CODEWITHSUNDEEP
c++c++11lambdallvmclang
user2015453
user2015453

Reputation: 5104

Is my book's discussion of lambda return types wrong?

My book says this:

Lambdas with function bodies that contain anything other than a single return statement that do not specify a return type return void.

but this:

auto f = []{
  int i=0; i++;
  return std::string("foo");
};
std::cout << f() << std::endl;

actually compiles and prints out "foo", but that lambda expr has more than just a single return statement so it should return void, because it does not manually specify "-> std::string" as a return type.

What's going on here?

I'm using Apple's compiler in the latest Xcode 4.6, based on Clang 3.2 it seems:

clang --version

Apple LLVM version 4.2 (clang-425.0.24) (based on LLVM 3.2svn) Target: x86_64-apple-darwin12.2.0 Thread model: posix

Upvotes: 13

Views: 657

Answers (5)

Richard Smith
Richard Smith

Reputation: 14158

Clang implements the proposed resolution to C++ core issue 975. That allows an arbitrary body for a lambda, with any number of return statements, and deduces the return value from the returned expression under the proviso that they must all produce the same type.

In C++14, this support is generalized further by N3638, which was voted into the working draft for the standard at the Bristol meeting of WG21.

Upvotes: 1

Spook
Spook

Reputation: 25927

If you use popular compilers (gcc, Visual Studio), you usually don't need to specify return type as long as the compiler is able to determine it unambiguously - like in your example.

The following example shows a lambda, which requires explicit return type information:

auto lambda = [](bool b) -> float
    { 
        if (b) 
            return 5.0f; 
        else 
            return 6.0; 
    };

I asked Bjarne Stroustrup regarding this matter, his comment:

I do not know if C++11 allows the deduction of the return type is there are several return statements with identical return type. If not, that's planned for C++14.

Upvotes: 3

Ben Voigt
Ben Voigt

Reputation: 283604

The book accurately reflects the rules in draft n3290 of the Standard. Perhaps your compiler implemented a different draft.

In section 5.1.2p4, the draft reads

If a lambda-expression does not include a trailing-return-type, it is as if the trailing-return-type denotes the following type:

  • if the compound-statement is of the form { attribute-specifier-seqopt return expression ; } the type of the returned expression after lvalue-to-rvalue conversion, array-to-pointer conversion, and function-to-pointer conversion;
  • otherwise, void.

The syntactic construct attribute-specifier-seq may be alignas or the double-bracketed attributes. Not variable declarations.

Draft n3485, which followed publication of C++11 (i.e. it is work in progress toward C++1y), contains the same wording. I don't know if there was a different rule in some draft earlier than n3290.

Upvotes: 6

David Rodr&#237;guez - dribeas
David Rodr&#237;guez - dribeas

Reputation: 208323

I am not sure of what to make from the quote in the question, but here is what the C++11 standard says about lambdas without declarator or return type:

If a lambda-expression does not include a lambda-declarator, it is as if the lambda-declarator were (). If a lambda-expression does not include a trailing-return-type, it is as if the trailing-return-type denotes the following type (5.1.2p4):

— if the compound-statement is of the form { attribute-specifier-seqopt return expression ; } the type of the returned expression after lvalue-to-rvalue conversion (4.1), array-to-pointer conversion (4.2), and function-to-pointer conversion (4.3);

— otherwise, void.

Upvotes: 1

user2033018
user2033018

Reputation:

Draft n3485 indicates that if the compiler can unambiguously determine the return type it will allow for the lambda to not specify it.

Upvotes: 0

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