CODEWITHSUNDEEP
carraysbit-manipulation
Yanlei Zhao
Yanlei Zhao

Reputation: 33

Mask or not mask when converting int to byte array?

Say you have a integer and you want to convert it to a byte array. After searching various places I've seen two ways of doing this, one with is shift only and one is shift then mask. I understand the shifting part, but why masking?

For example, scenario 1:

uint8 someByteArray[4];
uint32 someInt;

someByteArray[0] = someInt >> 24;
someByteArray[1] = someInt >> 16;
someByteArray[2] = someInt >> 8;
someByteArray[3] = someInt;

Scenario 2:

uint8 someByteArray[4];
uint32 someInt;

someByteArray[0] = (someInt >> 24) & 0xFF;
someByteArray[1] = (someInt >> 16) & 0xFF;
someByteArray[2] = (someInt >> 8) & 0xFF;
someByteArray[3] = someInt & 0xFF;

Is there a reason for choosing one over the other?

Upvotes: 3

Views: 1328

Answers (1)

Peter
Peter

Reputation: 36627

uint8 and uint32 are not standard types in C. I assume they represent 8-bit and 32-bit unsigned integral types, respectively (such as supported by Microsoft compilers as a vendor-specific extension).

Anyways ....

The masking is more general - it ensures the result is between 0 and 0xFF regardless of the actual type of elements someByteArray or of someInt.

In this particular case, it makes no difference, since the conversion of uint32 to uint8 is guaranteed to use modulo arithmetic (modulo 0xFF + 0x01 which is equal to 0x100 or 256 in decimal). However, if your code is changed to use variables or arrays of different types, the masking is necessary to ensure the result is between 0 and 255 (inclusive).

With some compilers the masking stops compiler warnings (it effectively tells the compiler that the expression produces a value between 0 and 0xFF, which can be stored in a 8 bit unsigned). However, some other compilers complain about the act of converting a larger type to an 8 bit type. Because of that, you will sometimes see a third variant, which truly demonstrates a "belts and suspenders" mindset.

uint8 someByteArray[4];
uint32 someInt;

someByteArray[0] = (uint8)((someInt >> 24) & 0xFF);
someByteArray[1] = (uint8)(someInt >> 16) & 0xFF);
someByteArray[2] = (uint8)((someInt >> 8) & 0xFF);
someByteArray[3] = (uint8)(someInt & 0xFF);

Upvotes: 2

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