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分享关于C# 加法的问题，想写一个方法，这个方法既可以执行double类型的相加，也可以执行int类型的相加等等
不知如何编程，好像要用到泛型还是什么别的知识，自己实在写不出来，谢谢
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jx315425246 2019-06-02
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基本概念，叫重载，基本操作叫效率 我就两个字符串 “1” + “2” =“12” 吗？
泡泡龙 2019-05-31
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C#是强类型语言，你只能用泛型或者装箱来实现这玩意
得鹿梦鱼、 2019-05-31
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Console.WriteLine(Get_Add(double.Parse("1.23"), double.Parse("2.32")));
Console.WriteLine(Get_Add(int.Parse("1"), int.Parse("2")));
Console.WriteLine(Get_Add("1","2"));
public dynamic Get_Add(dynamic d , dynamic a)
{
return a + d;
}
华芸智森 2019-05-31
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'//十几年前从一本NET的书上抄一来的,一直用到现在.
例如: MsgBox("1+2+3/5*pi+log(e)+max(1,2,3,4,5,6)".Evaluate)
#Region "字符串表达式计算"
Const Num As String = "(\-?\d+\.?\d*)"
Const Func1 As String = "(exp|log|log10|abs|sqr|sqrt|sin|cos|tan|asin|acos|atan|atan|round|tanh|truncate|floor|ceiling)"
Const Func2 As String = "(atan2)"
Const FuncN As String = "(min|max)"
Const Constants As String = "(e|pi)"
Private rePower As New Regex(String.Concat(Num, "\s*(\^)\s*", Num))
Private reAddSub As New Regex(String.Concat(Num, "\s*([-+])\s*", Num))
Private reMulDiv As New Regex(String.Concat(Num, "\s*([*/])\s*", Num))
Private reFunc1 As New Regex(String.Concat(Func1, "\(\s*", Num, "\s*\)"), RegexOptions.IgnoreCase)
Private reFunc2 As New Regex(String.Concat(Func2, "\(\s*", Num, "\s*,\s*", Num, "\s*\)"), RegexOptions.IgnoreCase)
Private reFuncN As New Regex(String.Concat(FuncN, "\((\s*", Num, "\s*,)+\s*", Num, "\s*\)"), RegexOptions.IgnoreCase)
Private reSign1 As New Regex("([-+/*^])\s*\+")
Private reSign2 As New Regex("\-\s*\-")
Private rePar As New Regex("(?<![A-Za-z0-9])\(\s*([-+]?\d+.?\d*)\s*\)")
Private reNum As New Regex("^\s*[-+]?\d+\.?\d*\s*$")
Private reConst As New Regex(String.Concat("\s*", Constants, "\s*"), RegexOptions.IgnoreCase)
''' <summary>
''' 字符串表达式计算.如:pi*(100+23)*1.2/35
''' 注:这个函数使用的是正则,所以速度是呵呵的,只能在前台界面.
''' (exp|log|log10|abs|sqr|sqrt|sin|cos|tan|asin|acos|atan|atan|round|tanh|truncate|floor|ceiling)
''' (atan2)
''' (min|max)
''' (e|pi)
''' 支持 括号,四则运算,EXP,LOG,LOG10,SQR,SQRT,三角函数.MIN,MAX,e,pi
''' ** 注：还没有测试线程安全，所以，先不要用在服务器上。
''' </summary>
''' <param name="expr"></param>
''' <returns></returns>
''' <remarks></remarks>
<Extension()> Public Function Evaluate(ByVal expr As String) As Double
If String.IsNullOrWhiteSpace(expr) Then Return 0
expr = reConst.Replace(expr, AddressOf DoConstants)
Do Until reNum.IsMatch(expr)
Dim saveExpr As String = expr
Do While rePower.IsMatch(expr)
expr = rePower.Replace(expr, AddressOf DoPower)
Loop
Do While reMulDiv.IsMatch(expr)
expr = reMulDiv.Replace(expr, AddressOf DoMulDiv)
Loop
Do While reFuncN.IsMatch(expr)
expr = reFuncN.Replace(expr, AddressOf DoFuncN)
Loop
Do While reFunc2.IsMatch(expr)
expr = reFunc2.Replace(expr, AddressOf DoFunc2)
Loop
Do While reFunc1.IsMatch(expr)
expr = reFunc1.Replace(expr, AddressOf DoFunc1)
Loop
expr = reSign1.Replace(expr, "$1")
expr = reSign2.Replace(expr, "+")
Do While reAddSub.IsMatch(expr)
expr = reAddSub.Replace(expr, AddressOf DoAddSub)
Loop
expr = rePar.Replace(expr, "$1")
'//表达式错误
If expr = saveExpr Then
'//Throw New SyntaxErrorException()
Return 0
End If
Loop
Return Val(expr)
End Function
Friend Function DoConstants(ByVal m As Match) As String
Dim V As String = String.Empty
Select Case m.Groups(1).Value.ToUpper
Case "PI"
V = Math.PI.ToString
Case "E"
V = Math.E.ToString
End Select
Return V
End Function
Friend Function DoPower(ByVal m As Match) As String
Dim n1 As Double = Val(m.Groups(1).Value)
Dim n2 As Double = Val(m.Groups(3).Value)
Return (n1 ^ n2).ToString
End Function
Friend Function DoMulDiv(ByVal m As Match) As String
Dim n1 As Double = Val(m.Groups(1).Value)
Dim n2 As Double = Val(m.Groups(3).Value)
Dim V As String = String.Empty
Select Case m.Groups(2).Value
Case "/"
V = (n1 / n2).ToString
