lambda：应该按引用捕获 const 引用会产生未定义的行为吗？

Question

I just found a nasty bug in my code because I captured a const reference to a string by reference. By the time the lambda was run the original string object was already long gone and the referenced value was empty whereas the purpose was that it would contains the value of the original string, hence the bug.

What baffles me is that this did not invoke a crash at runtime: after all, shouldn't this be undefined behaviour since afaik there is a dangling reference? Moreover when looking at id under the debugger, it doesn't even look like garbage but just like a properly constructed empty string.

Here's the test case; this just prints an empty line:

typedef std::vector< std::function< void() > > functions;

void AddFunction( const std::string& id, functions& funs )
{
  funs.push_back( [&id] ()
    {
        //the type of id is const std::string&, but there
        //is no object to reference. UB?
      std::cout << id << std::endl;
    } );
}

int main()
{
  functions funs;
  AddFunction( "id", funs );
  funs[ 0 ]();
}

Answer 1

Undefined behavior means there is no requirement what should happen. There is no requirement that it should crash. Whatever memory your dangling reference points at, there's no reason it shouldn't contain something that looks like an empty string, and it's plausible that the destructor of string leaves the memory in that state.

Answer 2

Capturing anything by reference means that you have to take care that it's alive long enough. If you don't the program may just work, but it might just call Domino's and order a double pepperoni. At least, according to the standard.

Answer 3

(as pointed out by dascandy) The problem has little or nothing to do with the const and reference syntax, more simply it's an abdication of the responsibility to ensure the existence of everything that is passed by reference at any time it is referenced. The literal in the function call is strictly temporary for that call and evaporates on return, so we are accessing a temporary - a flaw often detected by the compilers - just not in this case.

typedef std::vector<std::function<void()> > functions;

void AddFunction(const std::string& id, functions& funs) {
    funs.push_back([&id] ()
    {
        //the type of id is const std::string&, but there
        //is no object to reference. UB?
            std::cout <<"id="<< id << std::endl;
        });
}

int emain() {
    functions funs;

    std::string ida("idA");
           // let idB be done by the tenporary literal below
    std::string idc("idC");

    AddFunction(ida, funs);
    AddFunction("idB", funs);
    AddFunction(idc, funs);
    funs[0]();
    //funs[1](); // uncomment this for (possibly) bizarre results   
    funs[2]();
    std::cout<<"emain exit"<<std::endl;
    return 0;
}

int main(int argc, char* argv[]){
    int iret = emain();
    return 0;
}