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;
}
答案 0 :(得分：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答案 1 :(得分：0)
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：
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。