Re: C++ Syntax Confusion

On May 20, 9:46 am, coder_...@yahoo.com wrote:


hello,
according to my understanding ..In your first point..it means a
pointer of Array<int32> that points to Array<int32>(10). it's not
exactly pointing to array of 10 int32.
let take for exp..

 int *ptr = new int(10); here it's pointing to integer that's
initialized to 10.
 and
 int *ptr = new int[10]; but here points to array of 10 integer;

although explicit conversion is there..in second case "iarrayPtr[2] =
100"..and one can avoid it using explicit keyword with the class
constructor..

but coming on to your third point..


How can Array(int32 aSize) constructor be called with 100 as
aSize??..when you are passing 10 as value.?

now see code below..made default constructor as public. Set the aSize
as "10";

Array<int32> *iarrayPtr = new Array<int32>[10];

now iarryPtr is pointer that points to Array of 10 Array<int32>(10).

typedef int int32;
using namespace std;
template<class T> class Array { public:
         Array() {dataLen = 10;data = new T[dataLen];}
         Array(int32 aSize){dataLen = aSize;data = new T[dataLen];}
        virtual ~Array(){delete [] data;}
        int32 length() const{return(dataLen);}
        const T& operator[](int32 i) const {return(data[i]);}
        T& operator[](int32 i){return(data[i]);}
        T *getData() const {return(data);}
private:
       // Array();
private:
        T *data;
        int32 dataLen;
};

int main(){
Array<int32> ia(10); // Array that can contain 10 int32
ia[1] = 1; // Array element 1 set to 1
Array<int32> *iarrayPtr = new Array<int32>[10];
(*(iarrayPtr+9))[9] = 99;
cout <<(*(iarrayPtr+9))[9]<<endl;
iarrayPtr[9][9] = 100;
cout <<iarrayPtr[0].length()<<endl;
cout <<iarrayPtr[9][9]<<endl;
 return 0;
}

pls let me know if i'm wrong somewhere..:-).