编译技术第5次上机内容

编译技术第5次上机内容目的：充分理解语义分析的方法及相关语义计算的执行时机。

要求：1．以S属性的语法制导定义为基础，将下表的语义规则嵌套在语法分析的过程中，即实现语法制导的翻译过程。

即以LR 文法为基础。

当进行产生式归约时执行对应的语义动作。

3．输入：5+3+8*2输出：244．若输入有误，如：3++2则应提示：重新输入！5．由于输入串是具体的数值，因此应调用相应的词法分析的功能。

// Expression.cpp : Defines the entry point for the console application.//#include "stdafx.h"#include "conio.h"#include<string.h>#include<iostream>#include<stack>using namespace std;#define L 0#define E 1#define T 2#define E_ 3#define T_ 4#define F 6#define digit 7 // 数字#define add 8 // 左括号#define mul 9 // 右括号#define lb 10#define rb 11int nStackPtr;int Stack[100]; // 栈void guiyue(stack<int> &state,stack<int> &value,char nex); void Push(int n){nStackPtr ++;Stack[nStackPtr] = n;}void Pop(){nStackPtr--;}void PrintStack(){int i;for (i = nStackPtr; i >= 0; i--){if (Stack[i] == E) printf("E ");if (Stack[i] == E_ ) printf("E' ");if (Stack[i] == T ) printf("T ");if (Stack[i] == T_ ) printf("T' ");if (Stack[i] == F) printf("F ");if (Stack[i] == digit) printf("digit ");if (Stack[i] == add) printf("+ ");if (Stack[i] == mul) printf("* ");if (Stack[i] == lb) printf("( ");if (Stack[i] == rb) printf(") ");}printf("\n");}/////////////////////////////////////////////////////////////////// 利用栈来分析表达式串，判定表达式串是否正确///////////////////////////////////////////////////////////////////int main(int argc, char* argv[]){char strInput[100]; // 存放表达式串bool bResult;int nInputPtr;nStackPtr = -1;nInputPtr = 0;bResult = true;// 输入表达式串，存放在strInput中printf("请输入表达式串：");scanf("%s",strInput);Push(E);PrintStack();while (bResult && nStackPtr >= 0){switch(Stack[nStackPtr]){case E:if (strInput[nInputPtr] >= '0' && strInput[nInputPtr] <= '9'){Pop();Push(E_);Push(T);}else if (strInput[nInputPtr] == '('){Pop();Push(E_);Push(T);}elsebResult = false;PrintStack();break;case T:if (strInput[nInputPtr] >= '0' && strInput[nInputPtr] <= '9'){Pop();Push(T_);Push(F);}else if (strInput[nInputPtr] == '('){Pop();Push(T_);Push(F);}else bResult = false;PrintStack();break;case E_:if (strInput[nInputPtr] == '+'){Pop();Push(E_);Push(T);Push(add);}else if (strInput[nInputPtr] == ')' || strInput[nInputPtr] == 0) Pop();else bResult = false;PrintStack();break;case T_:if (strInput[nInputPtr] == '+' || strInput[nInputPtr] == ')' || strInput[nInputPtr] == 0) Pop();else if (strInput[nInputPtr] == '*'){Pop();Push(T_);Push(F);Push(mul);}else bResult = false;PrintStack();break;case F:if (strInput[nInputPtr] >= '0' && strInput[nInputPtr]<='9') {Pop();Push(digit);}else if (strInput[nInputPtr] == '('){Pop();Push(rb);Push(E);Push(lb);}else bResult = false;PrintStack();break;case digit:if (strInput[nInputPtr] >= '0' && strInput[nInputPtr]<='9') {Pop();nInputPtr++;}else bResult = false;PrintStack();break;case add:if (strInput[nInputPtr] == '+'){Pop();nInputPtr++;}else bResult = false;PrintStack();break;case mul:if (strInput[nInputPtr] == '*'){Pop();nInputPtr++;}else bResult = false;PrintStack();break;case lb:if (strInput[nInputPtr] == '('){Pop();nInputPtr++;}else bResult = false;PrintStack();break;case rb:if (strInput[nInputPtr] == ')'){Pop();nInputPtr++;}else bResult = false;PrintStack();break;default: bResult = false; break;}}if (bResult == false) printf("表达式有问题了\n");else printf("表达式没问题!!\n");//以上是词法、语法分析代码//------------------------------------------------------------------------------------------------------------//以下是语义分析代码stack<int> state;stack<int> value;int i=0;int size=strlen(strInput);for(i=0;i<size;i++){if(strInput[i]>='0'&&strInput[i]<='9'){state.push(digit);value.push(strInput[i]-'0');if(i!=(size-1))guiyue(state,value,strInput[i+1]);elseguiyue(state,value,'#');}else if(strInput[i]=='+'){state.push(add);}else if(strInput[i]=='*'){state.push(mul);}else if(strInput[i]=='('){state.push(lb);}else if(strInput[i]==')'){state.push(rb);if(i!=(size-1))guiyue(state,value,strInput[i+1]);elseguiyue(state,value,'#');}}// printf("Hello World!\n");return 0;}void guiyue(stack<int> &state,stack<int> &value,char nex) {int next;if(nex>='0'&&nex<='9')next=digit;else if(nex=='#')next=0;else{switch(nex){case '+':next=add;break;case '*':next=mul;break;case '(':next=lb;break;case ')':next=rb;break;}}stack<int> state1=state;stack<int> value1=value;int top_s1,top_s2;int top_v1;int top=state.top();bool flag=true;while(flag){switch(state.top()){case digit:state.pop();state.push(F);state1=state;break;case F:if(state.size()>=3){state1.pop();top_s1=state1.top();state1.pop();top_s2=state1.top();state1.pop();if(top_s1==mul&&top_s2==T){state1.push(T);state=state1;value1.pop();top_v1=value1.top();value1.pop();value1.push(top_v1*value.top());value=value1;}else{state.pop();state.push(T);state1=state;}}else{state.pop();state.push(T);state1=state;}break;case T:if(next==mul){flag=false;}else if(state.size()>=3){state1.pop();top_s1=state1.top();state1.pop();top_s2=state1.top();state1.pop();if(top_s1==add&&top_s2==E){state1.push(E);state=state1;value1.pop();top_v1=value1.top();value1.pop();value1.push(top_v1+value.top());value=value1;}else{state.pop();state.push(E);state1=state;}}else{state.pop();state.push(E);state1=state;}break;case E:if(next==add||next==rb){flag=false;}else{if(state.size()==1&&state.top()==E){state.pop();state.push(L);printf("%d\n",value.top());exit(0);}state1=state;}break;case rb:state1.pop();top_s1=state1.top();state1.pop();top_s2=state1.top();state1.pop();if(top_s1==E&&top_s2==lb){state1.push(F);state=state1;}state1=state;break;}}}。