C++ RPN计算器c++;错误处理和多个运算符
编辑:: 我已经对我的程序进行了更改。看看calc.cpp。特别是等运算符函数和第四个while循环 我认为问题在于char op没有设置为“/”。当我回到等运算符函数时C++ RPN计算器c++;错误处理和多个运算符,c++,linked-list,stack,calculator,rpn,C++,Linked List,Stack,Calculator,Rpn,编辑:: 我已经对我的程序进行了更改。看看calc.cpp。特别是等运算符函数和第四个while循环 我认为问题在于char op没有设置为“/”。当我回到等运算符函数时 int isOperator(char ch) { int count = 0; char ops[] = {'-','+','*','/', '^'}; for (int i = 0; i < 4; i++) { if (ch == ops[i]) cou
int isOperator(char ch)
{
int count = 0;
char ops[] = {'-','+','*','/', '^'};
for (int i = 0; i < 4; i++)
{
if (ch == ops[i])
count++;
}
return count;
}
到
op可以设置为“/”,但不能设置为“^”。一定有什么简单的东西我没看到
Dstack.h(堆栈类):
#ifndef DSTACK#H
#定义DSTACK_H
#包括
类数据堆栈
{
公众:
Dstack();
~Dstack();
无效推力(双值);
bool pop(双倍值);
双头();
int size();
bool empty();
bool print();
私人:
类节点
{
公众:
节点(双值,节点*下一个)
{m_value=value;m_next=next;}
双m_值;
Node*m_next;
};
节点*m_头;
};
#恩迪夫
Dstack.cpp(堆栈函数定义):
#包括“dstack.h”
#包括
Dstack::Dstack()
{
m_头=空;
}
void Dstack::push(双值)
{
m_头=新节点(值,m_头);
}
booldstack::pop(双倍值(&value)
{
如果(m_head==NULL)
返回false;
值=m_头->m_值;
节点*温度=m_头;
m_head=m_head->m_next;
删除临时文件;
返回true;
}
double-Dstack::top()
{
双值=m_头->m_值;
返回值;
}
int-Dstack::size()
{
整数计数=0;
节点*ptr=m_头;
while(ptr!=NULL)
{
计数++;
ptr=ptr->m_next;
}
返回计数;
}
booldstack::empty()
{
如果(m_head==NULL)
返回true;
返回false;
}
Dstack::~Dstack()
{
节点*ptr=m_头;
while(ptr!=NULL)
{
节点*temp=ptr;
ptr=ptr->m_next;
删除临时文件;
}
}
booldstack::print()
{
//如果(m_head->m_next==NULL)
//返回false;
标准::cout m_值>标准::ws;
while(isdigit(std::cin.peek())
{
std::cin>>num;
stack.push(num);
while(isspace(std::cin.peek())
{
std::cin.ignore();
std::cin.peek();
}
}
while(等运算符(std::cin.peek())
{
//确保有多个操作数要计算
如果(stack.size()解决您的问题,我建议您对输入处理进行全面重构
首先是to,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,只需丢弃角色并继续下一个
- 如果它是一个数字,则在它是一个数字时提取字符,并根据该数字构造数字
- 如果它是一个有效的操作符,那么处理它
- 如果是其他错误,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是上述方法中内置的
还可以很容易地扩展上述内容,以识别可用于变量或函数的符号
主循环在代码中可能类似于以下内容:
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]
}
}
while(std::getline(std::cin,line))
{
对于(大小为当前字符索引=0;当前字符索引
为了实用,我还建议将主代码拆分为函数
最后,除非你的练习是关于创建自己的堆栈类,否则首先使用。这不是一个。请,我也推荐。@Someprogrammerdude谢谢你告诉我正确的例子。我一直在努力,现在有一个更具体的问题。
char ops[] = {'^','-','+','*','/'};
char ops[] = {'-','+','*','/', '^'};
#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
#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;
}
#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 = '\0';
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;
}
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]
}
}