accessing unordered_maps as pointers

Question

For some reason, I am having trouble editing values in my unordered_map, and am wondering what I'm doing wrong.

In the following code, parameteris a struct. For some reason, the following code is throwing a syntax error, not liking the [.

void MyClass::setParameter(string name, parameter param) {
    if (this->param_name_to_data == nullptr) {
        //create it lazily
        this->param_name_to_data = new unordered_map<string, parameter>();
    }
    this->param_name_to_data->[name] = param;
}

The dictionary id declared in the corresponding .h file as:

private:
 std::unordered_map<std::string, parameter> * param_name_to_data = nullptr;

What am I doing wrong?

Is there some reason why you dynamically allocate the unordered_map? — Nicol Bolas
– Nicol Bolas, Commented Apr 3, 2016 at 22:04

David Haim · Accepted Answer · 2016-04-03 22:08:59Z

4

param_name_to_data->[name] = param; is not a valid syntax

when the compiler sees -> it looks for either a member variable or a member function. the statement ->[...] is meaningless since [] alone is neither a member variable nor a function.

you can write instead

(*param_name_to_data)[name] = param;

or turn the pointer to reference first

auto& map = *param_name_to_data;
map[name] = param;

you can also use the ugly literal form of

param_name_to_data->operator[] (name) = param

but the last one is discouraged.

edited Apr 3, 2016 at 22:08

answered Apr 3, 2016 at 22:04

David Haim

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4 Comments

user650261 Over a year ago

I had thought I had seen ->[] somewhere in code before, which is why I was trying to use it. Is there a reason why it's forbidden?

user650261 Over a year ago

Thanks. I still find it weird that it is unable to intelligently search for any other operator's abbreviation. It's not a big deal, I'm just legitimately curious if there's some syntax-tree-ambiguity related reason or something that the compiler can't do that step.

David Haim Over a year ago

it's basically like that because the standard didn't state that ->[] is a valid syntax for calling operator [] from a pointer. so no compiler implemented it. if the standard stated clearly that it should be valid, it would be correct

Nicol Bolas Over a year ago

@user650261: "* I still find it weird that it is unable to intelligently search for any other operator's abbreviation.*" That's because it's not an "abbreviation". operator[] is the function's name. The -> operator is grammatically intended to be followed by a member name of the type returned by operator->. [] is not a member name, so it fails.

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