c#计算器，支持+-*/()的混合运算，对负号不支持.
原理:将点击的按钮赋给inputStr，然后对inputStr分析、计算。看懂代码的重要前提是你有一定的算法基础（前缀、中缀、后缀表达式）和数据结构基础（链表、栈）。
当初写这份代码的时候还是小白，随着对C#的深入学习，个人觉得本文代码还有很多可以优化的地方。
wybing最后编辑于20190510
截图：
运行截图:
界面控制代码：
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using System;
-
using System.Collections.Generic;
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using System.ComponentModel;
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using System.Data;
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using System.Drawing;
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using System.Linq;
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using System.Text;
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using System.Threading.Tasks;
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using System.Windows.Forms;
-
-
namespace 计算器
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{
-
public
partial
class
FrmMain :
Form
-
{
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public static List<char> inputStr=
new List<
char>(
1000);
//用户的输入
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public FrmMain()
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{
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InitializeComponent();
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}
-
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private void Btn0_Click(object sender, EventArgs e)
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{
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inputStr.Add(
'0');
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textBox1.AppendText(
"0");
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}
-
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private void BtnDot_Click(object sender, EventArgs e)
-
{
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inputStr.Add(
'.');
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textBox1.AppendText(
".");
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}
-
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private void Btn1_Click(object sender, EventArgs e)
-
{
-
inputStr.Add(
'1');
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textBox1.AppendText(
"1");
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}
-
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private void Btn2_Click(object sender, EventArgs e)
-
{
-
inputStr.Add(
'2');
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textBox1.AppendText(
"2");
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}
-
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private void Btn3_Click(object sender, EventArgs e)
-
{
-
inputStr.Add(
'3');
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textBox1.AppendText(
"3");
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}
-
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private void Btn4_Click(object sender, EventArgs e)
-
{
-
inputStr.Add(
'4');
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textBox1.AppendText(
"4");
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}
-
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private void Btn5_Click(object sender, EventArgs e)
-
{
-
inputStr.Add(
'5');
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textBox1.AppendText(
"5");
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}
-
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private void Btn6_Click(object sender, EventArgs e)
-
{
-
inputStr.Add(
'6');
-
textBox1.AppendText(
"6");
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}
-
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private void Btn7_Click(object sender, EventArgs e)
-
{
-
inputStr.Add(
'7');
-
textBox1.AppendText(
"7");
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}
-
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private void Btn8_Click(object sender, EventArgs e)
-
{
-
inputStr.Add(
'8');
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textBox1.AppendText(
"8");
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}
-
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private void Btn9_Click(object sender, EventArgs e)
-
{
-
inputStr.Add(
'9');
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textBox1.AppendText(
"9");
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}
-
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private void BtnLeft_Click(object sender, EventArgs e)
-
{
-
inputStr.Add(
'(');
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textBox1.AppendText(
"(");
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}
-
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private void BtnRig_Click(object sender, EventArgs e)
-
{
-
inputStr.Add(
')');
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textBox1.AppendText(
")");
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}
-
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private void BtnEqual_Click(object sender, EventArgs e)
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{
-
//等号代码
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textBox1.AppendText(
"=");
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textBox2.Text = textBox1.Text;
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textBox1.Text = DataOp.DataMain();
-
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string temp= DataOp.DataMain();
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inputStr.Clear();
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for(
int i =
0; i < temp.Length; i++)
-
{
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inputStr.Add(temp[i]);
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}
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}
-
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private void BtnAdd_Click(object sender, EventArgs e)
-
{
-
inputStr.Add(
'+');
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textBox1.AppendText(
"+");
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}
-
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private void BtnSubt_Click(object sender, EventArgs e)
-
{
-
inputStr.Add(
'-');
-
textBox1.AppendText(
"-");
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}
-
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private void BtnMul_Click(object sender, EventArgs e)
-
{
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inputStr.Add(
'*');
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textBox1.AppendText(
"*");
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}
-
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private void BtnDivi_Click(object sender, EventArgs e)
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{
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inputStr.Add(
'/');
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textBox1.AppendText(
"/");
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}
-
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private void BtnCe_Click(object sender, EventArgs e)
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{
-
textBox1.Text =
"";
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inputStr.Clear();
//清空链表的所有元素
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}
-
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private void BtnC_Click(object sender, EventArgs e)
-
{
-
//界面撤销
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inputStr.RemoveAt(inputStr.Count
-1);
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textBox1.Text =
"";
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for (
int i=
0;i<inputStr.Count;i++)
-
{
-
textBox1.Text += inputStr[i];
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}
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}
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}
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}
数据操作代码:
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using System;
-
using System.Collections.Generic;
-
using System.Linq;
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using System.Text;
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using System.Threading.Tasks;
-
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namespace 计算器
-
{
-
//表达式存在inputStr中
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class
DataOp :
FrmMain
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{
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static Stack<double> m =
new Stack<
double>();
//数字栈
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static Stack<char> s =
new Stack<
char>();
//符号栈
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public static void Read() //Read()从inputStr输入流中读值
-
{
-
for (
int i =
0; i < inputStr.Count; i++)
-
{
-
if (!IsOperator(inputStr[i]))
//数字和小数点
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{
-
string s =
null;
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while (i < inputStr.Count && !IsOperator(inputStr[i]))
