Question

我正在尝试用C ++编写一个函数来评估后缀符号方程。我的一般策略是扫描一个字符串（格式正确，例如“10 20 + 30 - ”）我通过递增索引变量i来做到这一点。在每个增量处，我检查字符是否为数字，运算符或两者都不是。如果它是一个数字，我使用getNextNum（）函数获取所有后续数字，将其转换为浮点数，然后将其推送到堆栈。我也按照捕获的数字的长度递增i。如果字符是运算符，我得到堆栈的前两个元素，执行操作，然后将结果推回堆栈。

麻烦的是，我的while循环似乎只经历了一次。该函数仅返回字符串中的第一个数字。我无法弄清楚什么是错的，我将不胜感激任何帮助！我在while循环中插入了cout语句，而我只是在第一个数字之后递增到索引。

编辑：好的，我添加了getNextNum（）函数。此外，我用一个strLength的cout更新了evalPostfix（），以及在while循环的每次迭代之后的i。运行给定的代码时，我得到了这个：

Running…
Please enter an expression in postfix notation: 555 666+
3
555
3
Your expression evaluates to: 555

似乎strLength被设置为低于它应该的值。为什么会这样？

#include <iostream>
#include <string>
#include <vector>
#include <deque>
#include <stack>
using namespace std;

string getNextNum(string equation, int i);
float evalPostfix(string postfix);
float doOperation(float x, float y, char op);

float doOperation(float x, float y, char op)
{
    switch (op) {
        case '+':
            return x + y;
        case '-':
            return x - y;
        case '*':
            return x * y;
        case '/':
            return x / y;
        default:
            return 0.0;
    }
}


string getNextNum(string equation, int i)
{
    string num = "";
    const string DELIM = "+-*/%^ ";
    while (i<equation.length()) {
        // Iterate through equation until you hit a delimiter.
        if (DELIM.find(equation[i]) != -1) {
            break;
        }
        num += equation[i];
        i++;
    }
    return num;
}

float evalPostfix(string postfix)
{
    const string OPS = "+-*/%^";
    const string NUMS = "0123456789";
    int strLength = postfix.length();
    stack<float> numStack;
    int i = 0;
    cout << strLength << endl;
    while (i<strLength) {
        if (NUMS.find(postfix[i]) != -1) {
            // If a character is a digit, then you should get the 
            // value and push it to the stack (could be multiple characters long).
            string sNextNum = getNextNum(postfix, i);
            float fNextNum = atof(sNextNum.c_str());

            numStack.push(fNextNum);
            cout << sNextNum << endl;
            i += (sNextNum.length() - 1);
            }
        else if (OPS.find(postfix[i] != -1)) {
            // Otherwise, pop the top two elements of the stack, perform the
            // operation, then push the result back to the stack.
            char op = postfix[i];

            float x = numStack.top();
            numStack.pop();

            float y = numStack.top();
            numStack.pop();

            float z = doOperation(x, y, op);
            numStack.push(z);
            }
        i++;
        cout << i << endl;
    };

    // Once the entire string has been scanned through, there should be a float
    // left in the stack, simply return that.
    return numStack.top();
}

int main ()
{
    cout << "Please enter an expression in postfix notation: ";
    string postfix;
    cin >> postfix;

    float eval = evalPostfix(postfix);
    cout << "Your expression evaluates to: " << eval;
    return 0;
}

Answer 1

你有一些问题，其中一个主要问题是拼写错误，你错了)这个：

else if (OPS.find( postfix[i] != -1 ) ) {
                 ^                  ^

应该是：

else if (OPS.find( postfix[i] ) != std::string::npos) {
                 ^            ^

因此，您要将位置char的{{1}}与i进行比较，然后对布尔结果执行-1。接下来，您应该使用find来比较find的结果，但std::string::npos

正如乔纳森指出的那样：

-1

仅读取第一个黑色或换行符。使用cin >> postfix ;将解决该问题：

getline

Answer 2

一个主要问题是输入cin >> postfix;语句只读取第一个单词。回声输入以确保程序看到您认为它看到的内容。

Shafik Yaghmour指出另一个problem。

要学习的内容：

回显输入以确保程序看到您认为它看到的内容;
使用合适的打印消息跟踪关键变量;
发布SSCCE（Short, Self-Contained, Correct Example） - 可以编译的代码;
发布示例输入和您从中获取的输出。

此代码适用于输入555 666+：

#include <iostream>
#include <string>
#include <stack>
using namespace std;

static float doOperation(float x, float y, char op)
{
    cout << "doOp: x = " << x << ", y = " << y << ", op = " << op << endl;
    if (op == '+')
        x += y;
    return x;
}

string getNextNum(string equation, int i)
{
    string num = "";
    const string DELIM = "+-*/%^ ";
    while (i<equation.length()) {
        // Iterate through equation until you hit a delimiter.
        if (DELIM.find(equation[i]) != -1) {
            break;
        }
        num += equation[i];
        i++;
    }
    return num;
}

float evalPostfix(string postfix)
{
    const string OPS = "+-*/%^";
    const string NUMS = "0123456789";
    int strLength = postfix.length();
    stack<float> numStack;
    int i = 0;
    while (i<strLength) {
        cout << "Top - i: " << i << ", strLength: " << strLength << endl;
        if (NUMS.find(postfix[i]) != -1) {
            // If a character is a digit, then you should get the 
            // value and push it to the stack (could be multiple characters long).
            string sNextNum = getNextNum(postfix, i);
            float fNextNum = atof(sNextNum.c_str());

            numStack.push(fNextNum);
            cout << sNextNum << endl;
            i += (sNextNum.length() - 1);
            }
        else if (OPS.find(postfix[i])!= -1) {
            // Otherwise, pop the top two elements of the stack, perform the
            // operation, then push the result back to the stack.
            char op = postfix[i];

            float x = numStack.top();
            numStack.pop();

            float y = numStack.top();
            numStack.pop();

            float z = doOperation(x, y, op);
            numStack.push(z);
            }
        i++;
        cout << "End - i: " << i << ", strLength: " << strLength << endl;
    }
    cout << "After - i: " << i << ", strLength: " << strLength << endl;

    // Once the entire string has been scanned through, there should be a float
    // left in the stack, simply return that.
    return numStack.top();
}

int main ()
{
    cout << "Please enter an expression in postfix notation: ";
    string postfix;
    //cin >> postfix;
    if (getline(cin, postfix))
    {
        cout << "Evaluating: " << postfix << endl;
        float eval = evalPostfix(postfix);
        cout << "Your expression evaluates to: " << eval << endl;
    }
    return 0;
}

示例跟踪：

Please enter an expression in postfix notation: 555 666+
Evaluating: 555 666+
Top - i: 0, strLength: 8
555
End - i: 3, strLength: 8
Top - i: 3, strLength: 8
End - i: 4, strLength: 8
Top - i: 4, strLength: 8
666
End - i: 7, strLength: 8
Top - i: 7, strLength: 8
doOp: x = 666, y = 555, op = +
End - i: 8, strLength: 8
After - i: 8, strLength: 8
Your expression evaluates to: 1221

显然，一旦您解决的具体问题得到解决，您可能会丢失大部分诊断输出，但是准备按照显示的方式添加它可以大大加快解决它的过程。

while循环在一次迭代后终止

2 个答案: