std ::移动和地图分配

Question

I am bit puzzled about how standard governs this case:

struct Foo {
    Foo & operator = (std::string xxx)
    {
        x = std::move(xxx);
        return *this;
    }

    std::string x;
};

std::map<std::string, Foo> bar;

std::string baz = "some string";

bar[baz] = std::move(baz);

Can compilers produce code so that baz will be moved before it's used to initialise and get reference to element in bar (to initialise std::string xxx )? Or is this code safe and there's no undefined behaviour ?

Answer 1

Hell no. The expression is, yes, equivalent to

(bar.operator[](baz)).operator=(std::move(baz))

But there is no guaranteed order between the evaluation of (bar.operator[](baz)).operator= - formally, the postfix-expression designating the function to be called - and the evaluation of the initialization of the argument to operator= , which is what moves from baz .

In fact, this asserts on GCC :

std::map<std::string, Foo> bar;
std::string baz = "some string";
bar[baz] = std::move(baz);
assert(bar.count("some string"));