我正在尝试为RPN计算器编写代码，我正在努力使其正确计算多于一个＆＃34;左＆＃34;和＆＃34;对＆＃34;操作数。
到目前为止，我的代码适用于以下输入： 10 15 + = 25
我无法弄清楚如何正确输出这些输入：
输入：100 10 50 25 / * - -2 / =输出：-40
此外，现在我只是将堆栈大小除以2，这适用于某些错误检查，但是它会输出类似于上/下输入的错误。 100 10 50 25 / * - -2 / = -40
如何获取以下这两个输入的错误检查代码？
输入：10 20 * / =输出：错误：操作员太多
12 20 30 / = Error: Too many operand
非常感谢任何帮助，谢谢！
#include <iostream>
#include <stack>
#include <string>
#include <sstream> //make use of a class called istringstream
#include<iomanip>
using namespace std;
//Function prototype for isOperator
bool isOperator(const string& input);
//Function prototype for perforOperation
int performOperation(const string& input, stack<double>& calcStack);
int main()
{
cout << "RPN Calculator: " << endl;
cout << "Input\n";
stack<double> calcStack;
string input;
while(input != "0")
{
//Terminate program when 0 is entered by user
while(input != "=")
{
// get input
cin >> input;
// check for being numeric value
double num;
if(istringstream(input) >> num)
{
//use push function
calcStack.push(num);
}
// check for operator
else if(isOperator(input))
{
performOperation(input, calcStack);
}
// If user enters 0 on a line followed by a new line, the program exits ????????????
else if(input == "0\n")
{
return -1;
}
// invalid output check
//else
//{
//cout << "Invalid input" << endl;
//}
}
}
}
bool isOperator(const string& input)
{
string operators[] = {"-", "+", "*", "/"};
for(int i=0; i<4; i++)
{
if(input == operators[i])
{
return true;
}
}
return false;
}
int performOperation(const string& input, stack<double>& calcStack)
{
double firstOperand;
double secondOperand;
double result;
if( calcStack.size() > 2 ) //Error check gives a false error for last input ???
{
cout << "Error: too many operands" << endl;
return 1;
}
//Error chceck for too many operators ////STILL PRINT OUTPUT???
if( calcStack.size() < 2 )
{
cout << "Error: too many operators" << endl;
return 1;
}
secondOperand = calcStack.top();
calcStack.pop();
firstOperand = calcStack.top();
calcStack.pop();
if(input == "-")
{
result = firstOperand-secondOperand;
}
else if (input == "+")
{
result = firstOperand + secondOperand;
}
else if (input == "*")
{
result = firstOperand * secondOperand;
}
else if( input == "/")
{
result = firstOperand / secondOperand;
}
// If user enters 0 on a line followed by a new line, the program exits ???????????
else if(input == "0\n")
{
return -1;
}
//Division by zero error
if(secondOperand == 0)
{
cout << "Error: Division by 0.\n";
return -1;
}
cout << "Output\n";
cout << result << endl;
calcStack.push(result);
return 0;
}
答案 0 :(得分：0)
这里的代码
if( calcStack.size() > 2 )
{
cout << "Error: too many operands" << endl;
return 1;
}
需要转移到主要并稍微转换
else if(isOperator(input))
{
performOperation(input, calcStack);
}
else if(input == "=")
{
if (calcStack.size() != 1)
{
cout << "Error: too many operands" << endl;
return 1;
}
else
{
cout << "Result: " << calcStack.top();
// now decide whether or not you are preserving the result for
// the next computation
calcStack.pop(); // Assuming not keeping result
}
}
这意味着您需要重新考虑此循环while(input != "=")
你真的很亲密。
两个建议：
您可以优化isOperator功能。
bool isOperator(const string& input)
{
static const string operators ="-+*/";
if (input.length() == 1) // right size to be an operator.
{
return operators.find_first_of(input[0]) != string::npos;
// look in the operator string for the first (and only) character in input
}
return false;
}
由于您知道运算符中只有一个字符，因此您可以使用比if / else更优雅的字符：
switch (input[0])
{
case '-':
result = firstOperand - secondOperand;
break;
case '+':
result = firstOperand + secondOperand;
break;
case '*':
result = firstOperand * secondOperand;
break;
case '/':
if (secondOperand == 0)
{ // moved this test to here because it's the only place it matters.
cout << "Error: Division by 0.\n";
return -1;
}
result = firstOperand / secondOperand;
break;
}
到目前为止，您的代码应该已经发生了很大变化，您可能希望使用当前代码开始一个新问题。如果没有，这就是我在评论中所说的。
else if(isOperator(input))
{
if (performOperation(input, calcStack) != 0)
{
// empty out the stack and delete all pending user input.
calcStack.clear();
cin.ignore(numeric_limits<streamsize>::max(), '\n');
cout << "Bad input. Try again." << endl;
break; // exit input loop.
}
}
else if(input == "=")
{
...
答案 1 :(得分：0)
double helpRpnFunc(string arr[], int& size) {
if (size == 0)
throw "Error: Too many operators\n";
try {
return stoi(arr[size - 1]);
}
catch (invalid_argument) {
char operator1 = arr[size - 1][0];
double secondOperand = helpRpnFunc(arr, --size), firstOperand = helpRpnFunc(arr, --size);
switch (operator1) {
case '+':
return firstOperand + secondOperand;
case '-':
return firstOperand - secondOperand;
case '*':
return firstOperand * secondOperand;
case '/':`enter code here`
if (secondOperand == 0)
throw "Error: canot divide by 0!\n";
return firstOperand / secondOperand;
}
}
}
string rpnFunc(string arr[], int size) {
try {
string str = to_string(helpRpnFunc(arr, size));
if (size > 1)
throw "Error: too many operands";
return str.substr(0, str.find(".") + 3);
}
catch (const char* error) {
return error;
}
}
int main()
{
string arr1[] = { "100","10","50","25","/","*","-","-2","/" };
int size = 9;
cout << rpnFunc(arr1, size) << endl;
string arr2[] = { "10","20","*","/" };
size = 4;
cout << rpnFunc(arr2, size) << endl;
string arr3[] = { "12","20","30","/" };
cout << rpnFunc(arr3, size) << endl;
}