我正在制作一个后缀计算器，我想为它添加一个新功能。如何对堆栈中的所有项求和，然后从堆栈中删除用于求和的项，然后将和推回到堆栈？所有这些都应该通过按下命令的来实现。
我试过这样做，但它总是只打印0。
case 's':
if (numbers.top().first == underflow)
cout << "Stack empty" << endl;
else {
numbers.pop();
if (numbers.top().first == underflow) {
cout << "Stack has just one entry" << endl;
numbers.push(p);
}
else {
while (!numbers.empty()) {
sum += numbers.top().second;
}
numbers.push(sum);
cout << sum << endl;
}
}
break;
这是我的主.cpp.的一些代码
#include <iostream>
#include "Stack.h"
#include <cmath>
using namespace std;
void help()
{
cout
<< "[?]push to stack [=]print top [x]exchange [s]sum [a]average" << endl
<< "[+] [-] [*] [/] [%] [^] [v] are arithmetic operations" << endl
<< "[Q]uit." << endl;
}
char get_command()
{
char command;
bool waiting = true;
cout
<< "Select command and press <Enter>:" << endl;
while (waiting) {
cin >> command;
command = tolower(command);
if (command == '?' || command == '=' || command == '+' ||
command == '-' || command == '*' || command == '/' ||
command == 'q' || command == 'x' || command == 's' ||
command == 'a' || command == '%' || command == '^' ||
command == 'v') waiting = false;
else {
cout << "Please enter a valid command:" << endl;
}
}
return command;
}
bool do_command(char command, Stack& numbers)
{
double p, q, g, r;
double sum = 0;
switch (command) {
case '?':
cout << "Enter a real number: " << flush;
cin >> p;
if (numbers.push(p) == overflow)
cout << "Warning: Stack full, lost number" << endl;
break;
case '=':
if (numbers.top().first == underflow)
cout << "Stack empty" << endl;
else
cout << p << endl;
break;
case '+':
if (numbers.top().first == underflow)
cout << "Stack empty" << endl;
else {
numbers.pop();
if (numbers.top().first == underflow) {
cout << "Stack has just one entry" << endl;
numbers.push(p);
}
else {
numbers.pop();
if (numbers.push(q + p) == overflow)
cout << "Warning: Stack full, lost result" << endl;
}
}
break;
case '-':
if (numbers.top().first == underflow)
cout << "Stack empty" << endl;
else {
numbers.pop();
if (numbers.top().first == underflow) {
cout << "Stack has just one entry" << endl;
numbers.push(p);
}
else {
numbers.pop();
if (numbers.push(q - p) == overflow)
cout << "Warning: Stack full, lost result" << endl;
}
}
break;
case '*':
if (numbers.top().first == underflow)
cout << "Stack empty" << endl;
else {
numbers.pop();
if (numbers.top().first == underflow) {
cout << "Stack has just one entry" << endl;
numbers.push(p);
}
else {
numbers.pop();
if (numbers.push(q * p) == overflow)
cout << "Warning: Stack full, lost result" << endl;
}
}
break;
case '/':
if (numbers.top().first == underflow)
cout << "Stack empty" << endl;
else {
numbers.pop();
if (numbers.top().first == underflow) {
cout << "Stack has just one entry" << endl;
numbers.push(p);
}
else {
numbers.pop();
if (numbers.push(q / p) == overflow)
cout << "Warning: Stack full, lost result" << endl;
}
}
break;
case 'x':
if (numbers.top().first == underflow)
cout << "Stack empty" << endl;
else {
numbers.pop();
if (numbers.top().first == underflow) {
cout << "Stack has just one entry" << endl;
numbers.push(p);
}
else {
numbers.pop();
numbers.pop();
numbers.push(p);
numbers.push(q);
}
}
break;
case 's':
if (numbers.top().first == underflow)
cout << "Stack empty" << endl;
else {
numbers.pop();
if (numbers.top().first == underflow) {
cout << "Stack has just one entry" << endl;
numbers.push(p);
}
else {
while (!numbers.empty()) {
sum += numbers.top().second;
}
numbers.push(sum);
cout << sum << endl;
}
}
break;
这是我的Stack.cpp.
Error_code Stack::push(const Stack_entry& item)
/*
Pre: None.
Post: If the Stack is not full, item is added to the top
of the Stack. If the Stack is full,
an Error_code of overflow is returned and the Stack is left unchanged.
*/
{
Error_code outcome = success;
if (count >= maxstack)
outcome = overflow;
else
entry[count++] = item;
return outcome;
}
Error_code Stack::pop()
/*
Pre: None.
Post: If the Stack is not empty, the top of
the Stack is removed. If the Stack
is empty, an Error_code of underflow is returned.
*/
{
Error_code outcome = success;
if (count == 0)
outcome = underflow;
else --count;
return outcome;
}
pair<Error_code, int> Stack::top() const
{
pair<Error_code, int> p;
Error_code outcome = success;
if (count == 0)
p.first = underflow;
p.second = 0;
return p;
p.first = outcome;
p.second = entry[count - 1];
return p;
}
这是我的Stack.h
#include "Utility.h"
typedef double Stack_entry;
const int maxstack = 10;
class Stack {
public:
Stack();
bool empty() const;
Error_code pop();
//Error_code top(Stack_entry& item) const;
Error_code push(const Stack_entry& item);
//Error_code pop_top(Stack& s, Stack_entry& item);
double size() const;
std::pair<Error_code, int> top() const;
private:
int count;
Stack_entry entry[maxstack];
};
我希望它将堆栈中的所有数字相加，然后删除它们，然后将其推回堆栈。如果有任何帮助，我将不胜感激。提前谢谢。
pop不像C++中的其他语言那样返回值。您需要使用top函数来返回堆栈中的顶部元素。您可以将该元素添加到sum中，然后使用pop将其移除。您可以实现如下：
while (!numbers.empty())
{
sum += numbers.top();
numbers.pop()
}
对于Op的另一个请求：
将top的实现更改为：
std::pair<Error_code,int> top() const
{
std::pair<Error_code,int> p;
Error_code outcome = success;
if (count == 0)
p.first = underflow;
p.second = 0;
return p;
p.first = outcome;
p.second = entry[count -1];
return p;
}
你的第一个问题的解决方案是：
while (!numbers.empty())
{
sum += numbers.top().second;
numbers.pop()
}
如果你想要outcome，你可以做numbers.top().first。仅此而已。