我的C#计算器不会计算超过2个连续数字
我的C#计算器不会计算超过2个连续数字
提问于 2018-04-06 15:23:32
我最近刚刚完成了我的第一个Windows窗体计算器，它工作得很好，但我的问题是-如何在命中等号之前将第三个甚至第四个数字添加到计算中？我一直在尝试将ThirdNumber变量放在代码中的许多不同位置，如下所示--
private void bequal_Click(object sender, EventArgs e)
{
double SecondNumber;
double ThirdNumber;
double Result;
SecondNumber = Convert.ToDouble(textBox1.Text);
ThirdNumber = Convert.ToDouble(textBox1.Text);
if (Operation == "+")
{
Result = (FirstNumber + SecondNumber) + ThirdNumber;
textBox1.Text = Convert.ToString(Result);
FirstNumber = Result;
}然而，每次我得到的答案都是错误的(例如：6+6+8= 22，而我知道应该是20)。请告诉我如何修复这个缺陷的正确方向，并请记住，我还是一个初学者，所以任何建议和信息都将不胜感激。
谢谢
我应用程序的其余代码--
public partial class Form1 : Form
{
double FirstNumber;
string Operation;
public Form1()
{
InitializeComponent();
}
private void n1_Click(object sender, EventArgs e)
{
if (textBox1.Text == "0" && textBox1.Text != null) // If the textbox value is 0 or is not null then
{ // the number that corresponds to the button that is clicked
textBox1.Text = "1"; // will be subsituted in
}
else
{ // Else add the button value to the existing text box value
textBox1.Text = textBox1.Text + "1";
}
}
private void n2_Click(object sender, EventArgs e)
{
if(textBox1.Text == "0" && textBox1.Text != null)
{
textBox1.Text = "2";
}
else
{
textBox1.Text = textBox1.Text + "2";
}
}
private void n3_Click(object sender, EventArgs e)
{
if(textBox1.Text == "0" && textBox1.Text != null)
{
textBox1.Text = "3";
}
else
{
textBox1.Text = textBox1.Text + "3";
}
}
private void n4_Click(object sender, EventArgs e)
{
if(textBox1.Text == "0" && textBox1.Text != null)
{
textBox1.Text = "4";
}
else
{
textBox1.Text = textBox1.Text + "4";
}
}
private void n5_Click(object sender, EventArgs e)
{
if(textBox1.Text == "0" && textBox1.Text != null)
{
textBox1.Text = "5";
}
else
{
textBox1.Text = textBox1.Text + "5";
}
}
private void n6_Click(object sender, EventArgs e)
{
if(textBox1.Text == "0" && textBox1.Text != null)
{
textBox1.Text = "6";
}
else
{
textBox1.Text = textBox1.Text + "6";
}
}
private void n7_Click(object sender, EventArgs e)
{
if(textBox1.Text == "0" && textBox1.Text != null)
{
textBox1.Text = "7";
}
else
{
textBox1.Text = textBox1.Text + "7";
}
}
private void n8_Click(object sender, EventArgs e)
{
if(textBox1.Text == "0" && textBox1.Text != null)
{
textBox1.Text = "8";
}
else
{
textBox1.Text = textBox1.Text + "8";
}
}
private void n9_Click(object sender, EventArgs e)
{
if(textBox1.Text == "0" && textBox1.Text != null)
{
textBox1.Text = "9";
}
else
{
textBox1.Text = textBox1.Text + "9";
}
}
private void n0_Click(object sender, EventArgs e)
{
textBox1.Text = textBox1.Text + "0";
}
private void bad_Click(object sender, EventArgs e)
{
FirstNumber = Convert.ToDouble(textBox1.Text);
textBox1.Text = "0";
Operation = "+";
}
private void sub_Click(object sender, EventArgs e)
{
FirstNumber = Convert.ToDouble(textBox1.Text);
textBox1.Text = "0";
Operation = "-";
}
private void bmul_Click(object sender, EventArgs e)
{
FirstNumber = Convert.ToDouble(textBox1.Text);
textBox1.Text = "0";
Operation = "*";
}
private void bdiv_Click(object sender, EventArgs e)
{
FirstNumber = Convert.ToDouble(textBox1.Text);
textBox1.Text = "0";
Operation = "/";
}
private void bc_Click(object sender, EventArgs e)
{
textBox1.Text = "0";
}
private void bdot_Click(object sender, EventArgs e)
{
textBox1.Text = textBox1.Text + ",";
}
private void bequal_Click(object sender, EventArgs e)
{
double SecondNumber;
double ThirdNumber;
double Result;
SecondNumber = Convert.ToDouble(textBox1.Text);
ThirdNumber = Convert.ToDouble(textBox1.Text);
if (Operation == "+")
{
Result = (FirstNumber + SecondNumber) + ThirdNumber;
textBox1.Text = Convert.ToString(Result);
FirstNumber = Result;
}
else if (Operation == "-")
{
Result = (FirstNumber - SecondNumber);
textBox1.Text = Convert.ToString(Result);
FirstNumber = Result;
}
else if (Operation == "*")
{
Result = (FirstNumber * SecondNumber);
textBox1.Text = Convert.ToString(Result);
FirstNumber = Result;
}
else if (Operation == "/")
{
if (SecondNumber == 0)
{
textBox1.Text = "Cannot divide by zero";
}
else
{
Result = (FirstNumber / SecondNumber);
textBox1.Text = Convert.ToString(Result);
FirstNumber = Result;
}
}
}
}回答 2
Stack Overflow用户
发布于 2018-04-06 15:53:10
你得到6+6+8= 22，因为你基本上做的是：
- input 6，按+
- 6存储为
FirstNumber - input另外6，按+
-
FirstNumber被新的6覆盖。 - input 8，按=H212H113检索D14-即6-并将8加两次。
F216
另一种方法是创建一个既存储数字又存储相关操作的类型。
因此，工作流程如下所示：
在事件输入6中，按+.
- information，就像
{ op: multiply, value: 6 }存储在'events'.
- keep 的集合中一样。
- 在按
=时，迭代这些事件的集合并计算结果。
Stack Overflow用户
发布于 2018-04-06 15:42:51
private void bdiv_Click(object sender, EventArgs e)
{
FirstNumber = Convert.ToDouble(textBox1.Text);
textBox1.Text = "0";
Operation = "/";
}每次你按下'+‘键，第一个数字就会被重置，它的旧值就会丢失。
SecondNumber = Convert.ToDouble(textBox1.Text);
ThirdNumber = Convert.ToDouble(textBox1.Text);第二个和第三个数字总是相同的。
6+6+8在你的计算器中是22，因为它实际计算的是6+8+8，在第二次加法操作中丢失了一个6。和8是重复的，因为第二个和第三个数字都等于textBox1.Text。
解决方案：
使用+=操作
private void bdiv_Click(object sender, EventArgs e)
{
FirstNumber += Convert.ToDouble(textBox1.Text);
textBox1.Text = "0";
Operation = "/";
}并删除SecondNumber变量。
SecondNumber = Convert.ToDouble(textBox1.Text); // remove this line
ThirdNumber = Convert.ToDouble(textBox1.Text);不需要第二个数字，因为它现在在FirstNumber变量中。
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/49687507
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