RPN计算器c++错误处理和多个运算符

Question

编辑：我已经对我的程序进行了更改。看看calc.cpp。特别是isOperator函数，以及第四个while循环，即

while (isOperator(std::cin.peek()))
...
...
if ( op == '/')

我认为问题在于char-op没有设置为"/"。当我回到我的isOperator功能时

int isOperator(char ch)
{
int count = 0;
char ops[] = {'-','+','*','/', '^'};
for (int i = 0; i < 4; i++)
{   
if (ch == ops[i])
count++;
}
return count;
}

并更改

char ops[] = {'^','-','+','*','/'};

至

char ops[] = {'-','+','*','/', '^'};

op可以设置为"/"，但不能设置为"^"。一定有什么简单的事情我没看到。

Dstack.h(堆栈类(:

#ifndef DSTACK_H
#define DSTACK_H
#include <iostream>
class Dstack
{
public:
Dstack();
~Dstack();
void push(double value);
bool pop(double &value);
double top();
int size();
bool empty();
bool print();
private:
class Node
{
public:
Node(double value, Node* next)
{m_value = value; m_next = next;}
double m_value;
Node* m_next;
};
Node* m_head;
};

#endif 

Dstack.cpp(堆栈函数定义(:

#include "dstack.h"
#include <iostream>
Dstack::Dstack()
{
m_head = NULL;
}
void Dstack::push(double value)
{
m_head = new Node (value, m_head);
}
bool Dstack::pop(double &value)
{
if (m_head == NULL)
return false;
value = m_head->m_value;
Node *temp = m_head;
m_head = m_head->m_next;
delete temp;
return true;
}
double Dstack::top()
{
double value = m_head->m_value;
return value;
}         
int Dstack::size()
{
int count = 0;
Node *ptr = m_head;
while (ptr != NULL)
{
count++; 
ptr = ptr->m_next;
}
return count;
}   
bool Dstack::empty()
{
if (m_head == NULL)
return true;
return false;
}
Dstack::~Dstack()
{
Node* ptr = m_head;
while (ptr != NULL)
{
Node* temp = ptr;
ptr = ptr->m_next;
delete temp;
} 
}
bool Dstack::print()
{
//if (m_head->m_next == NULL)
//return false;
std::cout << m_head->m_value << std::endl;
return true;
}

Calc.cpp(计算器功能(

#include "dstack.h"
#include <iostream>
#include <sstream>
#include <string>
#include <cmath>
#include <sstream> 
int isOperator(char ch) 
{ 
int count = 0;  
char ops[] = {'-','+','*','^','/'};
for (int i = 0; i < 4; i++) 
{   
if (ch == ops[i])
count++;
}   
return count;
}

int main()
{ 
Dstack stack;
double num;
double result = 0;
char op = '';
double a = 0;
double b = 0;

while (std::cin.peek() != EOF)
{   
std::cin >> std::ws;
while (isdigit(std::cin.peek()))
{     
std::cin >> num;
stack.push(num);
while(isspace (std::cin.peek()))
{   
std::cin.ignore();
std::cin.peek();
}   
}   
while (isOperator(std::cin.peek())) 
{
//make sure there is more than one operand to calculate
if (stack.size() <2)
{
std::cerr << "Error: Invalid Expression." << std::endl;
exit(1);
}
//make sure there are enough operators
if (isOperator(std::cin.peek() +1 < stack.size() ))
{
std::cerr << "Error: Invalid Expression." << std::endl;
exit(1);
}
//clear white space for each cycle
std::cin >> std::ws; 
//operate!
std::cin >> op;
if (op == '+')
{
b = stack.top();
stack.pop(b);
a = stack.top();
stack.pop(a);
result = a + b;
stack.push(result);
}
if (op == '-')
{
b = stack.top();
stack.pop(b);
a = stack.top();
stack.pop(a);
result = a - b;
stack.push(result);
}
if (op == '*')
{
b = stack.top();
stack.pop(b);
a = stack.top();
stack.pop(a);
result = a * b;
stack.push(result);
}
if (op == '^')
{
b = stack.top();
stack.pop(b);
a = stack.top();
stack.pop(a);
result = pow(a,b);
stack.push(result);
}
if (op == '/')
{
b = stack.top();
stack.pop(b);
a = stack.top();
stack.pop(a);
//b cant equal 0!!!
if (b == 0)
{
std::cerr << "Error: Invalid expression." << std::endl;
exit(1);
}
result = a / b;
stack.push(result);
}
std::cout << op << std::endl;
//move char to next position    
std::cin.peek();
//ignore possible white spaces left in expression and repeat
while (isspace (std::cin.peek()))
{
std::cin.ignore();
std::cin.peek();
}
}
}
if (stack.size() == 1)
std::cout << result << std::endl;
else
{
std::cerr << "Error: Invalid expression." << std::endl;
exit(1);
}
return 0;
}

Answer 1

对于您的问题，我建议对您的输入处理进行全面重构。

首先读取整行，然后逐个字符解析输入行。这样实际上更容易识别和处理各种运算符。

例如：从输入行获取下一个字符，并检查它是什么类型的字符：

• 对于空白，只需丢弃字符并继续下一个
• 如果它是一个数字，则在它是数字时提取字符，并根据该字符构造数字
• 如果它是一个有效的运算符，则处理它
• 如果是其他内容，则处理错误(例如跳过当前行并读取下一行(

正如您所看到的，输入的一种错误处理在某种程度上内置于上述方法中。

扩展以上内容也很容易识别符号，这些符号可用于变量或函数。

---

主循环在代码中可能看起来像这样：

while (std::getline(std::cin, line))
{
for (size_t current_char_index = 0; current_char_index < line-size(); ++current_char_index)
{
// TODO: Handle current character at line[current_char_index]
}
}

为了实用，我还建议将主代码拆分为函数。

最后，除非您的练习是关于创建自己的堆栈类，否则请使用std::stack。