|
1195| 8
|
[项目分享] c++代码计算器 |
|
template <char...> struct char_vector { using vec_type = char_vector; }; template<typename t, t... cs>constexpr char_vector<cs...> operator ""_str() { return {}; }然后我们需要两个栈来保存操作数和操作符。第一个栈是 int 类型的，第二个栈是 char 类型的。我们把栈写好。 template <char...> struct char_vector { using vec_type = char_vector; };template <typename> struct pop_char {};template <char c, char... cs>struct pop_char<char_vector<c, cs...>> : std::integral_constant<char, c>, char_vector<cs...> {};template <char, typename> struct push_char {};template <char c, char... cs>struct push_char<c, char_vector<cs...>> : char_vector<c, cs...> {};template <int...> struct int_vector { using vec_type = int_vector; };template <typename> struct pop_int {};template <int i, int... is>struct pop_int<int_vector<i, is...>> : std::integral_constant<int, i>, int_vector<is...> {};template <int, typename> struct push_int {};template <int i, int... is>struct push_int<i, int_vector<is...>> : int_vector<i, is...> {};好了。接下来我们考虑做一个状态机。每一个状态都是 { 操作数栈，操作符栈，输入流 } 的集合。template <typename ns, typename os, typename s>struct state{ using num_stack = ns; using op_stack = os; using stream = s;}; 然后我们就可以 parse char 啦。这部分参照网上，很多资料都会讲如何用两个栈求中序表达式的值。我就不赘述啦。直接看代码吧。template <int... is, char... cs1, char c, char... cs2>constexpr auto eval_once(state<int_vector<is...>, char_vector<cs1...>, char_vector<c, cs2...>>){ if constexpr (c == ' ') { return state<int_vector<is...>, char_vector<cs1...>, char_vector<cs2...>>{}; } else if constexpr (c >= '0' && c <= '9') { using merge_type = decltype(merge(char_vector<c, cs2...>{})); using ops = typename push_int<merge_type::value, int_vector<is...>>::vec_type; return state<ops, char_vector<cs1...>, typename merge_type::type>{}; } else if constexpr (c == '(') { return state<int_vector<is...>, char_vector<c, cs1...>, char_vector<cs2...>>{}; } else if constexpr (c == ')') { return calc_brk(state<int_vector<is...>, char_vector<cs1...>, char_vector<c, cs2...>>{}); } else if constexpr (c == '+' || c == '-' || c == '*' || c == '/') { if constexpr (sizeof...(cs1) == 0) return state<int_vector<is...>, char_vector<c, cs1...>, char_vector<cs2...>>{}; else if constexpr (((pop_char<char_vector<cs1...>>::value == '+' || pop_char<char_vector<cs1...>>::value == '-') && (c == '*' || c == '/')) || pop_char<char_vector<cs1...>>::value == '(') return state<int_vector<is...>, char_vector<c, cs1...>, char_vector<cs2...>>{}; else return calc_op(state<int_vector<is...>, char_vector<cs1...>, char_vector<c, cs2...>>{}); }} 这个 eval once 来 parse 当前的字符，并且进行相应的操作。其中 merge 操作是合并字符变成一个数字的。template <int ret = 0, char c, char... cs>constexpr auto merge(char_vector<c, cs...>){ if constexpr (c >= '0' && c <= '9') return merge<10 * ret + (c - '0')>(char_vector<cs...>{}); else return merge_pair<ret, char_vector<c, cs...>>{};}template <int ret = 0>constexpr auto merge(char_vector<>){ return merge_pair<ret, char_vector<>>{};} 其中 merge pair 就是 { ret, stream } 的集合。这里的 stream 是合并完之后剩的流。template <int n, typename t>struct merge_pair{ static const int value = n; using type = t;}; 我们的 eval once 还用到了两个 calc 方法。其实就是遇到 ) 和优先级低的操作符的时候要进行的计算。 template <int... is, char... cs1, char c, char... cs2>constexpr auto calc_brk(state<int_vector<is...>, char_vector<cs1...>, char_vector<c, cs2...>>){ if constexpr (pop_char<char_vector<cs1...>>::value == '(') return state<int_vector<is...>, typename pop_char<char_vector<cs1...>>::vec_type, char_vector<cs2...>>{}; else return calc_brk(calc_once<pop_char<char_vector<cs1...>>::value>( state<int_vector<is...>, typename pop_char<char_vector<cs1...>>::vec_type, char_vector<c, cs2...>>{}));}template <int... is, char... cs1, char c, char... cs2>constexpr auto calc_op(state<int_vector<is...>, char_vector<cs1...>, char_vector<c, cs2...>>){ return calc_once<pop_char<char_vector<cs1...>>::value>( state<int_vector<is...>, typename push_char<c, typename pop_char<char_vector<cs1...>>::vec_type>::vec_type, char_vector<cs2...>>{});}其中 calc once 就是根据传入的操作符，从操作数栈里弹出两个操作数计算，然后把结果放入操作数栈里。 template <char op, int i1, int i2, int... is, char... cs1, char... cs2>constexpr auto calc_once(state<int_vector<i1, i2, is...>, char_vector<cs1...>, char_vector<cs2...>>){ if constexpr (op == '+') return state<int_vector<i2 + i1, is...>, char_vector<cs1...>, char_vector<cs2...>>{}; else if constexpr (op == '-') return state<int_vector<i2 - i1, is...>, char_vector<cs1...>, char_vector<cs2...>>{}; else if constexpr (op == '*') return state<int_vector<i2 * i1, is...>, char_vector<cs1...>, char_vector<cs2...