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Duncan_L
n00b
Joined: 06 Sep 2008
Posts: 42
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 Posted: Sun Sep 21, 2008 2:06 am Post subject: C++ passing class instance as another class's member - help |
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| Code: | #include <iostream>
using std::cout;
using std::cin;
using std::endl;
struct myClassA
{
int* my_array; //getting read to make a dynamically allocated array
myClassA()
{my_array=new int[5];} //dynamically allocates array, my_array, with arbitrary size 5
};//end of myClassA structure
struct myClassB
{
myClassA* pt; //I want to be able to use an instance of myClassA as a member in myClassB. I intend to pass it via the constructor.
myClassB(myClassA* myPt) //Constructor takes pointer to instance of myClassA
{pt=myPt;} //I set myClassB's member, pt, to the pointer passed as an argument in the constructor
void myMethod()
{cout<<*pt.my_array[2]<<endl;} //theoretically this should work right??
//error: request for member 'my_array' in '((myClassB*)this)->myClassB::pt', which is of non-class type 'myClassA*'
};
int main()
{return 0;}
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Can anybody help fix the error I'm getting? (It's a few lines from the end) |
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timeBandit
Bodhisattva
Joined: 31 Dec 2004
Posts: 2719
Location: here, there or in transit
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| Posted: Sun Sep 21, 2008 3:42 am Post subject: |
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It's a common C pitfall--a simple matter of operator precedence. You need to write either (*pt).my_array[2] or pt->my_array[2].
The dot operator has higher precedence than the dereference operator (*). As written, you've attempted to access a member (my_array) of a pointer, which is of course impossible. Use parentheses to force the pointer dereference to be evaluated first: (*pt). Then you can access members of the pointed-to structure. This is such a common idiom that the language provides an additional operator (->) for the purpose.
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Duncan_L
n00b
Joined: 06 Sep 2008
Posts: 42
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| Posted: Sun Sep 21, 2008 6:48 am Post subject: |
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| oh, cool thanks. |
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Portage & Programming
