CODEWITHSUNDEEP
c++visual-studio-2010lambda
Xeo
Xeo

Reputation: 131789

Different sizes of lambda expressions in VS2010?

Out of curiosity, I tested the size of a lamba expression. My first thought was, that they'd be 4 bytes big, like a function pointer. Strangely, the output of my first test was 1:

auto my_lambda = [&]() -> void {};
std::cout << sizeof(my_lambda) << std::endl;

Then I tested with some calculations inside the lambda, the output still being 1:

auto my_lambda2 = [&]() -> void {int i=5, j=23; std::cout << i*j << std::endl;};
std::cout << sizeof(my_lambda2) << std::endl;

My next idea was kinda random, but the output finally changed, displaying the awaited 4:

auto my_lambda3 = [&]() -> void {std::cout << sizeof(my_lambda2) << std::endl;};
std::cout << sizeof(my_lambda3) << std::endl;

At least in Visual Studio 2010. Ideone still display 1 as the output.
I know of the standard rule, that a lambda expression cannot appear in an unevaluated context, but afaik that only counts for direct lambda use like
sizeof([&]() -> void {std::cout << "Forbidden." << std::endl;})
on which VS2010 prompts me with a compiler error. Anyone got an idea what's going on?

Upvotes: 3

Views: 709

Answers (2)

Xeo
Xeo

Reputation: 131789

Thanks to @Hans Passant's comment under the question, the solution was found. My original approach was wrong in the fact that I thought every object would be captured by the lambda, but that isn't the case, only those in the enclosing scope are, and only if they are used.
And for everyone of those captured objects, 4 bytes are used (size of the reference).

Upvotes: 2

Edward Strange
Edward Strange

Reputation: 40859

Visual Studio probably doesn't implement lambda objects as functions. You're probably getting an object back. Who knows what it looks like. If you're truly interested you could always look at your variables with a debugger and see what they look like...if it'll let you.

Upvotes: 1

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