Transforming an expression template tree

Transforming an expression template tree

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Transforming an expression template tree
Tag : cpp , By : Reiner
Date : November 28 2020, 01:01 AM

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Passing an expression tree as a parameter to another expression tree in EntityDataModel

Tag : chash , By : unfool
Date : March 29 2020, 07:55 AM
wish help you to fix your issue Option 1
Give this a try first - I just tried it with LINQ-to-SQL and it works so it should work in EF. It's not a million miles from what you had, just saying invoke e1 with the result of invoking e2:
var e3 = Expression.Lambda<Func<Customer, bool>>(Expression.Invoke(e1, Expression.Invoke(e2, e2.Parameters)), e2.Parameters);
// replace parameter use anywhere in e1.Body with e2.Body
var remover = new ParameterReplaceVisitor(e2.Body);
var bb = remover.Visit(e1.Body);

// create a new lambda with our amended body but still with e2 parameters i.e. the Customer
var e3 = Expression.Lambda<Func<Customer, bool>>(bb, e2.Parameters);

public class ParameterReplaceVisitor : ExpressionVisitor
    Expression _replace;

    public ParameterReplaceVisitor(Expression replace)
        _replace = replace;

    protected override Expression VisitParameter(ParameterExpression node)
        // when we encounter a parameter replace it
        return _replace;

Transforming a Boost C++ Phoenix Expression Tree

Tag : cpp , By : thatotherguy
Date : March 29 2020, 07:55 AM
I hope this helps . In the Boost Phoenix article, "Transforming the Expression Tree", here, a set of specialisations of a custom invert_actions class, are used to invert binary arithmetic expressions. For example a+b becomes a-b; a*b becomes a/b; and vice versa for both. , This is how you do it with straight Proto:
#include <iostream>
#include <boost/phoenix.hpp>
#include <boost/proto/proto.hpp>
namespace proto = boost::proto;
using namespace boost::phoenix;
using namespace arg_names;

struct invrt:
      // Turn plus nodes into minus
      proto::plus<proto::_, proto::_>,
        invrt(proto::_left), invrt(proto::_right)
      // This recurses on children, transforming them with invrt
      proto::nary_expr<proto::_, proto::vararg<invrt> >

int main(int argc, char *argv[])
  auto f = invrt()(_1+_1&_2);
  std::cout << f(1,2) << std::endl;
  return 0;
#include <iostream>
#include <boost/phoenix.hpp>
#include <boost/proto/proto.hpp>

using namespace boost;
using namespace proto;
using namespace phoenix;
using namespace arg_names;

struct invrt {
  template <typename Rule>
  struct when :
    // NOTE!!! recursively transform children and reassemble
    nary_expr<_, vararg<proto::when<_, evaluator(_, _context)> > >

template <>
struct invrt::when<rule::plus> :
      evaluator(_left, _context), evaluator(_right, _context)

int main()
  auto f = phoenix::eval( _1+_1&_2 , make_context(make_env(), invrt()) );
  std::cout << f(1,2) << std::endl; // Prints 0. Huzzah!

Finding method call in expression tree / iterating expression tree

Tag : chash , By : GunnarHafdal
Date : March 29 2020, 07:55 AM
With these it helps As svick correctly points out: use the ExpressionVisitor base class:

Transforming a tree back to JSON using tree-model-js

Tag : javascript , By : markku
Date : March 29 2020, 07:55 AM
This might help you Is there perhaps a way in which you can convert a TreeModel to a JSON string. That way it can be stored and then later recreated using tree.parse()? , Use JSON.stringify(root.model) instead.

Transforming tree recursively in JS/ES6

Tag : javascript , By : KT.
Date : March 29 2020, 07:55 AM
Any of those help You could take a recursive approach and iterate all keys and build new objects and take either the array or take the object for the next recursive call.
function getParts(object) {
    return Array.isArray(object)
        ? object
        : Object.keys(object).map(function (k) {
            return { name: k, children: getParts(object[k]) };

var data = { Parent: { Child1: ["toy1"], Child2: { Nephew: ["toy2", "toy3"] } } },
    result = { name: 'root', children: getParts(data) };

.as-console-wrapper { max-height: 100% !important; top: 0; }
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