编译原理-语法制导翻译实现计算器

目录
  1. 1. 任务描述
  2. 2. 文法
  3. 3. LL(1) 预测分析表
  4. 4. SDD
  5. 5. 程序实现

任务描述

设计一个文法，匹配合法的计算式，并返回正确计算式的结果。

文法

\begin{align}
&A \rightarrow BC \
&C \rightarrow +BC \mid -BC \mid \varepsilon \
&B \rightarrow DE \
&E \rightarrow *DE \mid /DE \mid \varepsilon \
&D \rightarrow digit \mid (A) \
\end{align}

\begin{align}
FIRST(A) = \Big\lbrace digit,( \Big\rbrace &\quad FOLLOW(A) = \Big\lbrace $,) \Big\rbrace\
FIRST(B) = \Big\lbrace digit,( \Big\rbrace &\quad FOLLOW(B) = \Big\lbrace $,),+,- \Big\rbrace\
FIRST(C) = \Big\lbrace +,-,\varepsilon \Big\rbrace &\quad FOLLOW(C) = \Big\lbrace $,) \Big\rbrace\
FIRST(D) = \Big\lbrace digit,( \Big\rbrace &\quad FOLLOW(D) = \Big\lbrace $,),*,/,+,- \Big\rbrace\
FIRST(E) = \Big\lbrace *,/,\varepsilon \Big\rbrace &\quad FOLLOW(E) = \Big\lbrace $,),+,- \Big\rbrace\
\end{align}

LL(1) 预测分析表

$ $ digit ( ) + - * / $
A $A \rightarrow BC$ $A \rightarrow BC$
B $B \rightarrow DE$ $B \rightarrow DE$
C $C \rightarrow \varepsilon$ $C \rightarrow +BC$ $C \rightarrow -BC$ $C \rightarrow \varepsilon$
D $D \rightarrow digit$ $D \rightarrow (A)$
E $E \rightarrow \varepsilon$ $E \rightarrow \varepsilon$ $E \rightarrow \varepsilon$ $E \rightarrow *DE$ $E \rightarrow /DE$ $E \rightarrow \varepsilon$

SDD

$ $ 产生式 语义规则
0) $A \rightarrow BC$ \begin{align} &C.inh=B.syn \\ &A.syn = C.syn \end{align}
1) $C \rightarrow +BC_1$ \begin{align} &C_1.inh=C.inh + B.syn \\ &C.syn=C_1.syn \end{align}
2) $C \rightarrow -BC_1$ \begin{align} &C_1.inh=C.inh-B.syn \\ &C.syn=C_1.syn \end{align}
3) $C \rightarrow \varepsilon$ \begin{align} C.syn=C.inh \end{align}
4) $B \rightarrow DE$ \begin{align} &E.inh=D.syn \\ & B.syn=E.syn \end{align}
5) $E \rightarrow *DE_1$ \begin{align} &E_1.inh=E.inh \times D.syn \\ &E.syn=E_1.syn \end{align}
6) $E \rightarrow /DE_1$ \begin{align} &E_1.inh=E.inh / D.syn \\ & E.syn=E_1.syn \end{align}
7) $E \rightarrow \varepsilon$ \begin{align} &E.syn=E.inh \end{align}
8) $D \rightarrow digit$ \begin{align} &D.syn = digit.lexval \end{align}
9) $D \rightarrow (A)$ \begin{align} &D.syn = A.syn \end{align*}

程序实现

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#include <vector>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <exception>
#include <algorithm>
using namespace std;

struct node {
    int id, syn, inh;
    node(int id=0, int syn=0, int inh=0): id(id), syn(syn), inh(inh) {}
};
char key[128];
int M[10][10];
vector<int> v[10];

inline int idx(char c) {
    return key[c];
}

inline void initid() {
    memset(key, 0, sizeof key);

    for(int k='0'; k <= '9'; ++k) key[k] = 1;
    key['('] = 2;
    key[')'] = 3;
    key['+'] = 4;
    key['-'] = 5;
    key['*'] = 6;
    key['/'] = 7;
    key['$'] = 8;

    key['A'] = -1;
    key['B'] = -2;
    key['C'] = -3;
    key['D'] = -4;
    key['E'] = -5;
}

inline void inittable() {
#define pb push_back
    for(int i=0; i < 10; ++i) v[i].clear();

