也作一下装配脑袋的Expression习题【转】

一.习题

http://www.cnblogs.com/Ninputer/archive/2009/08/28/expression_tree1.html

二.参考

http://msdn.microsoft.com/zh-cn/library/system.linq.expressions.expression.aspx
http://msdn.microsoft.com/zh-cn/library/bb361179%28v=VS.100%29.aspx

三.思路

先写出lambda，参照lambda拼出Expression

四.做题

(1).-a

Expression<Func<int, int>> lambdaexpression = a => -a;

ParameterExpression p1 = Expression.Parameter(typeof(int), "a");
Expression body = Expression.Negate(p1);
LambdaExpression expression = Expression.Lambda(body, p1);

Console.WriteLine(lambdaexpression.ToString());
Console.WriteLine(expression.ToString());

Expression.Negate方法返回类型是UnaryExpression，一元运算表达式

(2).a + b * 2

Expression<Func<int, int, int>> lambdaexpression = (a, b) => a + b * ;

ParameterExpression p1 = Expression.Parameter(typeof(int), "a");
ParameterExpression p2 = Expression.Parameter(typeof(int), "b");
Expression body = Expression.Add(p1, Expression.Multiply(p2, Expression.Constant()));
LambdaExpression expression = Expression.Lambda(body, p1, p2);

Console.WriteLine(lambdaexpression.ToString());
Console.WriteLine(expression.ToString());

Expression.Add和Expression.Multiply都返回BinaryExpression，二元运算表达式，
Expression.Constant返回ConstantExpression，常量表达式

(3).Math.Sin(x) + Math.Cos(y)

Expression<Func<double, double, double>> lambdaexpression = (x, y) => Math.Sin(x) + Math.Cos(y);

ParameterExpression p1 = Expression.Parameter(typeof(double), "x");
ParameterExpression p2 = Expression.Parameter(typeof(double), "y");
Expression body = Expression.Add(
    Expression.Call(null, typeof(Math).GetMethod("Sin", BindingFlags.Static | BindingFlags.Public), p1)
    , Expression.Call(null, typeof(Math).GetMethod("Cos", BindingFlags.Static | BindingFlags.Public), p2));
LambdaExpression expression = Expression.Lambda(body, p1, p2);

Console.WriteLine(lambdaexpression.ToString());
Console.WriteLine(expression.ToString());

Expression.Call返回MethodCallExpression，方法调用表达式

(4).new StringBuilder(“Hello”)

Expression<Func<StringBuilder>> lambdaexpression = () => new StringBuilder("Hello");

Expression body = Expression.New(typeof(StringBuilder).GetConstructor(new Type[] { typeof(string) }),
    Expression.Constant("Hello"));
LambdaExpression expression = Expression.Lambda(body);

Console.WriteLine(lambdaexpression.ToString());
Console.WriteLine(expression.ToString());

Expression.New返回NewExpression，构造函数调用表达式

(5).new int[] { a, b, a + b}

Expression<Func<int, int, int[]>> lambdaexpression = (a, b) => new int[] { a, b, a + b };

ParameterExpression p1 = Expression.Parameter(typeof(int), "a");
ParameterExpression p2 = Expression.Parameter(typeof(int), "b");
Expression body = Expression.NewArrayInit(typeof(int), p1, p2, Expression.Add(p1, p2));
LambdaExpression expression = Expression.Lambda(body, p1, p2);

Console.WriteLine(lambdaexpression.ToString());
Console.WriteLine(expression.ToString());

Expression.NewArrayInit返回NewArrayExpression，创建数组表达式

(6).a[i – 1] * i

Expression<Func<int[], int, int>> lambdaexpression = (a, i) => a[i - ] * i;

ParameterExpression p1 = Expression.Parameter(typeof(int[]), "a");
ParameterExpression p2 = Expression.Parameter(typeof(int), "i");
Expression body = Expression.Multiply(Expression.ArrayIndex(p1, Expression.Subtract(p2, Expression.Constant())), p2);
LambdaExpression expression = Expression.Lambda(body, p1, p2);

Console.WriteLine(lambdaexpression.ToString());
Console.WriteLine(expression.ToString());

Expression.ArrayIndex返回也是BinaryExpression，这个有点意外，我以为会是MethodCallExpression呢

(7).a.Length > b | b >= 0

Expression<Func<int[], int, bool>> lambdaexpression = (a, b) => a.Length > b | b >= ;

ParameterExpression p1 = Expression.Parameter(typeof(int[]), "a");
ParameterExpression p2 = Expression.Parameter(typeof(int), "b");
Expression body = Expression.Or(Expression.GreaterThan(Expression.ArrayLength(p1), p2),
    Expression.GreaterThanOrEqual(p2, Expression.Constant()));
LambdaExpression expression = Expression.Lambda(body, p1, p2);

Console.WriteLine(lambdaexpression.ToString());
Console.WriteLine(expression.ToString());

Expression.GreaterThan、Expression.Or和Expression.ArrayLength都返回BinaryExpression，二元运算表达式，
Expression.ArrayLength和(6)中一样误解了

(8).（高难度）new System.Windows.Point() { X = Math.Sin(a), Y = Math.Cos(a) }

Expression<Func<double, System.Windows.Point>> lambdaexpression = a => new System.Windows.Point() { X = Math.Sin(a), Y = Math.Cos(a) };

ParameterExpression p1 = Expression.Parameter(typeof(double), "a");
Expression body = Expression.MemberInit(Expression.New(typeof(System.Windows.Point))
    , new MemberBinding[] {
        Expression.Bind(typeof(System.Windows.Point).GetProperty("X"),
        Expression.Call(null, typeof(Math).GetMethod("Sin", BindingFlags.Static | BindingFlags.Public), p1)),
        Expression.Bind(typeof(System.Windows.Point).GetProperty("Y"),
        Expression.Call(null, typeof(Math).GetMethod("Cos", BindingFlags.Static | BindingFlags.Public), p1))
    });
LambdaExpression expression = Expression.Lambda(body, p1);

Console.WriteLine(lambdaexpression.ToString());
Console.WriteLine(expression.ToString());

Expression.MemberInit返回MemberInitExpression，如方法名字一样是成员初始化表达式。
果然像题目所写，难度不低，一开始以为使用Expression.New后面的参数，后来还是看了评论才会写。

五.执行结果

a => -a
a => -a

(a, b) => (a + (b * 2))
(a, b) => (a + (b * 2))

(x, y) => (Sin(x) + Cos(y))
(x, y) => (Sin(x) + Cos(y))

() => new StringBuilder("Hello")
() => new StringBuilder("Hello")

(a, b) => new [] {a, b, (a + b)}
(a, b) => new [] {a, b, (a + b)}

(a, i) => (a[(i - 1)] * i)
(a, i) => (a[(i - 1)] * i)

(a, b) => ((ArrayLength(a) > b) Or (b >= 0))
(a, b) => ((ArrayLength(a) > b) Or (b >= 0))

a => new Point() {X = Sin(a), Y = Cos(a)}
a => new Point() {X = Sin(a), Y = Cos(a)}
Press any key to continue . . .

作完这些习题，感觉让我这个Expression新手真的进步了不少

转载自：http://www.cnblogs.com/llcto/archive/2011/08/12/2136334.html

谢谢浏览！