newbie question on vectors

[SirNitro]

I'm not having much luck with this one thing I'm trying to do with vectors (apvectors specifically. I've been using them since I first started). Basically, I have a pointer to a struct. One of the struct's members is a vector of pointers to another type of struct. What I want to do is access the members of the structs in the vector, but I can't figure out how, and I can't find much help on it. Here's basically what it's like:

struct somedata
{
	int x = 5;
	char y = '?';
};

typedef somedata *p_somedata;

struct example
{
	apvector<p_somedata> thisVector(;
	//other data members
};

typedef example *p_ex;
p_ex linkedList; 

void getMembers(p_ex &linkedList;
{
	int i = 0;
	p_somedata c;
	c = linkedList;
	
	//now, here's where my problem is
	cout << c->thisVector //now what?
}

The compiler wants me to do something like

	cout << c->thisVector::x[i];

but that doesn't seem right. I thought :: was only used for classes and the like, not vectors and their members. Plus, with

[i]

at the end, I think it would only refer to the x, not the vector. I'm probably overlooking something simple, but I don't know what, so could anyone help me out? I'd greatly apprectiate it.

 

[BoulderBum]

Here's one approach:

vector.at(?) will return the vector's object at '?' position. After such a declaration, you treat the returned object as you would a normal instantiation of an object. For instance, a vector of vectors might look like this:

int main()
{
std::vector< std::vector<int> > v;
std::vector< int > subV1;
std::vector< int > subV2;

subV1.push_back( 5 );
subV1.push_back( 7 );

subV2.push_back( 9 );
subV2.push_back( 3 );


v.push_back( subV1 );
v.push_back( subV2 );

//print the first value of [at(0)] of the second vector [v.at(1)]
//v.at(1) returns subV1, so v.at( 1 ).at( 0 ) == subV2.at( 0 )
cout << v.at( 1 ).at( 0 ) << endl;

return 0;
}

For a vector of structs, use .at() and treat the struct as you normally would with the dot or arrow operator.

 

[SirNitro]

Ah, ok. Thanks, that helped a lot.

 

[BoulderBum]

No prob.