写了一个中缀表达式计算器,要求对错误输入进行提示,比如计算后被除数是0、运算符多了、少了等等情况。我的思路是把中缀转后缀,然后计算,这部分已经写出来了,但是我不知道怎么提示错误TT
(我想知道思路,不过如果有现成的就可遇不可求啦😃)
怎么对错误输入进行提示
- 写回答
- 好问题 0 提建议
- 关注问题
- 邀请回答
-
1条回答 默认 最新
关注
#include <bits/stdc++.h> #include <stack> #include <cmath> #include <string> using namespace std; int main() { cout<<"Enter some infix expression,such as 1+1(must be integers,Ctrl+z to end):" <<endl; start: string line; //使用标号语句和跳转语句,当输入的表达式非法时可以便捷地跳回到此处, //重新输入数据 while(getline(cin,line)) { stack<double> numStack; stack<int> expStack; expStack.push(-2); string s=""; int prop=0,first=0,num=0; //prop表示运算符的优先级 double val1=0.0,val2=0.0,result=0.0; char ch; //getline(cin,line); line+='#'; //自己设置停止时刻 for(int i=0;line[i]!='#';++i) { ch=line[i]; if(isdigit(ch)) s+=ch; else { //当s不为空时才进行转换 if(s!="") { num=atoi(s.c_str()); numStack.push(num); s=""; //将s清空 } if(ch==')') { while(expStack.top()!=-2&&expStack.top()!=-7) //还没到栈底,并且不是左括号时 { first=expStack.top(); expStack.pop(); val1=numStack.top(); numStack.pop(); val2=numStack.top(); numStack.pop(); switch(first) { case 1: result=val2+val1; break; case 2: result=val2-val1; break; case 4: result=val2*val1; break; case 5: //当输入除数为0时可能存入val1的不为0,可能很接近0,只要val1足够接近0就认为是0 if(val1<0.0000001) { cerr<<"The divider can't be 0! input again:"<<endl; //exit(1); goto start; } result=val2/val1; break; } numStack.push(result); } //将左括号删去 if(expStack.top()==-7) expStack.pop(); else { cout<<"'(' and ')' can't match! input again:"<<endl; goto start; //exit(1); } } else { switch(ch) { case '+': prop=1; break; case '-': prop=2; break; case '*': prop=4; break; case '/': prop=5; break; case '(': prop=7; //左括号栈外的优先级最高,可直接入栈 } first=expStack.top(); //说明栈顶符号的优先级大于读入的优先级 if(prop-first<2) { expStack.pop(); //除去运算符栈栈顶元素 //取出两个操作数 val1=numStack.top(); numStack.pop(); val2=numStack.top(); numStack.pop(); switch(first) { case 1: result=val2+val1; break; case 2: result=val2-val1; break; case 4: result=val2*val1; break; case 5: if(0==val1) { cerr<<"The divider can't be 0! input again:"<<endl; goto start; //exit(1); } result=val2/val1; break; } numStack.push(result); //将运算的结果重新压到操作数栈中 } if(7==prop) expStack.push(-7); //-7是左括号的标志 else expStack.push(prop); //将新来的运算符压入运算符栈 } } } //最后一个输入的操作数单独处理 if(s!="") { num=atoi(s.c_str()); numStack.push(num); } //到了expStack的栈顶 while(expStack.top()!=-2) { first=expStack.top(); //如果还剩左括号时,输出错误信息 if(first==-7) { cout<<"'(' and ')' can't match! input again:"<<endl; goto start; //exit(1); } expStack.pop(); val1=numStack.top(); numStack.pop(); val2=numStack.top(); numStack.pop(); switch(first) { case 1: result=val2+val1; break; case 2: result=val2-val1; break; case 4: result=val2*val1; break; case 5: if(val1<0.0000001) { cerr<<"The divider can't be 0! input again:"<<endl; //除数不能为0 goto start; //exit(1); } result=val2/val1; break; } numStack.push(result); } cout<<"result:"<<numStack.top()<<endl; } return 0; }解决 无用评论 打赏 举报