Compiler warning: lambda return type cannot be ded

问题:

Consider this example:

#include <algorithm>
#include <iostream>

int main()
{
    std::string str = "abcde4fghijk4l5mnopqrs6t8uvwxyz";
    std::string str2;

    std::remove_copy_if(str.begin(), str.end(),
        std::back_inserter(str2),
        [](char& c) {
            if (std::isdigit(c))
                return true;      // <----- warning here
            else
                return false;
        }
    );

    std::cout << str2 << '\n';
}

With GCC 4.6.1, this compiles fine and prints expected output (the alphabet) but I get a warning saying "lambda return type can only be deduced when the return statement is the only statement in the function body".

Now, I know how to get rid of the warning (using trailing return type or simply saying return isdigit(c);), but I'm curious, since compiler doesn't warn for nothing (or so it should be): what could possibly go wrong in code like this? Does standard say anything about it?

回答1:

As @ildjarn says in his comment, your code is simply ill-formed according to the standard.

§5.1.2 [expr.prim.lambda] p4

[...] 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.

[...]

That's it, basically if the code inside the curly brackets (called compund-statement in the standard) is anything but return some_expr;, the standard says the return type is undeducible and you get a void return type.

回答2:

It is worth noting that this issue has been rectified in C++14. The language now correctly deduces the return type as the OP expected. From the draft standard [7.1.6.4.11]:

If the type of an entity with an undeduced placeholder type is needed to determine the type of an expression, the program is ill-formed. Once a return statement has been seen in a function, however, the return type deduced from that statement can be used in the rest of the function, including in other return statements. [ Example:

auto n = n; // error, n’s type is unknown
auto f();
void g() { &f; } // error, f’s return type is unknown
auto sum(int i) {
if (i == 1)
    return i; // sum’s return type is int
else
    return sum(i-1)+i; // OK, sum’s return type has been deduced
}

—end example ]