This seem just dummy question.
What is a problem. I have some old project, that works on 8 bit prosecos. Now I reforming for arm7 procesor.
Ok, I find out, that in old procesor is int 16 bit type, in arm is 32 bit type, no problem. In rtl.h are definitions like U32 as unsigned int, ... and on the and U8 as unsigned char. Sizeof(U8) is 1 - 1 byte (8 bit).
Than i define my own structure for exaple:
typedef struct { U8 byte_1; U16 bytes_2; } MY_STRUCTURE;
sizeof(MY_STRUCTURE) is 4!!! How could be this posible?
" in old procesor is int 16 bit "
To be precise, this is defined by the compiler, not the processor.
The compiler writers could have decided to make their 'int' 24 or 32 bits - or anything else that took their fancy!
That would, of course, have been a very unusual - even bizarre - decision; but they might have had their reasons, and there's nothing to stop them...!
IBM has an old system with 36-bit integers :)
This results in a char either having 7 or 9 bits, depending on if the compiler fints 4 or 5 characters into an int.
Your structure really is 4 bytes, so sizeof(x) should return 4. Four of your structures will consume 16 bytes, so if you copy 16 bytes from the memory card into an array of structures, all data will be completely messed up. You will take 5 1/3 structures from the memory card and copy into 4 structures in the ARM memory...
You are now learning that you should never use memcopy to import/export data between different systems. The only portable way is to read the data from the memory card as an array of bytes, and then explode the data into structures, field by field.
In some situations, you may use a packed structure - i.e. requesting that the compiler should ignore data align and store the information without internal padding. The problem with packed data is that the processor has to take extra steps whenever it is processing the misaligned fields. The compiler may generate code that read your 16-bit variable as two 8-bit values and then have one of the bytes left-shifted 8 bits and combined with the other byte. That hurts both code size and speed.
Even if you use packed arrays, you have to take byte order into consideration. Is the external data stored with least and most significant byte of multi-byte integers placed in the same order as your processor needs? If not, then you must either byte-swap all multi-byte fields in the struct after read or before write, or you must write your own code to pack/unpack from a byte array.
"Is the external data stored with least and most significant byte of multi-byte integers placed in the same order as your processor needs? If not, then you must either byte-swap all multi-byte fields in the struct after read or before write, or you must write your own code to pack/unpack from a byte array."
I did like you said.
part of code:
data->ip_1 = MK_U32_FROM8S(buf[counter], buf[counter+1], buf[counter+2], buf[counter+3]); #define MK_U32_FROM8S(high8,med_high8,med_low8,low8) \ MK_U32(MK_U16(high8,med_high8), MK_U16(med_low8, low8)) #define MK_U32(high16,low16) \ ((U32)( ((U32)(high16) << 16) | (U32)(low16) )) #define MK_U16(high8,low8) \ ((U16)( ((U16)(high8) << 8) | (U16)(low8) ))
This way I'm realy sure, that I get regular value.
Thanks four yours explanations, I learned o lot of new and useful things.
Peter