Given the code below:
double** vr(new double*[nx]);
vr[i]=new double[ny];
for(int i=0;i<nx;++i) { //Loop 200times
for(int j=0;j<ny;++j) { //Loop 200times
vr[i][j]=double(i*i*i*j*c1);
}
}
For vr[i][j] = double(i * i * i * j * c1);
why is it not required to address the value using * for example *(vr[i][j]).
It is just an address isnt it?
vr[i][j] is a double not an address.
Basically, if p is of type T* then p[i] means *(p+i) and is of type T, where T can be any type.
In your case vr is double**, so vr[i] is double* and vr[i][j] is double.
When you do vr[i] you're already dereferencing because vr[i] is equivalent to *(vr + i). This means that vr[i][j] is double dereferencing, and is equivalent to *(*(vr + i) + j).
See pointer arithmetic for further information.
From dcl.array/6
the subscript
operator []is interpreted in such a way thatE1[E2]is identical to*((E1)+(E2))([expr.sub]). Because of the conversion rules that apply to+, ifE1is an array andE2an integer, thenE1[E2]refers to theE2-thmember ofE1. Therefore, despite its asymmetric appearance, subscripting is a commutative operation. — end note ]
Thus, operator[], the subscript operator, works the same as using operator* (unary) for indirection.
int a[5] = {1, 2, 3, 4, 5}; // array
int* pa = a; // pointer to a
std::cout << pa[2]; // (ans. 3) using subscript operator
std::cout << *(pa + 2); // (ans. 3) using indirection, accessing A's 3rd element
Subscript operator ([i]) does the dereferencing. It says to shift on i "cells" and get the object located there.
+ summarize? You used c-style pointers. They are not classes. But, for example, unique_ptr has overloaded [] operator.