In my linker script file, I have defined two symbols
define symbol _region_RAM_start__ = 0xC0000000;
define symbol _region_RAM_end__ = 0xC00fffff;
Below code should works as expected:
extern const volatile unsigned int _region_RAM_start__;
extern const volatile unsigned int _region_RAM_end__;
....
int GetRAMSize()
{
int size = 0;
unsigned int address_1 = &_region_RAM_start__;
unsigned int address_2 = &_region_RAM_end__;
size = address_2 - address_1 + 1U;
return size;
}
Since they are generic address symbols you are trying to access and not necessarily pointers to a specific type, you don't want to declare them unsigned int, rather declare them as
extern void _region_RAM_START;
then &_region_RAM_START will have the appropriate type 'void *'.
That's a bit old but i will answer it anyways…
From the ld manual:
Accessing a linker script defined variable from source code is not intuitive. In particular a linker script symbol is not equivalent to a variable declaration in a high level language, it is instead a symbol that does not have a value.
Before going further, it is important to note that compilers often transform names in the source code into different names when they are stored in the symbol table. For example, Fortran compilers commonly prepend or append an underscore, and C++ performs extensive name mangling. Therefore there might be a discrepancy between the name of a variable as it is used in source code and the name of the same variable as it is defined in a linker script. For example in C a linker script variable might be referred to as:
extern int foo;
But in the linker script it might be defined as:
_foo = 1000;
In the remaining examples however it is assumed that no name transformation has taken place.
When a symbol is declared in a high level language such as C, two things happen. The first is that the compiler reserves enough space in the program's memory to hold the value of the symbol. The second is that the compiler creates an entry in the program's symbol table which holds the symbol's address. ie the symbol table contains the address of the block of memory holding the symbol's value. So for example the following C declaration, at file scope:
int foo = 1000;
creates a entry called "foo" in the symbol table. This entry holds the address of an int sized block of memory where the number 1000 is initially stored.
When a program references a symbol the compiler generates code that first accesses the symbol table to find the address of the symbol's memory block and then code to read the value from that memory block. So:
foo = 1;
looks up the symbol foo in the symbol table, gets the address associated with this symbol and then writes the value 1 into that address. Whereas:
int * a = & foo;
looks up the symbol foo in the symbol table, gets it address and then copies this address into the block of memory associated with the variable "a".
Linker scripts symbol declarations, by contrast, create an entry in the symbol table but do not assign any memory to them. Thus they are an address without a value. So for example the linker script definition:
foo = 1000;
creates an entry in the symbol table called @samp{foo} which holds the address of memory location 1000, but nothing special is stored at address 1000. This means that you cannot access the value of a linker script defined symbol - it has no value - all you can do is use the address of a linker script defined symbol.
Hence when you are using a linker script defined symbol in source code you should always take the address of the symbol, and never attempt to use its value. For example suppose you want to copy the contents of a section of memory called .ROM into a section called .FLASH and the linker script contains these declarations:
start_of_ROM = .ROM; end_of_ROM = .ROM + sizeof (.ROM); start_of_FLASH = .FLASH;
Then the C source code to perform the copy would be:
extern char start_of_ROM, end_of_ROM, start_of_FLASH; memcpy (& start_of_FLASH, & start_of_ROM, & end_of_ROM - & start_of_ROM);
Note the use of the "&" operators. They are correct.