
* 打印栈里面的元素
*/
public String toString() {
StringBuffer tempStack = new StringBuffer();
for (int i = 0; i < top; i++) {
tempStack.append(s[i]);
}
return tempStack.toString();
}
}
二、GrammarAnalyze.java
package grammarAnalyze;
public class GrapparAnalyze {
//分析表将E'替换为e,T'替换t
private String tab[][] = {
{ "$", "i", "+", "*", "(", ")", "#" },
{ "E", "Te", "$", "$", "Te", "$", "$" },
{ "e", "$", "+Te", "$", "$", "ε", "ε" },
{ "T", "Ft", "$", "$", "Ft","$", "$" },
{ "t", "$", "ε", "*Ft", "$", "ε", "ε" },
{ "F", "i", "$", "$", "(E)","$", "$" } };
private String input; //input中存放输入的表达式
private StringBuffer tempBuffer; //存放要输出的字符串
private int ptr, row, col, step; //指针,预测表中的行,列,当前步骤
private boolean symbol;
private Stack stack;
public GrapparAnalyze(){
stack = new Stack();
tempBuffer = new StringBuffer();
symbol=true;
input="";
stack.clear();
stack.push("#");
row=1;
ptr=0;
step=1;
}
public int column(char c) { //判断预测表中的列
switch (c) {
case 'i':
return 1;
case '+':
return 2;
case '*':
return 3;
case '(':
return 4;
case ')':
return 5;
case '#':
return 6;
default:
return -1;
}
}
public int line(char c) { //判定预测表中的行
switch (c) {
case 'E':
return 1;
case 'e':
return 2;
case 'T':
return 3;
case 't':
return 4;
case 'F':
return 5;
default:
return -1;
}
}
public void show(String str) {
tempBuffer.append(String.format("%-7d%-14s%-20s%-20s\r\n",
step, filter(stack.toString()), filter(input.substring(ptr)), filter(str)));
step++;
}
public void analyse() {
stack.push(tab[row][0]); //预测表中的第一个元素‘E’
show("初始化");
String tmp;
char ctmp; //栈顶元素
while (!(input.charAt(ptr) == '#' && stack.getTop() == '#')) {
ctmp = stack.getTop();//获取栈顶的元素
if (input.charAt(ptr) == ctmp) { //与栈顶元素比较
stack.pop();
ptr++;
show("" + ctmp + "匹配");
continue;
}
//判断ptr位置的终结符所在预测表的列位置
col = column(input.charAt(ptr));
if (col == -1) {
symbol = false;
show("未定义的字符");
ptr++;
break;
}
//判断栈顶位置所在预测表的行位置
row = line(ctmp);
if (row == -1) {
symbol = false;
show("出错");
stack.pop();
if (input.charAt(ptr) != '#') {
ptr++;
}
continue;
}
tmp = tab[row][col];
if (tmp.charAt(0) == '$') {
symbol = false;
show("出错");
stack.pop();
if (input.charAt(ptr) != '#') {
ptr++;
}
} else if (tmp.charAt(0) == 'ε') {
stack.pop();
show("" + ctmp + "->" + 'ε');
} else {
stack.pop();
stack.push(tmp);
show("" + ctmp + "->" + tmp);
}
}
} //过滤方法将打印的字符串中e和t替换为E'和T'
public String filter(String src) {
if(src.contains("e") || src.contains("t")) {
StringBuffer result = new StringBuffer();
char item;
for(int i = 0;i < src.length(); i++) {
item = src.charAt(i);
if(item == 'e') {
result.append("E'");
continue;
}else if(item == 't') {
result.append("T'");
continue;
}
result.append(item);
}
return result.toString();
}else {
return src;
}
}
public String work(String inputExpression) {
input = inputExpression + '#';
symbol = true;
tempBuffer.append(String.format("%-8s%-20s%-20s%-20s\r\n",
"步骤","分析栈","剩余输入栈","所用产生式"));
analyse();
if (symbol) {
tempBuffer.append("\r是正确的符号串\r");
return tempBuffer.toString();
} else {
tempBuffer.append("\r不是正确的符号串\r");
return tempBuffer.toString();
}
}
public StringBuffer getTempBuffer() {
return tempBuffer;
}
public void setTempBuffer(StringBuffer tempBuffer) {
this.tempBuffer = tempBuffer;
}
public Stack getStack() {
return stack;
}
public void setStack(Stack stack) {
this.stack = stack;
}
public String[][] getTab() {
return tab;
}
public void setTab(String[][] tab) {
this.tab = tab;
}
public String getInput() {
return input;
}
public void setInput(String ns) {
this.input = ns;
}
public int getPtr() {
return ptr;
}
public void setPtr(int ptr) {
this.ptr = ptr;
}
public int getRow() {
return row;
}
public void setRow(int row) {
this.row = row;
}
public int getCol() {
return col;
}
public void setCol(int col) {
this.col = col;
}
public int getStep() {
return step;
}
public void setStep(int step) {
this.step = step;
}
public boolean isBoo() {
return symbol;
}
public void setBoo(boolean boo) {
this.symbol = boo;
}
}
三、主程序GrammarMain.java
package grammarAnalyze;
import java.util.Scanner;
public class GrammarMain {
public static void main(String[] args){
boolean isContinue = true;
while(isContinue) {
GrapparAnalyze analyze = new GrapparAnalyze();
Scanner scan = new Scanner(System.in);
System.out.println("请输入你要翻译的表达式:");
String inputExpression = scan.nextLine();
String srcdata = inputExpression.trim();
if("".equals(srcdata) || srcdata == null) {
System.out.println("输入表达式为空,请重新输入");
}else {
String result = analyze.work(srcdata);
System.out.println(result);
System.out.println("是否继续?y/n");
String yn = scan.nextLine();
if("no".equals(yn) || "n".equals(yn)) {
isContinue = false;
}
}
}
}
}
四、测试运行
转载于:https://www.cnblogs.com/ya-qiang/p/9101883.html
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本文系转载,前往查看
如有侵权,请联系 cloudcommunity@tencent.com 删除。