c#计算器奇怪的现像

Question

主要方法 
       private void btnCalculate_Click(object sender, EventArgs e)
        {
            if (this.txtFirst.Text==""||this.txtSecond.Text==""||this.cboOperator.Text=="")
            {
                MessageBox.Show("操作数或运算符不能为空","操作提示",MessageBoxButtons.OK,MessageBoxIcon.Warning);
                return;
            }
            double firstNum = double.Parse(this.txtFirst.Text);
            double secondNum = double.Parse(this.txtSecond.Text);
            string operate = this.cboOperator.Text;
            if (secondNum == 0 && operate == "/")
            {
                MessageBox.Show("除数不能为0，请重新输入！", "操作提示", MessageBoxButtons.OK, MessageBoxIcon.Information);
            }
            else
            {
                this.txtResult.Text = Calculate(firstNum, secondNum, operate).ToString();
            }
        }        public double Calculate(double firstNum, double secondNum, string operate)
        {
            if (operate == "+")
            {
                return firstNum + secondNum;
            }
            else if (operate == "-")
            {
                return firstNum - secondNum;
            }
            else if (operate == "*")
            {
                return firstNum * secondNum;
            }
            else
            {
                return firstNum / secondNum;
            }
        }
当输入5.05 - 5 时结果是0.0499999999999998 ，而不是5.00感兴趣的朋友也可以到 http://download.csdn.net/source/2516203 下载来看看，谢谢大家

Answer 1

把 double 改为 deciaml 就没问题了。

Answer 2

        private void btnCalculate_Click(object sender, EventArgs e)
        {
            if (this.txtFirst.Text==""||this.txtSecond.Text==""||this.cboOperator.Text=="")
            {
                MessageBox.Show("操作数或运算符不能为空","操作提示",MessageBoxButtons.OK,MessageBoxIcon.Warning);
                return;
            }
            decimal firstNum = decimal.Parse(this.txtFirst.Text);
            decimal secondNum = decimal.Parse(this.txtSecond.Text);
            string operate = this.cboOperator.Text;
            if (secondNum == 0 && operate == "/")
            {
                MessageBox.Show("除数不能为0，请重新输入！", "操作提示", MessageBoxButtons.OK, MessageBoxIcon.Information);
            }
            else
            {
                this.txtResult.Text = Calculate(firstNum, secondNum, operate).ToString();
            }
        }        public decimal Calculate(decimal firstNum, decimal secondNum, string operate)
        {
            if (operate == "+")
            {
                return firstNum + secondNum;
            }
            else if (operate == "-")
            {
                return firstNum - secondNum;
            }
            else if (operate == "*")
            {
                return firstNum * secondNum;
            }
            else
            {
                return firstNum / secondNum;
            }
        }

Answer 3

double 类型精度问题
就如比较double类型是否为0一样，要允许一定的误差
if(Math.Abs(dblValue)<0.000001)
{
   MessageBox.Show("dblValue=0");
}
else
{
   MessageBox.Show("dblValue!=0");}

Answer 4

写错了，是0.05，谢谢大家，我试下，不过double类型为什么出现这种情况呢

Answer 5

int,float ,double...二进制存储的，除了2的幂外进行十进制的小数运算就会有精度丢失，