我正试图用C++编写一个函数来计算后缀表示法方程。我的一般策略是扫描字符串（以正确的格式，例如"10 20+30-"）
我通过递增索引变量I来实现这一点。每次递增时，我都会检查字符是数字、运算符还是两者都不是。如果它是一个数字，我使用getNextNum（）函数来获取以下所有数字，将其转换为浮点值，然后将其推送到堆栈中。我还将I乘以捕获的数字的长度。如果这个字符是一个运算符，我会得到堆栈的前两个元素，进行运算，然后将结果推回到堆栈。
问题是，我的while循环似乎只通过一次。函数只返回字符串中的第一个数字。我不知道出了什么问题，如果有任何帮助，我将不胜感激！我在while循环中插入了cout语句，并且我只递增到第一个数字之后的索引。
编辑：好的，我添加了getNextNum（）函数。此外，我用strLength的cout更新了evalPostfix（），并在while循环的每次迭代后更新了I。当运行给定的代码时，我得到的是：
Running…
Please enter an expression in postfix notation: 555 666+
3
555
3
Your expression evaluates to: 555
strLength似乎被设置为小于它应该设置的值。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;
}
您有一些问题，其中一个主要问题是打字错误，您有一个放错地方的) this:
else if (OPS.find( postfix[i] != -1 ) ) {
^ ^
应该是：
else if (OPS.find( postfix[i] ) != std::string::npos) {
^ ^
所以您将位置i处的char与-1进行比较，然后对布尔结果执行find。接下来，您应该使用-1来比较find和std::string::npos 的结果
正如Jonathan所指出的：
cin >> postfix ;
只读到第一个黑色或换行符。使用getline可以解决这个问题：
if (getline(cin, postfix))
一个主要问题是输入cin >> postfix;语句只读取第一个单词。回显输入，以确保程序看到您认为它正在看到的内容。
Shafik Yaghmour指出了另一个问题。
学习要点：
- echo输入，以确保程序正在看到您认为它正在看到的内容
- 用合适的打印消息跟踪关键变量
- 后SSCCE（简短、自包含、正确示例）——可以编译的代码
- 张贴示例输入和您从中得到的输出
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;
}
样品痕迹：
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
很明显，一旦你正在解决的特定问题得到解决，你可能会失去很多诊断输出，但准备好按照所示的方式添加它可以大大加快解决过程。