实验三LR1分析报告法实验报告材料三点检验法检验表.docx
《实验三LR1分析报告法实验报告材料三点检验法检验表.docx》由会员分享，可在线阅读，更多相关《实验三LR1分析报告法实验报告材料三点检验法检验表.docx（9页珍藏版）》请在冰豆网上搜索。
实验三LR1分析报告法实验报告材料三点检验法检验表
实验三LR
（1）分析报告法实验报告材料_三点检验法检验表
一、实验目的构造LR
（1）分析程序，利用它进行语法分析，判断给出的符号串是否为该文法识别的句子，了解LR（K）分析方法是严格的从左向右扫描，和自底向上的语法分析方法。
二、LR
（1）分析法实验设计思想及算法
（1）总控程序，也可以称为驱动程序。
对所有的LR分析器总控程序都是相同的。
（2）分析表或分析函数，不同的文法分析表将不同，同一个文法采用的LR分析器不同时，分析表将不同，分析表又可以分为动作表（ACTION）和状态转换（GOTO）表两个部分，它们都可用二维数组表示。
（3）分析栈，包括文法符号栈和相应的状态栈，它们均是先进后出栈。
分析器的动作就是由栈顶状态和当前输入符号所决定。
LR分析器由三个部分组成：
其中:
SP为栈指针，S[i]为状态栈，X[i]为文法符号栈。
状态转换表用GOTO[i，X]=j表示，规定当栈顶状态为i，遇到当前文法符号为X时应转向状态j，X为终结符或非终结符。
ACTION[i，a]规定了栈顶状态为i时遇到输入符号a应执行。
动作有四种可能：
（1）移进：
action[i，a]=Sj：
状态j移入到状态栈，把a移入到文法符号栈，其中i,j表示状态号。
（2）归约：
action[i，a]=rk：
当在栈顶形成句柄时，则归约为相应的非终结符A，即文法中有A-B的产生式，若B的长度为R（即|B|=R），则从状态栈和文法符号栈中自顶向下去掉R个符号，即栈指针SP减去R，并把A移入文法符号栈，j=GOTO[i,A]移进状态栈，其中i为修改指针后的栈顶状态。
（3）接受acc:
当归约到文法符号栈中只剩文法的开始符号S时，并且输入符号串已结束即当前输入符是'#'，则为分析成功。
（4）报错:
当遇到状态栈顶为某一状态下出现不该遇到的文法符号时，则报错，说明输入端不是该文法能接受的符号串。
三、程序结构描述
1、定义的变量
stringaction[12][6]={
{"s5","error","error","s4","error","error"},
{"error","s6","error","error","error","acc"},
{"error","r2","s7","error","r2","r2"},
{"error","r4","r4","error","r4","r4"},
{"s5","error","error","s4","error","error"},
{"error","r6","r6","error","r6","r6"},
{"s5","error","error","s4","error","error"},
{"s5","error","error","s4","error","error"},
{"error","s6","error","error","s11","error"},
{"error","r1","r7","error","r1","r1"},
{"error","r3","r3","error","r3","r3"},
{"error","r5","r5","error","r5","r5"}
};
stringgo[12][3]={
{"1","2","3"},
{"error","error","error"},
{"error","error","error"},
{"error","error","error"},
{"8","2","3"},
{"error","error","error"},
{"error","9","3"},
{"error","error","10"},
{"error","error","error"},
{"error","error","error"},
{"error","error","error"},
{"error","error","error"},
};//初始化预测分析表
charVt[6]={'i','+','*','（','）','#'};终结符表
stringLR[6]={"E->E+T","E->T","T->T*F","T->F","F->（E）","F->i"};//LR文法
stackS;状态栈
stackX;符号栈
charinput[10];输入字符
2、定义的函数
intnum（strings）判断字符串中的数字
voidprint（inti,char*c）剩余输入串的输出
intsame（chara）用于查找终结符
voidanalyse（）分析程序
四、程序源代码及运行结果
#include
#include
#include
#include
usingnamespacestd;
//初始化预测分析表
stringaction[12][6]={
{"s5","error","error","s4","error","error"},
{"error","s6","error","error","error","acc"},
{"error","r2","s7","error","r2","r2"},
{"error","r4","r4","error","r4","r4"},
{"s5","error","error","s4","error","error"},
{"error","r6","r6","error","r6","r6"},
