How to declare linker symbol in C files

David,

This works for me.

// declare the external.

extern unsigned char *__bss_start;

// somewhere in a function.

printf("%u\n",(unsigned int)&__bss_start);

The print shows my correct start of bss.

Maybe you are trying to access the 'content of' rather than 'address of'.

Banx.

Thanks very much for your reply...

I am still a little confuse...

__bss_start is a pointer, it's value equal to the start address of bss section

why

printf("%u\n",(unsigned int)&__bss_start); but not

printf("%u\n",(unsigned int)__bss_start);

Thanks again,

David

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originally posted by david_cai@Aug 17 2006, 06:10 AM

thanks very much for your reply...

i am still a little confuse...

__bss_start is a pointer, it's value equal to the start address of bss section

why

printf("%u\n",(unsigned int)&__bss_start); but not

printf("%u\n",(unsigned int)__bss_start);

thanks again,

david

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You&#39;ve told the compiler that __bss_start is a pointer, with space allocated elsewhere. But there is no space allocated, so the only way to use your new pointer is to take the address of it, which gives you the value of __bss_start.

I do it this way when the symbol is within range of the global pointer:

extern unsigned char __some_near_symbol;
unsigned int NearSymbolAddr;
NearSymbolAddr = (unsigned int) &__some_near_symbol;

or if not in range of the global pointer:

extern unsigned char  __some_far_symbol;  // Use  to force compiler to not use global pointer.
unsigned int FarSymbolAddr;
FarSymbolAddr = (unsigned int) __some_far_symbol;

Thanks very much...

I almost got it... But I&#39;m still confusing that how to understand the symbol in linker script. It seems that we can take it as a variable in source files, but in fact, there is no space allocated, why don&#39;t we take it as a constant and why operator& which will return its address can get its actual value???

Thanks again,

david

David,

__bss_start isn&#39;t a pointer, nor does it hold anything. It&#39;s just a number generated by the linker that states where the bss starts. i.e. __bss_start does not occupy any memory space, it is a label. You could just as easily declare it as a short or a char.

extern char __bss_start;

In this case, if you printf("%d\n", __bss_start); then you will be printing the CONTENTS OF the first byte of bss, which you have fooled the compiler into thinking is a char that actually occupies a memory location. Just as if you had :

char fred;

fred = 42;

printf("%d\n", fred);

Using the &#39;&&#39; just makes the compiler print the ADDRESS OF (i.e. the value of the label) instead of CONTENTS OF what the label is pointing to.

I hope that helps.

Banx.

Hi,

I have done a further study, I find that no matter how you declare the __bss_start,such as:

extern alt_u8 __bss_start;

extern alt_u8* __bss_start;

extern alt_u16 __bss_start;

extern alt_u16* __bss_start;

extern alt_u32 __bss_start;

extern alt_u32* __bss_start;

Then

printf("\nThe value of __bss_start: %d\n", (unsigned int)&__bss_start);

You&#39;ll get the same answer which equal to the value of __bss_start in linker script.

So I think in order to use symbols defined in linker script, we should declare them with extern keyword, then use operator& to get their value.

Am I right?

Thanks for any reply,

David

Yeah,

Since it isn&#39;t really a pointer to anything I guess most folk would say that it should be declared as

extern void* __bss_start;

Banx.