Typecasting pointers

Question

I have a variable

UInt32 PageAddr = 0x80000000;

and a function

UInt8 Blank_Check(const UInt8 *Address)

I am calling the function after typecasting variable.

Blank_Check((UInt8 *)&PageAddr);

What will be the value that will be passed to the blank check. I tried to analyze but I am unable to understand how it works?

Answer 1

Short version: remove the & character.

Longer version:

It will be exactly the same pointer value. The type is only there to help the C compiler know how to index and dereference the pointer.

To clarify the difference between value and pointer, try:

UInt32 PageAddr = 0x80000000;
printf("%d,   %p\n", PageAddr, &PageAddr);

By the name of variable, I guess you really want to do something like:

Uint32* PageAddr = 0x80000000;   /* as a pointer */

Blank_Check((UInt8 *)PageAddr);  /* cast pointer to another kind of pointer */

Or if you want to keep the pointer represented as an integer (works on almost all platforms, but is not 100% portable), like this:

Uint32 PageAddr = 0x80000000;   /* as an integer */

Blank_Check((UInt8 *)PageAddr); /* cast integer to pointer */

Answer 2

The passed value is the address where the variable PageAddr has been put in memory by the compiler. That's the passed value.

Actually, C is a pass-by-value only language. When you pas a variable, the variable's value is passed. When you pass a pointer, like &PageAddr the address of PageAddr becomes the value that is passed.

Later, in the called function, you can use the address-value to look or changed what's stored in that address.

This is in contrast to languages that allow passing also by reference, like C++ or Java, where a reference to the value is passed, in which case anything you do to or with the reference, is actually done to the original variable.