[英]C# Expression.Lambda cannot be compiled at runtime
I have build expression parser CreateExpression()
which return constructed Expression Tree 我有构建表达式解析器CreateExpression()
,它返回构造的表达式树
Expression rule = CreateExpression(_BuyRuleString);
LambdaExpression lambda = Expression.Lambda(rule, _ParameterExpressions);
var func = lambda.Compile();
but it failed when I call lambda.Compile()
with the error 但是当我用lambda.Compile()
调用错误时失败了
variable 't1' of type 'System.Int32' referenced from scope '', but it is not defined
So I print out expression lambda
所以我打印出表达式lambda
.Lambda #Lambda1<System.Func`9[System.Int32,System.Int32,System.Int32,System.Int32,System.Int32,System.Int32,System.Double,System.Double,System.Boolean[]]>(
System.Int32 $t1,
System.Int32 $t2,
System.Int32 $t3,
System.Int32 $t4,
System.Int32 $t5,
System.Int32 $t6,
System.Double $r1,
System.Double $r2) {
.Call SwarmTrader.ExpressionParser.SeriesOperatorFunc.GTZ(.Call SwarmTrader.Indicator.RSI(
$t1,
"p"))
}
which equivalent to 相当于
Expression<Func<int, int, int, int, int, int, double, double, bool[]>> test = (t1, t2, t3, t4, t5, t6, r1, r2) => SwarmTrader.ExpressionParser.SeriesOperatorFunc.GTZ(SwarmTrader.Indicator.RSI(t1, "p"));
But var func = test.Compile();
但var func = test.Compile();
works. 作品。 So I try resolve it in combination ... 所以我尝试组合解决它...
lambda = Expression.Lambda(rule, _ParameterExpressions); // lambda.Compile() failed
lambda = Expression.Lambda(test.Body, _ParameterExpressions); // lambda.Compile() failed
lambda = Expression.Lambda(rule, test.Parameters); // lambda.Compile() failed
lambda = Expression.Lambda(test.Body, test.Parameters); // lambda.Compile() works
Can anyone point out why lambda.Compile()
does work only from test
? 任何人都可以指出为什么lambda.Compile()
只能从test
中起作用吗?
Most likely your CreateExpression()
does not reference parameters that are in _ParameterExpressions
, but defines its own instead. 很可能你的CreateExpression()
不引用_ParameterExpressions
参数,而是定义它自己的参数。 You have to use same ParameterExpression
in expression tree you're compiling and in lambda arguments. 您必须在正在编译的表达式树和lambda参数中使用相同的 ParameterExpression
。
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