When can we omit the return type in a C++11 lambda?

Question

As far as I know, in standard C++11 (not C++14), when omitting the return type of a lambda, its return type is deduced to be:

1. The type of the

Answer 1

This is what I find in the C++ Draft Standard N3337:

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:

— if the compound-statement is of the form

{ attribute-specifier-seq^opt 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.

[ Example:

auto x1 = [](int i){ return i; }; // OK: return type is int
auto x2 = []{ return { 1, 2 }; }; // error: the return type is void (a
                                  // braced-init-list is not an expression)

— end example ]

The standard seems to indicate that:

• When only one statement is present, and
• It is a return statement, and
• The object being returned is an expression

Then the return type is deduced from the expression. Otherwise the return type is void.