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Auto calculation of array size

Suppressor Chan

Dear all,

For some reason, I need to declare an array like that

#define STRING2_DESCRIPTOR_LENGTH 16
BYTE code product_string_descriptor[] = {
	STRING2_DESCRIPTOR_LENGTH,
	0x03,
	'U', 0x00,
	'S', 0x00,
	'B', 0x00,
	' ', 0x00,
	'H', 0x00,
	'U', 0x00,
	'B', 0x00
};
When I add characters to the array, I also need to increase STRING_DESCRIPTOR_LENGTH manually.
Do you know some way to calculate the size of the array automatically?
I know in assembly, I can do something like that
product_string_descriptor:
	DB	end - $
	DB	0x03,
	DB	'U', 0x00
	DB	'S', 0x00
	DB	'B', 0x00
	DB	' ', 0x00
	DB	'H', 0x00
	DB	'U', 0x00
	DB	'B', 0x00
end:
How about in C?

Parents
  • 0

    I do not know of a way to do this purely at compile time. Perhaps I'm overlooking something this morning, but this is as close as I got:

    typedef struct
	{
	U8 length;
	U8 val;
	} StrDescHdr;


U8 code pstr[] =
	{ 'U', 0, 'S', 0, 'B', 0, ' ', 0, 'H', 0, 'U', 0, 'B', 0};

typedef struct
	{
	StrDescHdr Hdr;
	U8 code* value;
	} StrDesc;

StrDesc code psd =
	{
	{ sizeof(pstr) + sizeof(StrDesc), 3 },
	pstr
	};

    It costs two extra bytes of space for the pointer to the value, though.

    The problem is that the compiler won't know the size of something until it's seen the complete definition.

    Incidentally, I've found it better practice to explicitly declare the size of such fixed-length arrays with preprocessor symbols for size, rather than using empty square brackets. Instead of

    #define STRING2_DESCRIPTOR_LENGTH 16
    BYTE code product_string_descriptor[] = {

    prefer

    #define STRING2_DESCRIPTOR_LENGTH 16
    BYTE code product_string_descriptor[STRING2_DESCRIPTOR_LENGTH] = {

    This way, the compiler can generate a warning if you have too many or too few bytes in the initializer. You'll catch the error at compile time. With the first form, the compiler will accept however many entries the initializer happens to have, and you can't be sure the array is the correct length.

Reply
  • 0

    I do not know of a way to do this purely at compile time. Perhaps I'm overlooking something this morning, but this is as close as I got:

    typedef struct
	{
	U8 length;
	U8 val;
	} StrDescHdr;


U8 code pstr[] =
	{ 'U', 0, 'S', 0, 'B', 0, ' ', 0, 'H', 0, 'U', 0, 'B', 0};

typedef struct
	{
	StrDescHdr Hdr;
	U8 code* value;
	} StrDesc;

StrDesc code psd =
	{
	{ sizeof(pstr) + sizeof(StrDesc), 3 },
	pstr
	};

    It costs two extra bytes of space for the pointer to the value, though.

    The problem is that the compiler won't know the size of something until it's seen the complete definition.

    Incidentally, I've found it better practice to explicitly declare the size of such fixed-length arrays with preprocessor symbols for size, rather than using empty square brackets. Instead of

    #define STRING2_DESCRIPTOR_LENGTH 16
    BYTE code product_string_descriptor[] = {

    prefer

    #define STRING2_DESCRIPTOR_LENGTH 16
    BYTE code product_string_descriptor[STRING2_DESCRIPTOR_LENGTH] = {

    This way, the compiler can generate a warning if you have too many or too few bytes in the initializer. You'll catch the error at compile time. With the first form, the compiler will accept however many entries the initializer happens to have, and you can't be sure the array is the correct length.

Children
  • 0 in reply to Drew Davis

    Hi Drew,

    Thanks for the suggestion of explicitly declare the size. But I think the compiler will not warn me if I have less initializer than the actual size. Most of the compiler will fill the rest of the array with zero. Am I right? Anyway, the compiler will still warn me if I supply more intializer than the actual size. So, I consider it as a better solution than before. Thanks again.