从size_t转换为int，或者用size_t迭代？

Question

Is it better to cast the iterator condition right operand from size_t to int , or iterate potentially past the maximum value of int ? Is the answer implementation specific?

int a;
for (size_t i = 0; i < vect.size(); i++)
{
    if (some_func((int)i))
    {
        a = (int)i;
    }
}

int a;
for (int i = 0; i < (int)vect.size(); i++)
{
    if (some_func(i))
    {
        a = i;
    }
}

Answer 1

I almost always use the first variation, because I find that about 80% of the time, I discover that some_func should probably also take a size_t.

If in fact some_func takes a signed int, you need to be aware of what happens when vect gets bigger than INT_MAX . If the solution isn't obvious in your situation (it usually isn't), you can at least replace some_func((int)i) with some_func(numeric_cast<int>(i)) (see Boost.org for one implementation of numeric_cast). This has the virtue of throwing an exception when vect grows bigger than you've planned on, rather than silently wrapping around to negative values.

Answer 2

I'd just leave it as a size_t , since there's not a good reason not to do so. What do you mean by "or iterate potentially up to the maximum value of type_t"? You're only iterating up to the value of vect.size() .

Answer 3

For most compilers, it won't make any difference. On 32 bit systems, it's obvious, but even on 64 bit systems, both variables will probably be stored in a 64-bit register and pushed on the stack as a 64-bit value.

If the compiler stores int values as 32 bit values on the stack, the first function should be more efficient in terms of CPU-cycles.

But the difference is negligible (although the second function "looks" cleaner)