Endian-ness in a char array containing binary characters
I'm building some code to read a RIFF wav file and I've bumped into something odd.
The first 4 bytes of the file header are the word RIFF in big-endian ascii coding:
0x5249 0x4646
I read this first element using:
char *fileID = new char[4];
filestream.read(fileID,4);
When I write this to screen the results are as expected:
std::cout << fileID << std::endl;
>> RIFF
Now, the next 4 bytes give the size 开发者_开发技巧of the file, but crucially they're little-endian.
So, I write a little function to flip the bytes, based on a union:
int flip4bytes(char* input){
union flip {int flip_int; char flip_char[4];};
flip.flip_char[0] = input[3];
flip.flip_char[1] = input[2];
flip.flip_char[2] = input[1];
flip.flip_char[3] = input[0];
return flip.flip_int;
}
This looks good to me, except when I call it, the value returned is totally wrong. Interestingly, the following code (where the bytes are not reversed!) works correctly:
int flip4bytes(char* input){
union flip {int flip_int; char flip_char[4];};
flip.flip_char[0] = input[0];
flip.flip_char[1] = input[1];
flip.flip_char[2] = input[2];
flip.flip_char[3] = input[3];
return flip.flip_int;
}
This has thoroughly confused me. Is the union somehow reversing the bytes for me?! If not, how are the bytes being converted to int correctly without being reversed?
I think there's some facet of endian-ness here that I'm ignorant to..
You are simply on a little-endian machine, and the "RIFF" string is just a string and thus neither little- nor big-endian, but just a sequence of chars. You don't need to reverse the bytes on a little-endian machine, but you need to when operating on a big-endian.
You need to figure of the endianess of your machine. #include <sys/param.h>
will help you do that.
You could also use the fact that network byte order is big ended (if my memory serves me correctly - you need to check). In which case convert to big ended and use the ntohs
function. That should work on any machine that you compile the code on.
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