size_t vs int in C++ and/or C

Question

Why is it that in C++ containers, it returns a size_type rather than an int? If we\'re creating our own structures, should we also be encouraged to

Answer 1

void f1(size_t n) {
    if (n <= myVector.size()) { assert(false); }
    size_t n1 = n - myVector.size(); // bug! myVector.size() can be > n       
    do_stuff_n_times(n1);
}

void f2(int n) {
    int n1 = n - static_cast(myVector.size());
    assert(n1 >= 0);
    do_stuff_n_times(n1);
}

f1() and f2() both have the same bug, but detecting the problem in f2() is easier. For more complex code, unsigned integer arithmetic bugs are not as easy to identify.

Personally I use signed int for all my sizes unless unsigned int should be used. I have never run into situation where my size won't fit into a 32 bit signed integer. I will probably use 64 bit signed integers before I use unsigned 32 bit integers.

The problem with using signed integers for size is a lot of static_cast from size_t to int in your code.