I am trying to write some processor independent code to write some files in big endian. I have a sample of code below and I can't understand why it doesn't work. All it is supposed to do is let byte store each byte of data one by one in big endian order. In my actual program I would then write the individual byte out to a file, so I get the same byte order in the file regardless of processor architecture.
#include <iostream>
int main (int argc, char * const argv[]) {
long data = 0x12345678;
long bitmask = (0xFF << (sizeof(long) - 1) * 8);
char byte = 0;
for(long i = 0; i < sizeof(long); i++) {
byte = data & bitmask;
data <<= 8;
}
return 0;
}
For some reason byte always has the value of 0. This confuses me, I am looking at the debugger and see this:
data = 00010010001101000101011001111000 bitmask = 11111111000000000000000000000000
I would think that data & mask would give 00010010, but it just makes byte 00000000 every time! How can his be? I have written some code for the little endian order and this works great, see below:
#include <iostream>
int main (int argc, char * const argv[]) {
long data = 0x12345678;
long bitmask = 0xFF;
char byte = 0;
for(long i = 0; i < sizeof(long); i++) {
byte = data & bitmask;
data >>= 8;
}
return 0;
}
Why does the little endian one work and the big endian not? Thanks for any help :-)