C++ - adding a hashmap to a struct

Question

I want a struct that has an unordered_map as a member.
I also don't want to type out the type all the time, so I typedef it.

typedef std::unordered_map<std::string, int> HM;
struct A {
    HM myHm;
}

A *myA = new A();
HM hashmap = new HM(); // empty hashmap
hashmap["1"] = 1;
myA->myHm = hashmap;

Of course this isn't correct.

I see examples of people doing std::unordered_map<std::string, int> HM;
And then immediately HM["1"] = 1;

First of all, I'm used to Java where std::unordered_map<std::string, int> HM; is just a type and if I want to add stuff to it, I need to instantiate it with ().

It seems that in C++, it's instantiated the moment I write std::unordered_map<std::string, int> HM;. Can anyone offer some insight into this?

Second, what is the correct way of doing what I want?

That code compiles and does what you would expect (well, if you put the code inside a function it would). — James Picone
– James Picone, Commented Nov 27, 2015 at 4:32
A mA; HM hashmap; are sufficient and preferred in C++ to create fully-fledged objects. No new means no memory leaks (usually). Also you are creating a hash map as a local variable and then copying it into the hash map in your struct. Why not just use that one directly? — Neil Kirk
– Neil Kirk, Commented Nov 27, 2015 at 4:32
@JamesPicone I get conversion from ‘HM*’ to non-scalar type ‘HM’ requested when I assign HM() to HM hashmap. — pushkin
– pushkin, Commented Nov 27, 2015 at 4:43
I have that code in a compiler right now; it definitely behaves. Are you sure you haven't written new HM(), or some equivalent? — James Picone
– James Picone, Commented Nov 27, 2015 at 4:45

John Zwinck · Accepted Answer · 2015-11-27 04:38:23Z

4

In C++ you would normally not write this:

HM hashmap = HM();

That's because what you're doing there is to create an anonymous temporary with HM() (the default constructor), then assign it to another one called hashmap. Instead, you can simply do this:

HM hashmap;

That will equally default-construct a hash map, and avoids any potential unnecessary assignment or double-construction.

Another issue with your existing code is that you use new and store the result in a "raw" pointer, but never call delete on it. Instead, you can skip using new altogether, and avoid redundancy, like this:

A myA; // contains HM
myA.myHm["1"] = 1;

This does everything you need in the minimal number of steps.

answered Nov 27, 2015 at 4:38

John Zwinck

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James Picone · Accepted Answer · 2015-11-27 04:43:58Z

A way of thinking about things that might help is to think of all variables in C++ as being 'value types', in C# lingo (I don't know what the appropriate Java term is).

This Java is perfectly legal:

int foo;
foo = 2;
foo *= 17

because 'foo' is just a value, it lives entirely on the stack and doesn't need to be allocated.

In Java, when you declare a variable of a complex type (say, a HashMap) it doesn't live on the stack, and it has to be allocated. In C++, when you declare a variable of a complex type, that's exactly what you get, and it lives on the stack. If you want a reference type, you have to explicitly ask for it.

The term to look for is 'auto variables'. Variables declared inside a function are 'auto variables', they have their constructors called when execution gets to their declaration, and they have their destructors called when their scope is left. Here's a link to the appropriate Wikipedia page.

It's a really useful behavior, and it lies behind a common C++ idiom called RAII.

If you want to call a non-default constructor of the type, you can just add parentheses:

SomeType foo(p1, p2);

That calls SomeType::SomeType(p1, p2) to construct foo.

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