Undefined Behavior with the C++0x Closure: I

问题:

Consider the example:

#include <iostream>
#include <functional>     // std::function
#include <vector>        // std::vector
#include <algorithm>    // std::for_each

int main(){

    auto adder = [](int x) {
        return [&](int y) { 
            return x+=y; 
        }; 
    };

    std::vector < std::function<int(int)> > vec;

    vec.push_back(adder(1));
    vec.push_back(adder(10));

    std::for_each(vec.begin(), vec.end(), [](std::function<int(int)> f){std::cout << f(33) << " ";});
    std::cout << std::endl;
}

One expects the integers 34 and 43 43 and 76, but instead gcc 4.6.0 produces "internal compiler error: Segmentation fault". What is wrong with the code?

Edit: Several other examples are discussed here.

回答1:

(Edit: this certainly does not explain the ICE; I read the original question too hastily.)

The One problem in that code is that the lambdas returned from the adder function contain dangling references to the x variable that no longer exists. Capture by copy ([=] or [i]) instead of a reference ([&]) and everything should work.

回答2:

It seems, that in your example trailing-return-type cannot be omitted. Here is excerpt from standard (5.1.2 Lambda expressions):

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 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.

Returned value in your example cannot be used for conversions mentioned above. Following code with explicitely added return type compiles in VS 2010:

auto adder = [] (int x) -> std::function<int (int)> {
  return [=]( int y ) {
    return x + y;
  };
};

回答3:

You're just completely missing the point. The need for std::function is very, very obvious.

1. All lambdas have a unique type at compile-time
2. You want the vector to hold any functional object at run-time.
3. Therefore, some sort of type erasure is required, which is the job std::function does.

How on earth could you ever create a vector that varies at run-time a compile-time fact, like the type contained within it? That's just logically impossible- unless you use an abstraction such as std::function.

Of course, if you only ever want one lambda type within, then you don't need std::function at all. This is relatively rare though.

int main() {
    auto adder = [](int x) {
        return [=](int y) {
            return x + y;
        };
    };
    // alternatively- you MUST copy the argument as it will cease to exist
    // but once it's in the lambda, you can use "mutable" to allow you to
    // modify the copy that each lambda has.
    /*
    auto adder = [](int x) {
        return [=](int y) mutable {
            return x += y;
        };
    };
    */
    std::vector<decltype(adder(0))> adders;
    adders.emplace_back(adder(0));
    adders.emplace_back(adder(1));

    std::for_each(adders.begin(), adders.end(), [](decltype(*adders.begin())& ref) {
        std::cout << ref(33);
    });
    std::cin.get();
}

MSVC won't actually compile this little snippet, but I think that's a bug and judging by the reports of your compiler, I expect that it will compile there and indeed work correctly.