Why do random access iterator's arithmetic operators accept / return int and not size_t?

Question

Since most operations on std::vector require / return size_t - that\'s the type I use for indexing. But now I\'ve enabled all compiler warnings to

Answer 1

I've got a lot of other similar messages suggesting that iterator's arithmetic operators accept and return int.

Not necessarily int. It's the (signed) difference_type defined by the iterator type's iterator_traits. For most iterator types, this defaults to ptrdiff_t.

Why not size_t?

Because arithmetic needs to work correctly with signed values; one would expect it + (-1) to be equivalent to it - 1.

Answer 2

It allows for things like it += index; where index can be both positive or negative (according to some logic).

Comparing with the following:

if (some_condition)
    it += index;
else
    it -= index;

Which would be needed if we could only pass unsigned values.