{"s5","error","error","s4","error","error"},
{"s5","error","error","s4","error","error"},
{"error","s6","error","error","s11","error"},
{"error","r1","r7","error","r1","r1"},
{"error","r3","r3","error","r3","r3"},
{"error","r5","r5","error","r5","r5"}
};
stringgo[12][3]={
{"1","2","3"},
{"error","error","error"},
{"error","error","error"},
{"error","error","error"},
{"8","2","3"},
{"error","error","error"},
{"error","9","3"},
{"error","error","10"},
{"error","error","error"},
{"error","error","error"},
{"error","error","error"},
{"error","error","error"},
};
charVt[6]={'i','+','*','（','）','#'};//终结符表
stringLR[6]={"E->E+T","E->T","T->T*F","T->F","F->（E）","F->i"};//LR文法
stackS;//状态栈
stackX;//符号栈
charinput[10];//输入字符
intnum（strings）{//判断字符串中的数字
inti;
stringstr="";
for（intj=0;jif（s[j]>='0'&&s[j]<='9'）str=str+s[j];
}
i=atoi（str.c_str（））;
returni;
}
voidprint（inti,char*c）//剩余输入串的输出
{
for（intj=i;j<10;j++）
cout<cout<<'\t';
}
intsame（chara）{//用于查找终结符
for（inti=0;i<6;i++）{
if（a==Vt[i]）returni;
}
}
///**********分析程序*******************
voidanalyse（）{
boolflag=true;//循环条件
intstep=1,point=0,state=0;//步骤、指针、状态
charch1,ch2;
intm,n,l;//用于判断终结符,分析表,表达式右部的长度
stringstr1;//用于判断对应分析表中的符号
stringstr2="#",str3="0";//记录符号栈的所有元素
cout<<"请输入要规约的字符串：
"<cin>>input;
X.push（'#'）;
S.push（0）;
cout<<"步骤"<<'\t'<<"状态栈"<<'\t'<<"符号栈"<<'\t'<<"输入串"<<'\t'<<"动作"<cout<print（point,input）;
cout<<"初始化"<//************进入循环
while（flag）{
state=S.top（）;
ch1=input[point];
m=same（ch1）;
str1=action[state][m];
//***********移进动作
if（str1[0]=='s'）{
n=num（str1）;
S.push（n）;
X.push（ch1）;
str2=str2+ch1;
ch2=n+48;
str3=str3+ch2;
point++;
cout<print（point,input）;
cout<'<<"移进"<}
//**********归约动作
elseif（str1[0]=='r'）{
n=num（str1）;
l=LR[n-1].length（）-3;
for（inti=1;i<=l;i++）{S.pop（）;
str3=str3.substr（0,str3.length（）-1）;
X.pop（）;
str2=str2.substr（0,str2.length（）-1）;
}
X.push（LR[n-1][0]）;
str2=str2+LR[n-1][0];
state=S.top（）;
if（LR[n-1][0]=='E'）{S.push（num（go[state][0]））;
ch2=num（go[state][0]）+48;
str3=str3+ch2;
}
elseif（LR[n-1][0]=='T'）{S.push（num（go[state][1]））;
ch2=num（go[state][1]）+48;
str3=str3+ch2;
}
elseif（LR[n-1][0]=='F'）{S.push（num（go[state][2]））;
ch2=num（go[state][2]）+48;
str3=str3+ch2;
}
cout<print（point,input）;
cout<'<}
//*********出错
elseif（str1=="error"）{
cout<<"ERROR"<<"程序错误，分析结束！
！
！
"<flag=false;
}
//**********分析成功
elseif（str1=="acc"）{cout<<"分析成功，终止程序"<flag=false;
}
}
}
intmain（）{
analyse（）;
return0;
}
测试：
i*i+i
结果：
五、实验总结
升级会员 