Confused about the use of ptrdiff_t in C++

Question

I need to translate this line of code in Java and I am not sure what to do about ptrdiff_t. Not sure what it does here. By the way, mask_block is of type size_t.

size_t lowest_bit = mask_block & (-(ptrdiff_t)mask_block);

Thanks

Answer 1

Beware! This is bit magic!

( x & ~(x-1) ) returns the lowest set bit in an expression. The author of the original code decided to use ( x & (-x) ) which is effectively the same due to the two's comlement representation of integers. But (the original author thought that) to get -x you need to use signed types and, as pointed out earlier, ptrdiff_t is signed, size_t is unsigned.

As Java does not have unsigned types, mask_block will be int and mask_block & (-mask_block) will work without any issue.

Note that due to the interoperability between signed and unsigned types, the cast is superfluous in C++ as well.

Answer 2

ptrdiff_t is the type that should be used for the (integer) difference between two pointers. That is, the result of subtracting one pointer from another. It is a signed integer, and should be large enough to stroe the size of largest possible array (so in Java, that would simply be an int, I'd guess)