C++ lambda syntax

Question

I have a function that searches a vector of iterators and returns the iterator if its names matches a string passed as an argument.

koalaGraph::PVertex lookUpByName(std::string Name, std::vector<koalaGraph::PVertex>& Vertices) {

    for (size_t i = 0; i < Vertices.size(); i++) {

        if(Vertices[i]->info.name == Name) 
            return Vertices[i];
    }
}

My question is how can I implement this as a lambda, to use it in connection with std::find_if?

I'm trying this:

std::vector<koalaGraph::PVertex> V;
std::string Name;
std::find_if(V.begin(), V.end(), [&Name]() {return Name == V->info.name;})

But it says that V

an enclosing-function local variable cannot be referenced in a lambda body unless it is in the capture list.

No `return` if nothing found? – Deduplicator Mar 19 '19 at 21:26 — Deduplicator, Mar 19 '19 at 21:26

NathanOliver · Answer 1 · 2019-03-19T21:28:06.777

find_if is going to pass the elements of the vector into your lambda. That means you need

std::find_if(V.begin(), V.end(), [&Name](auto const& V) {return Name == V->info.name;})

so that the V in the lambda body is the element of the vector, not the vector itself.

Ideally you'd give it a different name than V so you keep the vector and local variables separate like

std::find_if(V.begin(), V.end(), [&Name](auto const& element) {return Name == elememt->info.name;})

So now it is clear you are working on a element of the vector, instead of the vector itself.

Guillaume Racicot · Answer 2 · 2019-03-19T14:40:56.940

8

First, V->info.name is ill formed, inside or outside of the lambda.

The function object sent to the algoritm std::find_if must be a unary function. It must take the current element to check as a parameter.

auto found = std::find_if(
    V.begin(), V.end(), 
    [&Name](koalaGraph::PVertex const& item_to_check) {
        return Name == item_to_check->info.name;
    }
);

The type of found is an iterator to the element that has been found. If none is found, then it returns V.end()

If you use C++14 or better, you can even use generic lambdas:

auto found = std::find_if(
    V.begin(), V.end(), 
    [&Name](auto const& item_to_check) {
        return Name == item_to_check->info.name;
    }
);

edited Mar 19 '19 at 14:40

answered Mar 19 '19 at 14:29

Guillaume Racicot

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`V->info.name` is valid syntax. It just won't typecheck since vectors don't overload `->`. – Silvio Mayolo Mar 19 '19 at 14:31
3

`It just won't typecheck since vectors don't overload ->` so... it won't compile? It's an invalid syntax to use `->` on object of types that don't overload `operator->` – Guillaume Racicot Mar 19 '19 at 14:33
1

Not all compile errors are syntax errors. See [syntax vs. semantics](https://stackoverflow.com/questions/17930267/what-is-the-difference-between-syntax-and-semantics-of-programming-languages). – Silvio Mayolo Mar 19 '19 at 14:34
2

@SilvioMayolo The answer you linked me says that semantics is the meaning of the program, usually in it's runtime behaviour. The code I quoted is simply ill formed, there is no semantic yet. Or maybe I don't understand the concept correctly – Guillaume Racicot Mar 19 '19 at 14:40

score 4 · Answer 3 · answered Mar 19 '19 at 14:28

4

std::find_if's predicate will receive a reference to each element of the range in turn. You need:

std::find_if(
    V.begin(), V.end(),
    [&Name](koalaGraph::PVertex const &v) { return Name == v->info.name; }
);

answered Mar 19 '19 at 14:28

Quentin

62,093
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score 2 · Answer 4 · answered Mar 19 '19 at 14:29

2

Get V as parameter to the lambda.

std::find_if(V.begin(), V.end(), [&Name](type& V) {return Name == V->info.name;)

answered Mar 19 '19 at 14:29

Petar Velev

2,305
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score 1 · Answer 5 · answered Mar 19 '19 at 14:42

Use const auto & to access the individual elements from your vector in the lambda expression. Since the vector is an lvalue, auto will be deduced to const vector<PVertex> &. Then, you can use std::distance to find the element location of the object in the vector.

struct PVertex
{
    std::string name;
};

int main()
{
    std::vector<PVertex> V = {{"foo"},{"bar"},{"cat"},{"dog"}};

    std::string Name = "cat";

    auto found = std::find_if(std::begin(V), std::end(V), [&Name](const auto &v){return (Name == v.name);});

    std::cout<< "found at: V["<<std::distance(std::begin(V),found)<<"]" <<std::endl;
}

Result is:

found at: V[2]

Example: https://rextester.com/IYNA58046

C++ lambda syntax

5 Answers5