C++11 – Change `auto` Lambda to a Different Lambda in C++11

autoc++c++11lambda

Say I have the following variable containing a lambda:

auto a = [] { return true; };

And I want a to return false later on. Could I do something along the lines of this?

a = [] { return false; };

This syntax gives me the following errors:

binary '=' : no operator found which takes a right-hand operand of type 
'main::<lambda_a7185966f92d197a64e4878ceff8af4a>' (or there is no acceptable conversion)

IntelliSense: no operator "=" matches these operands
        operand types are: lambda []bool ()->bool = lambda []bool ()->bool

Is there any way to achieve something like this? I would like to change the auto variable to a different lambda. I'm relitively a beginner so I may be missing some knowledge about auto or lambdas. Thanks.

Best Answer

Every lambda expression creates a new unique type, so the type of your first lambda is different from the type of your second (example). Additionally the copy-assignment operator of a lambda is defined as deleted (example) so you're doubly-unable to do this. For a similar effect, you can have a be a std::function object though it'll cost you some performance

std::function<bool()> a = [] { return true; };
a = [] { return false; };

Related Solutions

C++11 Lambda Functions – Using Auto in a Lambda Function

auto keyword does not work as a type for function arguments, in C++11. If you don't want to use the actual type in lambda functions, then you could use the code below.

 for_each(begin(v), end(v), [](decltype(*begin(v)) it ){
       foo( it + 5);         
 });

The code in the question works just fine in C++ 14.

How to Use Lambda Auto Parameters in C++11

C++11 doesn't support generic lambdas. That's what auto in the lambda's parameter list actually stands for: a generic parameter, comparable to parameters in a function template. (Note that the const isn't the problem here.)

Note: C++14 does support lambdas with auto, const auto, etc. You can read about it here.

You have basically two options:

Type out the correct type instead of auto. Here it is the element type of X, which is pair<double, vector<int>>. If you find this unreadable, a typedef can help.

std::stable_sort(X.rbegin(), X.rend(),
                 [](const pair<double, vector<int>> & lhs,
                    const pair<double, vector<int>> & rhs)
                 { return lhs.first < rhs.first; });

Replace the lambda with a functor which has a call operator template. That's how generic lambdas are basically implemented behind the scene. The lambda is very generic, so consider putting it in some global utility header. (However do not using namespace std; but type out std:: in case you put it in a header.)
```
struct CompareFirst {
    template <class Fst, class Snd>
    bool operator()(const pair<Fst,Snd>& l, const pair<Fst,Snd>& r) const {
        return l.first < r.first;
    }
};
```
```
std::stable_sort(X.rbegin(), X.rend(), CompareFirst());
```

Best Answer

Related Solutions

C++11 Lambda Functions – Using Auto in a Lambda Function

How to Use Lambda Auto Parameters in C++11

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