CODEWITHSUNDEEP
c++functionc++11lambda
Anon
Anon

Reputation: 1354

Wrapped lambda expressions

Here's a lamba wrapper expression defined:

function <int(double)> f = 
     [](double x) -> int{ return static_cast <int> (x*x); };

It is used like this:

f(someintvalue);

What is the difference between normal functions and wrapped lambdas?

Upvotes: 1

Views: 625

Answers (2)

sehe
sehe

Reputation: 392893

Lambdas can capture surrounding scope ([=] or [&]) resulting in anonymous structs that contain a member function.

Certain tasks, like accessing lambda (types) across Translation Units, or taking standard pointer-to-memberfunction addresses from lambdas may prove hard/useless because the actual type of the automatically generated 'anonymous' type cannot be known (and is in fact implementation defined).

std::function<> was designed to alleviate exactly that problem: being able to bind a function pointer (semantically) to a lambda (or indeed, whatever callable object)

1: in fact the type can have external linkage, but you cannot portably refer to that from another translation unit because the actual type is implementation defined (and need not even be expressible in C++ code; within the original TU you'd be able to use decltype to get the magic type. Thanks, Luc for the good precisions about this)

Upvotes: 1

Sarfaraz Nawaz
Sarfaraz Nawaz

Reputation: 361302

question is - what is the difference between normal function and wrapped lambda?

Normal function is a normal function and what you call "wrapped lambda" is actually a function object.

By the way, why use std::function? You could simply write this:

auto f = [](double x) { return static_cast <int> (x*x); };

//call
int result = f(100.0);

Also, I omitted the return type, as it is implicitly known to the compiler from the return expression. No need to write -> int in the lambda expression.

Upvotes: 2

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