现在只在linux平台下运行成功 win10运行不成功
#include <iostream> #include <string> #include <list> using namespace std; // 中缀 转后缀 然后求值 int pri(const char c) { switch (c) { case '#': return 0; case ')': return 5; case '+': return 2; case '-': return 3; case '*': return 5; case '/': return 5; case '(': return 5; default: return -1; case ' ': return 1; } } float Compute(float a, float b, string c) { float res = 0; if (c == "+") { res = a + b; } else if (c == "-") { res = a - b; } else if (c == "*") { res = a * b; } else if (c == "/") { res = a / b; } else { res = -1; } return res; } int main() { string s1 = "50.6+2*((-1)*2+10+(-1)*5)#"; list<float> num; num.push_back(0); string s2 = ""; string tmp = ""; list<char> op; for (float i = 0; i < s1.size(); i++) { if (pri(s1[i]) == -1) { // is number tmp += s1[i]; cout << "tmp=" << tmp << endl; } else { if (tmp != "") { s2 += tmp + "#"; } cout << "s2=" << s2 << endl; tmp = ""; if (s1[i] == '-' && s1[i - 1] == '(') { s2 += "0#"; cout << "s2 first=" << s2 << endl; } if (s1[i] == ')') { while (op.back() != '(') { s2 += op.back(); op.pop_back(); } op.pop_back(); } else if (s1[i] == '(') { op.push_back(s1[i]); } else if (pri(s1[i]) != -1 && pri(s1[i]) != 0) { while (op.back() != '(' && pri(op.back()) > pri(s1[i]) && !op.empty()) { s2 += op.back(); op.pop_back(); } op.push_back(s1[i]); } else { while (!op.empty()) { s2 += op.back(); op.pop_back(); } } } } while (!op.empty()) { if (op.back() != '#') { s2 += op.back(); } op.pop_back(); } cout << "last s2=" << s2 << endl; tmp = ""; list<float> arrL; for (int i = 0; i < s2.size(); i++) { if (s2[i] == '#') { tmp.reserve(); arrL.push_back(atof(tmp.c_str())); tmp = ""; continue; } if (pri(s2[i]) == -1) { tmp += s2[i]; } else { float tmp1 = arrL.back(); //cout<<"tmp1 = "<<tmp1<<endl; arrL.pop_back(); float tmp2 = arrL.back(); //cout<<"tmp2 = "<<tmp2<<endl; arrL.pop_back(); string tmps = ""; tmps += s2[i]; float tmp3 = Compute(tmp2, tmp1, tmps); cout << "\ttmp3=" << tmp3 << endl; arrL.push_back(tmp3); } } cout << "原来的string = " << s1 << endl; cout << "result=" << arrL.back() << endl; return 0; }