    v[0].pb(idx('B')); v[0].pb(idx('C')); // A --> BC
    v[1].pb(idx('+')); v[1].pb(idx('B')); v[1].pb(idx('C')); // C --> +BC
    v[2].pb(idx('-')); v[2].pb(idx('B')); v[2].pb(idx('C')); // C --> -BC
    // C --> \varepsilon
    v[4].pb(idx('D')); v[4].pb(idx('E')); // B --> DE
    v[5].pb(idx('*')); v[5].pb(idx('D')); v[5].pb(idx('E')); // E --> *DE
    v[6].pb(idx('/')); v[6].pb(idx('D')); v[6].pb(idx('E')); // E --> /DE
    // E --> \varepsilon
    v[8].pb(idx('0')); // D --> digit
    v[9].pb(idx('(')); v[9].pb(idx('A')); v[9].pb(idx(')')); // D --> (A)

//    for(int i=0; i < 10; ++i) reverse(v[i].begin(), v[i].end());
#undef pb

    memset(M, -1, sizeof M);
    M[1][1] = 0; M[1][2] = 0;
    M[2][1] = 4; M[2][2] = 4;
    M[3][3] = 3; M[3][4] = 1; M[3][5] = 2; M[3][8] = 3;
    M[4][1] = 8; M[4][2] = 9; 
    M[5][3] = 7; M[5][4] = 7; M[5][5] = 7; M[5][6] = 5; M[5][7] = 6; M[5][8] = 7;
}

int getnum(const char*& s) {
    int num = 0;
    for(; isdigit(*s); ++s)
        num = num*10 + *s-'0';
    return num;
}

const int endsym = idx('$');
void work(const char* &s, node& sy, int cur) {
    int id = idx(*s);
    if( !id ) throw "不识别的符号";

    if( sy.id != endsym ) {
        if( sy.id == id ) {
            if( id == 1 ) {
                sy.syn = getnum(s);
            } else ++s;
            return;
        }    
        if( sy.id > 0 ) throw "error!";
        int Mid = M[-sy.id][id];
        if( Mid < 0 ) throw "error!";
        node sym[4];
        for(int i=0; i < v[Mid].size(); ++i) 
            sym[i].id = v[Mid][i];
        if( v[Mid].size() )work(s, sym[0], cur+1);
        switch( Mid ) {
            case 0: 
            case 4:
                sym[1].inh = sym[0].syn; 
                work(s, sym[1], cur+1);
                sy.syn = sym[1].syn; 
                break; // A --> BC 或 B --> DE
            case 1: 
                work(s, sym[1], cur+1);
                sym[2].inh = sy.inh+sym[1].syn; 
                work(s, sym[2], cur+1);
                sy.syn = sym[2].syn; 
                break; // C --> +BC
            case 2: 
                work(s, sym[1], cur+1);
                sym[2].inh = sy.inh-sym[1].syn; 
                work(s, sym[2], cur+1);
                sy.syn = sym[2].syn; 
                break; // C --> -BC
            case 3: 
            case 7:
                sy.syn = sy.inh; 
                break; // C --> \epsilon 或 E --> \epsilon
            case 5: 
                work(s, sym[1], cur+1);
                sym[2].inh = sy.inh*sym[1].syn; 
                work(s, sym[2], cur+1);
                sy.syn = sym[2].syn; 
                break; // E --> *DE
            case 6: 
                work(s, sym[1], cur+1);
                sym[2].inh = sy.inh/sym[1].syn; 
                work(s, sym[2], cur+1);
                sy.syn = sym[2].syn; 
                break; // E --> /DE
            case 8: sy.syn = sym[0].syn; 
                break; // D --> digit
            case 9: 
                work(s, sym[1], cur+1);
                sy.syn = sym[1].syn; 
                work(s, sym[2], cur+1);
                break; // D --> (A)
        }
    }
    if( cur == 0 ) throw sy.syn;
}

int main()
{
    string in;
    string coin;
    initid();
    inittable();
    while( getline(cin, in) ) {
        in.push_back('$');
        int len = in.length();
        const char* s = in.c_str();
        coin.clear();
        for(int i=0; i < len; ++i) 
            if( s[i] !=' ' && s[i] != '\t' && s[i] != '\n' ) 
                coin.push_back(s[i]);
        coin.push_back('\0');
        /* 以上去除空格 */

        try {
            node sy = node(idx('A'));
            s = coin.c_str();
            work(s, sy, 0);
        }
        catch(const char* str) {
            puts(str);
        }
        catch(int ans) {
            for(int i=0; i < len-1; ++i) putchar(in[i]);
            printf(" = %d\n", ans);
            printf("succuss!\n");
        }
    }
    return 0;    
}