C中的指针铸造

Question

Can someone please explain me what these 2 lines mean?

short * two_byte_pointer = (short*) 10;

int * four_byte_pointer = (int*) 10;

I thought that we can only use int* before the equal to sign. But how is it possible to cast a number with short* ? I'm sorry if this is a dumb question. I just started coding in C so I'm very confused. Can you please explain me how pointers are also casted in C?

Answer 1

Your code (for example first line) means:

Dear compiler, here is a pointer to short short * .
It is named two_byte_pointer .
I would like it to have a value assigned = .
The value I am thinking of is something else, but I want you to trust me that it is a pointer to short - honest (short*) .
The value I am using is 10.

And I would not believe a word of it if I were your compiler or your colleague reading the code. Not without some explanatory comments and a quote from the memory part of the description of your environment.

It might make sense in some embedded environments with memory mapped special registers at an unusual location. Maybe not, see the comments by the busybee below.
I am pretty sure that this is just a very theoretical example and "10" is just "a number example".

The naming of the variables with two_byte and four_byte is based on assumptions on the sizes of types in your environment. Apart from that I would name it pointer_to_two_byte , slightly longer, much more precise (not discussing naming style otherwise here).

Answer 2

If you cast something the compiler just takes whatever value you provide and interprets it as the type you cast it to. So in this example it will interpret 10 as a pointer to a short . The value of a pointer is the memory address it points to. Using this pointer will almost certainly fail because most systems don't allow you to access memory at such a low address.

Casting a number to a pointer can be useful if you need to do calculations on an address and use the result as a pointer.