>>{}; else if constexpr (op == '/') return state<int_vector<i2 / i1, is...>, char_vector<cs1...>, char_vector<cs2...>>{};}嗯。我们的 eval once 就解释完了。接下来我们需要一个递归函数启动 eval once。template <int... is, char... cs1, char... cs2>constexpr auto eval_all(state<int_vector<is...>, char_vector<cs1...>, char_vector<cs2...>>){ if constexpr (sizeof...(cs2) == 0) return state<int_vector<is...>, char_vector<cs1...>, char_vector<cs2...>>{}; else return eval_all(eval_once(state<int_vector<is...>, char_vector<cs1...>, char_vector<cs2...>>{}));} 这个 eval all 函数计算完之后，流就空了。只剩下操作数和操作符两个栈了。我们最后写一个 calc last 函数计算这两个栈的最后结果。 template <int... is, char... cs>constexpr int calc_last(state<int_vector<is...>, char_vector<cs...>, char_vector<>>){ if constexpr (sizeof...(cs) == 0) return pop_int<int_vector<is...>>::value; else return calc_last(calc_once<pop_char<char_vector<cs...>>::value>( state<int_vector<is...>, typename pop_char<char_vector<cs...>>::vec_type, char_vector<>>{}));}好啦。最后我们写一个 eval 函数把它们组合起来。 template <typename t>constexpr int eval(t){ if constexpr (std::is_same_v<t, char_vector<>>) return 0; else return calc_last(eval_all(state<int_vector<>, char_vector<>, t>{}));}现在我们可以写这样的代码了。 #include <boost/spirit/home/x3.hpp>#include <boost/fusion/include/adapt_struct.hpp>#include <boost/fusion/include/io.hpp>namespace x3 = boost::spirit::x3;using x3::double_;using x3::_attr;using x3::space;using x3::phrase_parse;using x3::_val; 然后，定义赋值、加减乘除的action： auto assign_ = [](auto& ctx) {_val(ctx) = _attr(ctx);};auto add_ = [](auto& ctx) { _val(ctx) += _attr(ctx);};auto sub_ = [](auto& ctx) { _val(ctx) -= _attr(ctx);};auto mul_ = [](auto& ctx) { _val(ctx) *= _attr(ctx);};auto div_ = [](auto& ctx) { _val(ctx) /= _attr(ctx);};很自然对不对，然后根据ebnf定义，做个猴子都会的填空题： auto const expression_def = term[assign_] >> *( ('+' >> term [add_]) | ('-' >> term [sub_]) ) ;auto const term_def = factor[assign_] >> *( ('*' >> factor [mul_]) | ('/' >> factor [div_]) ) ;auto const factor_def = double_ | '(' >> expression >> ')' ;BOOST_SPIRIT_DEFINE( expression , term , factor);最后补充个主函数就OK了 int main(){ std::string line; while (true) { std::getline(std::cin, line); if (line == "q") { break; } auto it = line.begin(); auto const end = line.end(); double result = 0.0; bool ret = phrase_parse(it, end, expression, space, result); if (ret && it == end) { std::cout << result << std::endl; } else { std::cout <<std::string("failed to parse at: \"").append(std::string(it, end)).append("\"") << std::endl;; } } return 0;}#include<cctype>#include<iostream>#include<map>#include<string>using namespace std;enum Token_value{ NAME,NUMBER,END,PLUS='+',MINUS='-',MUL='*',DIV='/',PRINT=';',ASSIGN='=',LP='(',RP=')'};Token_value curr_tok=PRINT;map<string,double> table;double number_value;string string_value;int no_of_errors;double expr(bool get);double term(bool get);double prim(bool get);Token_value get_token();double error(const string& s){ no_of_errors++; cerr<<"error:"<<s<<endl; return 1;}Token_value get_token(){ char ch=0; cin>>ch; switch (ch) { case 0: return curr_tok=END; case ';':case '*':case '/':case '+':case '-':case '(':case ')':case '=': return curr_tok=Token_value(ch); case '0':case '1':case '2':case '3':case '4':case '5':case '6':case '7':case '8':case '9':case '.': cin.putback(ch); cin>>number_value; return curr_tok=NUMBER; default: if (isalpha(ch)) { cin.putback(ch); cin>>string_value; return curr_tok=NAME; } error("bad token"); return curr_tok=PRINT; }}double prim(bool get){ if (get) get_token(); switch (curr_tok) { case NUMBER: { double v=number_value; get_token(); return v; } case NAME: { double& v=table[string_value]; if (get_token()==ASSIGN) v=expr(true); return v; } case MINUS: return -prim(true); case LP: { double e=expr(true); if (curr_tok!=RP) return error(") expected"); get_token(); return e; } default: return error("primary expected"); }}double term(bool get){ double left=prim(get); for (;;) switch (curr_tok) { case MUL: left*=prim(true); break; case DIV: if (double d=prim(true)) { left/=d; break; } return error("divide by 0"); default: return left; }}double expr(bool get){ double left=term(get); for(;;) switch(curr_tok) { case PLUS: left+=term(true); break; case MINUS: left-=term(true); break; default: return left; }}int main(){ table["pi"]=3.1415926535897932385; table["e"]=2.718284590452354; while (cin) { get_token(); if (curr_tok==END) break; if (curr_tok==PRINT) continue; cout<<expr(false)<<endl; } return no_of_errors;} |
© 2013-2024 Comsenz Inc. Powered by Discuz! X3.4 Licensed
置顶卡
变色卡
千斤顶
