CODEWITHSUNDEEP
c++
Toby
Toby

Reputation: 11

expression must have class type: c++ passing instance variable object as a pointer (dot notation parameter)

I've been looking for an answer for 2 days; I'm at a complete loss.

I've got...

struct shipDef
{ 
    int mBaseHealth; 

    b2BodyDef mBodyDef;
};

...and in another file...

b2Body* b2World::CreateBody(const b2BodyDef* def)
{
   //...
}

...and also...

shipDef * Game::getShipDef(const char* shipDefID)    
{    
    std::pair<std::multimap<std::string,shipDef*>::iterator, std::multimap<std::string,shipDef*>::iterator> iteratorPair = mShipDefMap.equal_range(shipDefID);

    std::size_t shipDefIDSize = std::distance( iteratorPair.first, iteratorPair.second );
    std::size_t randomSoundIndex = std::rand() % shipDefIDSize;
    std::advance( iteratorPair.first, randomSoundIndex );

    return iteratorPair.first->second;    
}

I'm calling...

mPhysicsBody = physicsWorld->CreateBody(Game::getInstance()->getShipDef(shipType).mBodyDef);

I'm getting an error:

expression must have class type

on the call to CreateBody.

I've tried every arrangement of the 'address of' operator I can think of and I've been all up and down google and stack overflow.

edited to include "b2BodyDef mBodyDef;"

Upvotes: 1

Views: 239

Answers (1)

filmor
filmor

Reputation: 32182

If getShipDef returns a pointer you have to use the -> operator to access its members:

mPhysicsBody = physicsWorld->CreateBody(
     Game::getInstance()->getShipDef(shipType)->mBodyDef
);

If a is a pointer the expression a->b is equivalent to (*a).b so in your case you could have also written

mPhysicsBody = physicsWorld->CreateBody(
     (*(Game::getInstance()->getShipDef(shipType))).mBodyDef
);

Upvotes: 3

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