Question

In the bit-shifting example shown
here:

 unsigned long int longInt = 1234567890;
 unsigned char byteArray[4];

 // convert from an unsigned long int to a 4-byte array
 byteArray[0] = (int)((longInt >> 24) & 0xFF) ;
 byteArray[1] = (int)((longInt >> 16) & 0xFF) ;
 byteArray[2] = (int)((longInt >> 8) & 0XFF);
 byteArray[3] = (int)((longInt & 0XFF));

Three questions:

1. Why is it (int) instead of (unsigned char)? I tried it with unsigned char and it seems to compile just fine.
2. Is 0XFF necessary? Isn’t the new bit shifted-in 0 because Wikipedia says C uses logical shifting and logical shifting shifts in 0? (EDIT: at least it doesn’t seem necessary on one with >> 24?)
3. Can’t I just do a memcpy() to copy longInt to a unsigned char buffer? Is it not so because of issue with Endianness? Is there any other reason?

Answer 1

1.

((longInt >> 24) & 0xFF) expression is of type unsigned long int. With the cast to int the expression is first converted to int then to unsigned char. If you don’t cast to int the expression is not first converted to int. There are no difference in the two situations and the cast is superfluous.

2.

The 0xff is not necessary. The conversion to unsigned char actually performs the same.

3.

You can use memcpy but it is not portable because it depends on the endianness of the system. It will give different results if the system is big endian or little endian while the bitwise shift solution will give the same results.