Should I use INT64_C as "int64_t literal" ...?

Question

How can I write portable code to compare a constant integer value with an int64_t variable across different platforms (MacOS and Ubuntu)?

int64_t a = 2;
std::min(1, a);

• Fails to compile on MacOS when I use 1L as the constant value.
• Using 1LL also fails to compile on Ubuntu 20.04.

I found INT64_C as a potential solution, but the documentation is unclear.

Answer 1

int64_t a = 2;
std::min(1, a); // error (unless int64_t is int)

Replacing 1 by INT64_C(1) is likely to "work", but it's not guaranteed. The INT64_C macro expands to an expression of type int_least64_t. That's likely to be the same type as int64_t, but it's not guaranteed. For example, if long and long long are both 64 bits, it's possible that int_least64_t could be long and int64_t could be long long. There's not much reason for an implementation to define them differently, but you shouldn't rely on that.

It's defined that way because, in most contexts, a constant of type int_least64_t is usable in a context where you need an int64_t value, since it will likely be promoted. Arguments to std::min are an exception; you need an expression that's actually of type int64_t.

There is no syntax for defining a literal of type int64_t. But you can have a constant expression of that type by using a conversion:

int64_t a = 2;
std::min(int64_t(1), a);

Or you can define a constant of the right type and use that instead of the literal:

std::int64_t a = 2;
constexpr int64_t one = 1;
std::min(one, a);

Answer 2

The first answer provided quite exotic solutions. What you need is just setting the parameters type:

int64_t a = 2;
std::min<int64_t>(1, a);