C语言四则运算计算器(带错误检测)

Java小作业，用C写了一遍

输入格式随意，程序自带去空格，欢迎检测bug。

#include <iostream>
#include <cctype>
#include <stack>
#include <string>
#include <vector>
#include <map>
using namespace std;
vector<string>s;
map<string, int>mm;
stack<string>st;
string t, alls, all;
int main()
{

	//优先级
	mm["+"] = 1;
	mm["-"] = 1;
	mm["*"] = 2;
	mm["/"] = 2;
	mm["("] = 3;
	getline(cin, alls);



	//删空格判非法符号
	for (int i = 0; i < alls.size(); i++)
	{
		if (!isdigit(alls[i]) && alls[i] != '+'&&alls[i] != '-'&&alls[i] != '*'&&alls[i] != '/'&&alls[i] != '('&&alls[i] != ')'&&alls[i] != '.'&&alls[i] != ' ')
		{
			printf("#ERROR\n");
			printf("Unknown symbol\n");
			return 0;
		}
		if (alls[i] != ' ')
			all += alls[i];
	}



	//判断表达式是否正确
	int left_bracket = 0;
	for (int i = 0; i < all.size(); i++)
	{
		//判括号
		if (all[i] == '(')
			left_bracket++;
		else if (all[i] == ')')
		{
			if (left_bracket == 0)
			{
				printf("#ERROR\n");
				printf("Brackets do not match\n");
				return 0;
			}
			left_bracket--;
		}
		//特判第一个
		if (i == 0)
		{
			if (all[i] != '+'&&all[i] != '-'&&all[i] != '(' && !isdigit(all[i]))
			{
				printf("#ERROR\n");
				printf("Operator error\n");
				return 0;
			}
		}
		//特判最后一个
		else if (i == all.size() - 1)
		{
			if (all[i] != ')' && !isdigit(all[i]))
			{
				printf("#ERROR\n");
				printf("Operator error\n");
				return 0;
			}
			if (all[i] == ')'&&all[i - 1] == '(')
			{
				printf("#ERROR\n");
				printf("Operator error\n");
				return 0;
			}
		}
		//中间字符，枚举ERROR情况
		else
		{
			if (isdigit(all[i]))
			{
				if (all[i - 1] == ')' || all[i + 1] == '(')
				{
					printf("#ERROR\n");
					printf("Operator error\n");
					return 0;
				}
			}
			else if (all[i] == '.')
			{
				if (!isdigit(all[i - 1]) || !isdigit(all[i + 1]))
				{
					printf("#ERROR\n");
					printf("radix point error\n");
					return 0;
				}
			}
			else if (all[i] == '*' || all[i] == '/')
			{
				if ((!isdigit(all[i - 1]) && all[i - 1] != ')') || (!isdigit(all[i + 1]) && all[i + 1] != '+'&&all[i + 1] != '-'&&all[i + 1] != '('))
				{
					printf("#ERROR\n");
					printf("Operator error\n");
					return 0;
				}
			}
			else if (all[i] == '+' || all[i] == '-')
			{
				if (isdigit(all[i - 1]) || all[i - 1] == ')')
				{
					if (!isdigit(all[i + 1]) && all[i + 1] != '+'&&all[i + 1] != '-')
					{
						printf("#ERROR\n");
						printf("Operator error\n");
						return 0;
					}
				}
				else
				{
					if (!isdigit(all[i + 1]))
					{
						printf("#ERROR\n");
						printf("Operator error\n");
						return 0;
					}
				}
			}
			else if (all[i] == '(')
			{
				if (isdigit(all[i - 1]) || all[i - 1] == ')' || all[i + 1] == '*' || all[i + 1] == '/' || all[i + 1] == ')')
				{
					printf("#ERROR\n");
					printf("Operator error\n");
					return 0;
				}
			}
			else if (all[i] == ')')
			{
				if ((!isdigit(all[i - 1]) && all[i - 1] != ')') || (all[i + 1] != '+'&&all[i + 1] != '-'&&all[i + 1] != '*'&&all[i + 1] != '/'&&all[i + 1] != ')'))
				{
					printf("#ERROR\n");
					printf("Operator error\n");
					return 0;
				}
			}
		}
	}



	//中缀转后缀
	for (int i = 0; i < all.size();)
	{
		if (!isdigit(all[i]))
		{
			//当前符号是加减号并且前一个字符是符号且不是右括号，则判定为正负号
			if ((i == 0 || !isdigit(all[i - 1])) && (all[i] == '+' || all[i] == '-') && (i == 0 || all[i - 1] != ')'))
			{
				if (all[i] == '-')
					t = all[i];
				i++;
				goto qwe;
			}
			t = all[i];
			i++;
		}
		else
		{
			t.clear();
		qwe:
			for (; i < all.size(); i++)
			{
				if (!isdigit(all[i]) && all[i] != '.')
					break;
				t += all[i];
			}
		}
		//如果是数字
		if (isdigit(t[0]) || t.size() > 1)
			s.push_back(t);
		else
		{
			//如果是右括号，弹出遇到左括号的所有符号
			if (t == ")")
			{
				while (!st.empty())
				{
					if (st.top() == "(")
					{
						st.pop();
						break;
					}
					s.push_back(st.top());
					st.pop();
				}
			}
			//否则，正常操作
			else
			{
				while (!st.empty())
				{
					if (st.top() == "(")
						break;
					if (mm[st.top()] >= mm[t])
					{
						s.push_back(st.top());
						st.pop();
					}
					else
						break;
				}
				st.push(t);
			}
		}
	}
	//剩下的符号除了括号全部入栈
	while (!st.empty())
	{
		if (st.top() != "("&&st.top() != ")")
			s.push_back(st.top());
		st.pop();
	}

	for (int i = 0; i < s.size(); i++)
	{
		if (i != 0)
			cout << " ";
		cout << s[i];
	}
	cout << endl;



	//后缀表达式求值
	stack<double>num;
	for (int i = 0; i < s.size(); i++)
	{
		if (s[i].size() > 1 || isdigit(s[i][0]))
		{
			double nums = 0;
			double mul = 1;
			for (int j = s[i].size() - 1; j >= 0; j--)
			{
				if (isdigit(s[i][j]))
				{
					nums += (s[i][j] - '0')*mul;
					mul *= 10;
				}
				else if (s[i][j] == '.')
				{
					nums /= mul;
					mul = 1;
				}
				else if (s[i][j] == '-')
					nums = -nums;
			}
			num.push(nums);
		}
		else
		{
			double a, b, sums;
			b = num.top();
			num.pop();
			a = num.top();
			num.pop();
			switch (s[i][0])
			{
			case '+':
				sums = a + b;
				break;
			case '-':
				sums = a - b;
				break;
			case '*':
				sums = a * b;
				break;
			case '/':
			{
				if (b == 0)
				{
					printf("#ERROR\n");
					printf("Can not be divided into zero.\n");
					return 0;
				}
				sums = a / b;
			}
			}
			num.push(sums);
		}
	}
	printf("%.1lf\n", num.top());
}