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using lambda expression and find_if on a collection

Tags:

c++

lambda

stl

I have a container object :

R Container;

R is of type list<T*> or vector<T*>

I am trying to write the following function:

template<typename T, typename R>
T& tContainer_t<T, R>::Find( T const item ) const
{   
typename R::const_iterator it = std::find_if(Container.begin(), Container.end(),  [item](const R&v) { return item == v; });
if (it != Container.end())
    return (**it);
else
    throw Exception("Item not found in container");
}

When trying the method (v is an object of my class)

double f = 1.1;
v.Find(f);

I get binary '==' : no operator found which takes a left-hand operand of type 'const double' (or there is no acceptable conversion)

I am confused with the lambda expression syntax and what i should write there and couldn't find any friendly explanation.

What is wrong ? 10x.

like image 919
Michael Avatar asked Dec 21 '25 23:12

Michael


1 Answers

Missing a bit of context, but I note:

  • You return **it so you possibly want to compare *v==itemt
  • You pass const R&v where I suspect you meant const T&v into the lambda
  • You used a const_iterator, but returned a non-const reference. That was a mismatch
  • I made some params const& for efficiency (and to support non-copyable/non-movable types)

Here is working code, stripped of the missing class references:

#include <vector>
#include <algorithm>
#include <iostream>

template<typename T, typename R=std::vector<T> >
T& Find(R& Container, T const& item ) 
{   
    typename R::iterator it = std::find_if(Container.begin(), Container.end(),  [&item](const T&v) { return item == v; });
    if (it != Container.end())
        return *it;
    else
        throw "TODO implement";
}

int main(int argc, const char *argv[])
{
    std::vector<double> v { 0, 1, 2, 3 };
    Find(v, 2.0); // not '2', but '2.0' !
    return 0;
}
like image 73
sehe Avatar answered Dec 24 '25 11:12

sehe