C++, class as parameter to a method, not template

Question

So, I came across an interesting method signature that I don't quite understand, it went along the lines of:

void Initialize(std::vector< std::string > & param1, class SomeClassName * p);

what I don't understand is the "class" keyword being used as the parameter, why is it there? Is it necessary to specify or it is purely superficial?

I saw this as well in a class member declaration and I was going crazy trying to find out what it meant. `class Foo{ private: class Bar* pointerToBar; }` — Vicro, Feb 26 '15 at 09:20

score 16 · Accepted Answer · edited May 23 '17 at 12:33

16

It is a forward declaration of the class. It is the effectively the same as

class SomeClassName;
void Initialize(std::vector< std::string > & param1, SomeClassName * p);

There are many situations in which a forward declaration is useful; there's a whole list of things that you can and can't do with a forward declaration in this answer to another question).

edited May 23 '17 at 12:33

Community

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answered May 14 '10 at 18:48

James McNellis

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does it act as forward declaration then? – ra170 May 14 '10 at 18:50
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"but you need to use a pointer or reference to an instance of the class" -> If you just have a non-definition function declaration like in this case, you can go fine with value-parameters, though. – Johannes Schaub - litb May 14 '10 at 23:30
interesting, I have never seen a forward declaration inside a function parameter before. – Remy Lebeau May 15 '10 at 01:33
@Remy: To tell the truth, I've _never_ seen this done before in real code either. @Johannes: Yeah; I posted the link to the list of what can and can't be done because I knew I'd forget something. – James McNellis May 15 '10 at 03:25

score 1 · Answer 2 · answered May 14 '10 at 18:48

1

It's not superficial. It allows you to specify "SomeClassName" as a parameter even if "SomeClassName" hasn't been defined yet.

answered May 14 '10 at 18:48

Peter Ruderman

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does it act as forward declaration then? – ra170 May 14 '10 at 18:51

score 1 · Answer 3 · answered May 14 '10 at 18:52

Most superficial. In C, when you define a struct:

struct x { 
    int member;
    char member2;
};

To use that, you have to use struct x whatever. If you want to use x as a name by itself (without the struct preceding it) you have to use a typedef:

typedef struct x { 
    int member;
    char member2;
} x;

In C++, however, the first (without the typedef) has roughly the same effect as the second (both define x so you can use it by itself, without struct preceding it). In C++, a class is essentially the same as a struct, except that it defaults to private instead of public (for both members and inheritance).

The only place there's a difference is if there is no definition of the class in scope at the point of the function declaration.

score 0 · Answer 4 · edited Feb 09 '12 at 16:45

0

it is the same as you write:

for(int i=0;i<10;i++)

as

int i; 
for(i=0;i<10;i++)

the class declaration is a definition and a pointer to it

edited Feb 09 '12 at 16:45

musefan

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answered May 07 '11 at 04:57

justCurious

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the scope of the variable i is different. – iOS Nov 22 '12 at 21:43
Does not answer the original question. – prapin Nov 22 '12 at 21:45

C++, class as parameter to a method, not template

4 Answers4