定义uint64_t常数的最佳/正确方法 [英] Best/proper way to define uint64_t constants

问题描述

constexpr auto v = static_cast<std::uint64_t>(1) << 32;并不理想，因为语法繁琐且强制转换在语义上是间接的.从此线程中，我了解到constexpr auto v = UINT64_C(1) << 32;，但是，

constexpr auto v = static_cast<std::uint64_t>(1) << 32; is not ideal, because of the tedious syntax and the cast which is semantically indirect. From this thread, I learned constexpr auto v = UINT64_C(1) << 32; However, the precise semantics of the macro is

扩展为具有指定值的整数常量表达式 根据其参数和类型uint_least64_t.

expands to an integer constant expression having the value specified by its argument and the type uint_least64_t.

因此，它不完全是uint64_t.我想知道定义uint64_t常数的最佳/正确方法是什么.

Therefore, it's not exactly uint64_t. I'm wondering what is the best/proper way to define uint64_t constants.

请注意，unsigned long long不一定映射到uint64_t.

Note that unsigned long long doesn’t necessarily map to uint64_t.

更新

我不愿意使用函数式/C风格的强制转换，因为有些人(例如Google C ++编码风格)说它是C语言，现代C ++应该避免使用它们.看来我应该对此有自己的见解，而不是一味地跟从别人.

I'm reluctant to use functional/C-style cast, because some (e.g., Google C++ Coding Style) say it's from C, and modern C++ should avoid using them. Looks like I should have my own opinion on this, instead of blindly following others.

推荐答案

因此，它不完全是uint64_t.我想知道定义uint64_t常数的最佳/正确方法是什么.

Therefore, it's not exactly uint64_t. I'm wondering what is the best/proper way to define uint64_t constants.

如果您想绝对确定，可以写uint64_t(1)(可能是std::合格的)来代替static_cast.

If you want to be absolutely sure, you can write uint64_t(1) (possibly std::-qualified) instead as a shorter alternative to your static_cast.

出于实际目的，UINT64_C很好.如果实现不提供满足其要求的任何类型，则不需要定义uint64_t.这样做很有意义，因为可以根据uint_least64_t定义UINT64_C.

For practical purposes, UINT64_C is fine though. Implementations aren't required to define uint64_t if they don't provide any type which meets its requirements. It makes sense because of that to define UINT64_C in terms of uint_least64_t.

但是在确实存在uint64_t的实现上-并且您已经隐含地假设您正在使用这样的实现-uint64_t和uint_least64_t必须具有完全相同的范围，这是否非常需要实施者的恶意行为使其无法成为完全相同的类型.

But on implementations where uint64_t does exist -- and you're implicitly already assuming you're using such an implementation -- uint64_t and uint_least64_t must have the exact same range, are it would pretty much require malice on the implementor's part to not make them the exact same type.

P.S.由于这是标记为Language-lawyer的标记，因此无法保证uint64_t(1) << 32仍具有类型uint64_t.假设，一个实现可以提供大于64位的int类型，在这种情况下，uint64_t将被提升为int.

P.S. Since this is tagged language-lawyer: there is no guarantee that uint64_t(1) << 32 still has type uint64_t. Hypothetically, an implementation may provide a larger-than-64-bit int type, in which case uint64_t would be promoted to int.

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