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{
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s += inputStr[i];
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i++;
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}
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i--;
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double mm = Convert.ToDouble(s);
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m.Push(mm);
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}
-
else
if (IsOper(inputStr[i]))
//+ - * /
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{
-
if (s.Count.Equals(
0) || s.Peek().Equals(
'('))
-
{
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s.Push(inputStr[i]);
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}
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else
if (OperatorPrecedence(inputStr[i]) > OperatorPrecedence(s.Peek()))
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{
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s.Push(inputStr[i]);
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}
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else
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{
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double n1, n2;
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char s1;
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n2 = m.Pop();
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n1 = m.Pop();
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s1 = s.Pop();
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double sum = Operat(n1, n2, s1);
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m.Push(sum);
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s.Push(inputStr[i]);
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}
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}
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else
//（和）
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{
-
if (inputStr[i].Equals(
'('))
-
{
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s.Push(inputStr[i]);
-
}
-
else
if (inputStr[i].Equals(
')'))
-
{
-
while (!s.Peek().Equals(
'('))
-
{
-
double n1, n2;
-
char s1;
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n2 = m.Pop();
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n1 = m.Pop();
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s1 = s.Pop();
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double sum = Operat(n1, n2, s1);
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m.Push(sum);
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}
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s.Pop();
-
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}
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}
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}
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}
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public static double PopStack()
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{
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double sum =
0;
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while (s.Count !=
0)
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{
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double n1, n2;
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char s1;
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n2 = m.Pop();
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n1 = m.Pop();
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s1 = s.Pop();
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sum = Operat(n1, n2, s1);
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m.Push(sum);
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}
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return sum;
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}
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public static bool IsOperator(char c) //是否是操作符
-
{
-
if (c.Equals(
'+') || c.Equals(
'-') || c.Equals(
'*') || c.Equals(
'/') || c.Equals(
'(') || c.Equals(
')'))
-
return
true;
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return
false;
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}
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public static bool IsOper(char c) //是否是运算符符
-
{
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if (c.Equals(
'+') || c.Equals(
'-') || c.Equals(
'*') || c.Equals(
'/'))
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return
true;
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return
false;
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}
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public static int OperatorPrecedence(char a) //操作符优先级
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{
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int i =
0;
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switch (a)
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{
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case
'+': i =
3;
break;
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case
'-': i =
3;
break;
-
case
'*': i =
4;
break;
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case
'/': i =
4;
break;
-
}
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return i;
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}
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public static double Operat(double n1, double n2, char s1)
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{
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double sum =
0;
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switch (s1)
-
{
-
case
'+': sum = n1 + n2;
break;
-
case
'-': sum = n1 - n2;
break;
-
case
'*': sum = n1 * n2;
break;
-
case
'/': sum = n1 / n2;
break;
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}
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return sum;
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}
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public static string DataMain()
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{
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Read();
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return PopStack().ToString();
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}
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}
-
}
程序源代码下载地址：https://wybing.lanzoui.com/is107ijyrha
【Bug修复】
感谢Xuuuton反馈的Bug，3*2-2*2+1出来的答案是1 。
20201119修改如下：
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public static double PopStack()
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{
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/*
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* 修复Bug 3*2-2*2+1结果为1
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* 将两个栈逆序后再运算
-
*/
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Stack<
double> tepM =
new Stack<
double>();
//数字栈
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Stack<
char> tepS =
new Stack<
char>();
//符号栈
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while (s.Count!=
0)
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{
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char temp;
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temp = s.Pop();
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tepS.Push(temp);
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}
-
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while (m.Count !=
0)
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{
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double temp;
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temp = m.Pop();
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tepM.Push(temp);
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}
-
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double sum =
0;
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while (tepS.Count !=
0)
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{
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double n1, n2;
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char s1;
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n2 = tepM.Pop();
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n1 = tepM.Pop();
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s1 = tepS.Pop();
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sum = Operat(n2, n1, s1);
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tepM.Push(sum);
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}
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return sum;
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}
关于非法表达式导致的程序异常，可以做异常处理，也可以简单判断数字栈和符号栈的Count，很显然，数字栈至少要两位以上数字，符号栈至少要一位以上，但前提要确保栈内操作数和操作符合法。
转载：https://blog.csdn.net/qq_41138935/article/details/82750974
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