Case "*"
V = (n1 * n2).ToString
End Select
Return V
End Function
Friend Function DoAddSub(ByVal m As Match) As String
Dim n1 As Double = Val(m.Groups(1).Value)
Dim n2 As Double = Val(m.Groups(3).Value)
Dim V As String = String.Empty
Select Case m.Groups(2).Value
Case "+"
V = (n1 + n2).ToString
Case "-"
V = (n1 - n2).ToString
End Select
Return V
End Function
Friend Function DoFunc1(ByVal m As Match) As String
Dim V As String = String.Empty
Dim n1 As Double = Val(m.Groups(2).Value)
Select Case m.Groups(1).Value.ToUpper
Case "EXP"
V = Math.Exp(n1).ToString
Case "LOG"
V = Math.Log(n1).ToString
Case "LOG10"
V = Math.Log10(n1).ToString
Case "ABS"
V = Math.Abs(n1).ToString
Case "SQR", "SQRT"
V = Math.Sqrt(n1).ToString
Case "SIN"
V = Math.Sin(n1).ToString
Case "COS"
V = Math.Cos(n1).ToString
Case "TAN"
V = Math.Tan(n1).ToString
Case "ASIN"
V = Math.Asin(n1).ToString
Case "ACOS"
V = Math.Acos(n1).ToString
Case "ATAN"
V = Math.Atan(n1).ToString
Case "ROUND"
V = Math.Round(n1).ToString
Case "TANH"
V = Math.Tanh(n1).ToString
Case "Truncate".ToUpper
V = Math.Truncate(n1).ToString
Case "Floor".ToUpper
V = Math.Floor(n1).ToString
Case "Ceiling".ToUpper
V = Math.Ceiling(n1).ToString
End Select
Return V
End Function
Friend Function DoFunc2(ByVal m As Match) As String
Dim V As String = String.Empty
Dim n1 As Double = Val(m.Groups(2).Value)
Dim n2 As Double = Val(m.Groups(3).Value)
Select Case m.Groups(1).Value.ToUpper
Case "ATAN2"
V = Math.Atan2(n1, n2).ToString
End Select
Return V
End Function
Friend Function DoFuncN(ByVal m As Match) As String
Dim V As String = String.Empty
Dim args As New ArrayList()
Dim i As Integer = 2
Do While m.Groups(i).Value <> String.Empty
args.Add(StrToDec(m.Groups(i).Value.Replace(","c, " "c)))
i += 1
Loop
Select Case m.Groups(1).Value.ToUpper
Case "MIN"
args.Sort()
V = args(0).ToString
Case "MAX"
args.Sort()
V = args(args.Count - 1).ToString
End Select
Return V
End Function
#End Region
jx315425246 2019-05-25
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public string Add(int a,int b)
{
return (a+b).ToString();
}
public string Add(int a,float b)
{
return ((float)a+b).ToString();
}
........
改参数类型就可以
好奇都是要学的 2019-05-25
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这个是重载问题吧 还好没想 INT+double
private string abc(obj a boj b)
{
return (a+b).tostring(); //可以实现 int、 double 、 int +double 和 double+int
}
jx315425246 2019-05-24
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写同函数名，返回值不同，参数不同，这叫重载
keaidepangzi 2019-05-24
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能进行，就是想用一个方法，调用这个方法，既可以用double类型的a,b，又可以用int的a,b作为参数进行相加运算
wid999 2019-05-24
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要方法的话，直接并列写，编译器直接认为是重载。
double Add(double op1, double op2) { return op1 + op2; }
int Add(int op1, int op2) { return op1 + op2; }
wid999 2019-05-24
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弱弱地问下，“+”不能计算这两种类型？
bloodish 2019-05-24
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需求比较奇怪。。。
public T Add<T>(T t1, T t2)
{
if(typeof(T).IsPrimitive
&& typeof(T) != typeof(bool)
&& typeof(T) != typeof(char))
{
dynamic d1 = t1;
dynamic d2 = t2;
return (d1 + d2);
}
else
{
return default;
}
}
正怒月神 2019-05-24
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你直接 都是转 Decmial类型算了。
别搞这么复杂。
stherix 2019-05-24
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最简单的就是方法重载
想要一个方法做到的,都会很别扭
Hi-Jimmy 2019-05-24
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public string Add<T>(T a, T b) where T:struct
{
if (typeof(T) == typeof(int))
{
return (int.Parse(a + "") + int.Parse(b + "")).ToString();
}
else if (typeof(T) == typeof(double))
{
return (double.Parse(a + "") + double.Parse(b + "")).ToString();
}
//float,long.....
return default(T).ToString();
}
智者知已应修善业 2019-05-24
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写错了跑到C++去了!
template <class 形参>
形参 相加(形参 a, 形参b){ return a + b; }
智者知已应修善业 2019-05-24
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template <class 形参>
void 相加(形参 A, 形参 B){return A+B;}
