How can I let a pointer assigned with a two dimensional array?
The following code won't work.
float a1[2][2] = { {0,1},{2,3}};
float a2[3][2] = { {0,1},{2,3},{4,5}};
float a3[4][2] = { {0,1},{2,3},{4,5},{6,7}};
float** b = (float**)a1;
//float** b = (float**)a2;
//float** b = (float**)a3;
cout << b[0][0] << b[0][1] << b[1][0] << b[1][1] << endl;
a1 is not convertible to float**. So what you're doing is illegal, and wouldn't produce the desired result.
Try this:
float (*b)[2] = a1;
cout << b[0][0] << b[0][1] << b[1][0] << b[1][1] << endl;
This will work because two dimensional array of type float[M][2] can convert to float (*)[2]. They're compatible for any value of M.
As a general rule, Type[M][N] can convert to Type (*)[N] for any non-negative integral value of M and N.
If all your arrays will have final dimension 2 (as in your examples), then you can do
float (*b)[2] = a1; // Or a2 or a3
The way you do this is not legit in c++. You need to have an array of pointers.
The problem here is that the dimensions of b are not known to the compiler. The information gets lost when you cast a1 to a float**. The conversion itself is still valid, but you cannot reference the array with b[][].
You can do it explicitly:
float a1[2][2] = { {0,1},{2,3}};
float* fp[2] = { a1[0], a1[1] };
// Or
float (*fp)[2] = a1;
Try assigning b to be directly equals to a1, that's mean that the pointer b is pointing to the same memory location that pointer a1 is pointing at, they carry the same memory reference now, and you should be able to walk through the array.
