Syntax problem

[Jay03]

This syntax with all those brackets is scaring me.... Can someone please tell me what is going on? for ( unsigned int j = 0; j < ms_numInstances; j++ ) { ( *( (Ball*)(ms_instances[j]) ) ).m_stateTimeInterval = timeInterval; } Thank You! :)

[Kharfax]

Is casting whatever there is inside ms_instances array to a Ball object pointer, and then is refering to that object that is beign pointed, you can resume it like this: Ball b; b.m_stateTimeInterval = timeInterbal; Obviously dont replace the code, is just for you to see the same in a more simpler way Hope it helps

[David Crow]

It could also be written as:

for (...)
{
Ball *pBall = ms_instances[j];
( *pBall ).m_stateTimeInterval = timeInterval;
// or
pBall->m_stateTimeInterval = timeInterval;
}

---

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[Viorel]

I would prefer this:

for ( unsigned int j = 0; j < ms_numInstances; j++ )
{
    Ball * const ball = ms_instances[j];
    ball->m_stateTimeInterval = timeInterval;
}

Now it seems much clearer, I think. The ms_instances array contains pointers to Ball objects. This loop initializes the m_stateTimeInterval field of all of Ball objects.

[Chris Losinger]

ms_instances is apparently an array of pointers to instances of some kind of object. this code gets element 'j' (a pointer) out of the array, then casts that pointer to a pointer-to-a-Ball: "compiler, treat this pointer as if it was a pointer to a Ball, instead of a pointer to whatever it is actually pointing at. then it dereferences the pointer and sets the m_state* member variable of what we all hope is an actual Ball object.

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[Zac Howland]

It is nothing more than the equivalent of doing the following:

for (unsigned int j = 0; j < ms_numInstances; j++)
{
	Ball* pBall = (Ball*)ms_instances[j];
	pBall->m_stateTimeInterval = timeInterval;
}

If you decide to become a software engineer, you are signing up to have a 1/2" piece of silicon tell you exactly how stupid you really are for 8 hours a day, 5 days a week Zac

[Jay03]

Viorel. wrote:

The ms_instances array contains pointers to Ball objects

What happens when j = 0; Ball * const ball = ms_instance[0]; Does this mean it has no pointers because ms_instance[0] ? Do you need atleast j = 1 to create a pointer ........ like ms_instance[1] means 1 pointer .... ms_instance[2] means array of two pointers......... so what does 0 mean? ball->m_stateTimeInterval = timeInterval; -------------------------------------------------------------------------------------------- So for each j value, there will be an array of pointers created?

[Zac Howland]

Jay03 wrote:

Ball * const ball = ms_instance[0];

Arrays in C/C++ are 0-based. This line will access the first element in the ms_instance array and cast it to a pointer-to-a-Ball.

If you decide to become a software engineer, you are signing up to have a 1/2" piece of silicon tell you exactly how stupid you really are for 8 hours a day, 5 days a week Zac