如何從（）=> foo.Title表達式獲取對象實例

Question

如何從（）=> foo.Title表達式獲取對象實例

.net-3.5 c# data-binding expression-trees lambda

題

我有一個帶有屬性的簡單類

class Foo 
{ 
    string Title { get; set; } 
}

我試圖通過調用類似的函數來簡化數據綁定

BindToText(titleTextBox, ()=>foo.Title );

聲明像

void BindToText<T>(Control control, Expression<Func<T>> property)
{
    var mex = property.Body as MemberExpression;
    string name = mex.Member.Name;

    control.DataBindings.Add("Text", ??? , name);
}

所以我要放什麼???作為我的Foo類的實例。如何從lambda表達式獲得對調用foo實例的引用？

編輯：該實例應該在某個地方，因為我可以調用property.Compile()並創建一個在我的BindToText函數中使用foo實例的委託。所以我的問題是，是否可以在函數參數中不添加對實例的引用的情況下完成此操作。我呼籲Occum的Razor提供最簡單的解決方案。

編輯2：許多人沒有註意到的是，如果我編譯lambda，則在訪問函數中的foo實例時存在閉包。編譯器怎麼知道在哪裡可以找到實例，而我卻不知道呢？我堅持必須有一個答案，而不必通過額外的爭論。

解

感謝VirtualBlackFox ，解決方案是這樣的：

void BindText<T>(TextBoxBase text, Expression<Func<T>> property)
{
    var mex = property.Body as MemberExpression;
    string name = mex.Member.Name;
    var fex = mex.Expression as MemberExpression;
    var cex = fex.Expression as ConstantExpression;            
    var fld = fex.Member as FieldInfo;
    var x = fld.GetValue(cex.Value);
    text.DataBindings.Add("Text", x, name);            
}

這使我只需鍵入BindText(titleText, () => foo.Title); 。

ja72

熱門答案

別。只需修改方法以獲取另一個參數，如＃3444294中所述。對於您的示例，它可能是這樣的：

void BindToText<T>(Control control, T dataSource, Expression<Func<T>> property)
{
    var mex = property.Body as MemberExpression;
    string name = mex.Member.Name;

    control.DataBindings.Add("Text", dataSource, name);
}

並會被稱為

BindToText(titleTextBox, foo, ()=>foo.Title );

還不錯，但很容易理解。沒有魔法發生。 ;）

hangy

在 Stack Overflow 上查看更多

Julien Roncaglia · Accepted Answer · 2/23/2011 4:47:53 PM

您想要的小LINQPad樣本：

void Foo<T>(Expression<Func<T>> prop)
{
    var propertyGetExpression = prop.Body as MemberExpression;

    // Display the property you are accessing, here "Height"
    propertyGetExpression.Member.Name.Dump();

    // "s" is replaced by a field access on a compiler-generated class from the closure
    var fieldOnClosureExpression = propertyGetExpression.Expression as MemberExpression;

    // Find the compiler-generated class
    var closureClassExpression = fieldOnClosureExpression.Expression as ConstantExpression;
    var closureClassInstance = closureClassExpression.Value;

    // Find the field value, in this case it's a reference to the "s" variable
    var closureFieldInfo = fieldOnClosureExpression.Member as FieldInfo;
    var closureFieldValue = closureFieldInfo.GetValue(closureClassInstance);

    closureFieldValue.Dump();

    // We know that the Expression is a property access so we get the PropertyInfo instance
    // And even access the value (yes compiling the expression would have been simpler :D)
    var propertyInfo = propertyGetExpression.Member as PropertyInfo;
    var propertyValue = propertyInfo.GetValue(closureFieldValue, null);
    propertyValue.Dump();
}

void Main()
{
    string s = "Hello world";
    Foo(() => s.Length);
}

如何從（）=> foo.Title表達式獲取對象實例

題

解

Fastest Entity Framework Extensions

一般承認的答案

熱門答案

Performance

Expression Evaluator