我在使用 C++ 类型特征时遇到了一些奇怪的行为，并将我的问题缩小到这个古怪的小问题，我将对此进行大量解释，因为我不想留下任何误解.
I experienced some odd behavior while using C++ type traits and have narrowed my problem down to this quirky little problem for which I will give a ton of explanation since I do not want to leave anything open for misinterpretation.
假设你有一个这样的程序:
Say you have a program like so:
#include <iostream>
#include <cstdint>
template <typename T>
bool is_int64() { return false; }
template <>
bool is_int64<int64_t>() { return true; }
int main()
{
std::cout << "int: " << is_int64<int>() << std::endl;
std::cout << "int64_t: " << is_int64<int64_t>() << std::endl;
std::cout << "long int: " << is_int64<long int>() << std::endl;
std::cout << "long long int: " << is_int64<long long int>() << std::endl;
return 0;
}
在使用 GCC(以及使用 32 位和 64 位 MSVC)的 32 位编译中，程序的输出将是:
In both 32-bit compile with GCC (and with 32- and 64-bit MSVC), the output of the program will be:
int: 0
int64_t: 1
long int: 0
long long int: 1
但是，64 位 GCC 编译产生的程序将输出:
However, the program resulting from a 64-bit GCC compile will output:
int: 0
int64_t: 1
long int: 1
long long int: 0
这很奇怪，因为 long long int 是一个有符号的 64 位整数，并且就所有意图和目的而言，与 long int 和 相同int64_t 类型，因此逻辑上，int64_t、long int 和 long long int 将是等效类型 - 使用这些时生成的程序集类型相同.一看 stdint.h 就会告诉我原因:
This is curious, since long long int is a signed 64-bit integer and is, for all intents and purposes, identical to the long int and int64_t types, so logically, int64_t, long int and long long int would be equivalent types - the assembly generated when using these types is identical. One look at stdint.h tells me why:
# if __WORDSIZE == 64
typedef long int int64_t;
# else
__extension__
typedef long long int int64_t;
# endif
在 64 位编译中，int64_t 是 long int，而不是 long long int(显然).
In a 64-bit compile, int64_t is long int, not a long long int (obviously).
解决这种情况非常简单:
The fix for this situation is pretty easy:
#if defined(__GNUC__) && (__WORDSIZE == 64)
template <>
bool is_int64<long long int>() { return true; }
#endif
但这是非常骇人听闻的，并且不能很好地扩展(实体的实际功能，uint64_t 等).所以我的问题是: 有没有办法告诉编译器一个 long long int 也是一个 int64_t，就像 long int 是?
But this is horribly hackish and does not scale well (actual functions of substance, uint64_t, etc). So my question is: Is there a way to tell the compiler that a long long int is the also a int64_t, just like long int is?
我最初的想法是这是不可能的，因为 C/C++ 类型定义的工作方式.没有一种方法可以为编译器指定基本数据类型的类型等效性，因为这是编译器的工作(并且允许这样做可能会破坏很多事情)并且 typedef 只能以一种方式进行.
My initial thoughts are that this is not possible, due to the way C/C++ type definitions work. There is not a way to specify type equivalence of the basic data types to the compiler, since that is the compiler's job (and allowing that could break a lot of things) and typedef only goes one way.
我也不太在意在这里得到答案，因为这是一个超级骗子的边缘案例，我不怀疑任何人会在示例不是精心设计的时候关心(这是否意味着这应该是社区维基?).
I'm also not too concerned with getting an answer here, since this is a super-duper edge case that I do not suspect anyone will ever care about when the examples are not horribly contrived (does that mean this should be community wiki?).
Append:我使用部分模板特化而不是更简单的例子的原因:
Append: The reason why I'm using partial template specialization instead of an easier example like:
void go(int64_t) { }
int main()
{
long long int x = 2;
go(x);
return 0;
}
那个例子仍然可以编译，因为 long long int 可以隐式转换为 int64_t.
is that said example will still compile, since long long int is implicitly convertible to an int64_t.
Append:目前唯一的答案是假设我想知道类型是否为 64 位.我不想误导人们认为我关心这个问题，并且可能应该提供更多关于这个问题表现出来的例子.
Append: The only answer so far assumes that I want to know if a type is 64-bits. I did not want to mislead people into thinking that I care about that and probably should have provided more examples of where this problem manifests itself.
template <typename T>
struct some_type_trait : boost::false_type { };
template <>
struct some_type_trait<int64_t> : boost::true_type { };
在这个例子中，some_type_trait 将是一个 boost::true_type，但 some_type_trait 不会是.虽然这在 C++ 的类型概念中是有道理的，但并不可取.
In this example, some_type_trait<long int> will be a boost::true_type, but some_type_trait<long long int> will not be. While this makes sense in C++'s idea of types, it is not desirable.
另一个例子是使用像 same_type 这样的限定符(这在 C++0x 概念中很常见):
Another example is using a qualifier like same_type (which is pretty common to use in C++0x Concepts):
template <typename T>
void same_type(T, T) { }
void foo()
{
long int x;
long long int y;
same_type(x, y);
}
该示例无法编译，因为 C++(正确地)看到类型不同.g++ 将无法编译并出现如下错误:没有匹配的函数调用 same_type(long int&, long long int&).
That example fails to compile, since C++ (correctly) sees that the types are different. g++ will fail to compile with an error like: no matching function call same_type(long int&, long long int&).
我想强调的是，我理解为什么会发生这种情况，但我正在寻找一种不会强迫我到处重复代码的解决方法.
I would like to stress that I understand why this is happening, but I am looking for a workaround that does not force me to repeat code all over the place.
您无需转到 64 位即可看到此类内容.在常见的 32 位平台上考虑 int32_t.它可能是 typedef 的 int 或 long，但显然一次只能是两者之一.int 和 long 当然是不同的类型.
You don't need to go to 64-bit to see something like this. Consider int32_t on common 32-bit platforms. It might be typedef'ed as int or as a long, but obviously only one of the two at a time. int and long are of course distinct types.
不难看出，在 32 位系统上没有使 int == int32_t == long 的变通方法.出于同样的原因，在 64 位系统上没有办法使 long == int64_t == long long.
It's not hard to see that there is no workaround which makes int == int32_t == long on 32-bit systems. For the same reason, there's no way to make long == int64_t == long long on 64-bit systems.
如果可以，对于重载 foo(int)、foo(long) 和 foo(long long) 的代码，可能的后果将是相当痛苦的 - 突然他们对同一个重载有两个定义?！
If you could, the possible consequences would be rather painful for code that overloaded foo(int), foo(long) and foo(long long) - suddenly they'd have two definitions for the same overload?!
正确的解决方案是你的模板代码通常不应该依赖于一个精确的类型，而应该依赖于该类型的属性.对于特定情况，整个 same_type 逻辑仍然可以:
The correct solution is that your template code usually should not be relying on a precise type, but on the properties of that type. The whole same_type logic could still be OK for specific cases:
long foo(long x);
std::tr1::disable_if(same_type(int64_t, long), int64_t)::type foo(int64_t);
即，重载 foo(int64_t) 在 完全 与 foo(long) 相同时未定义.
I.e., the overload foo(int64_t) is not defined when it's exactly the same as foo(long).
使用 C++11，我们现在有一个标准的方法来写这个:
[edit] With C++11, we now have a standard way to write this:
long foo(long x);
std::enable_if<!std::is_same<int64_t, long>::value, int64_t>::type foo(int64_t);
或 C++20
long foo(long x);
int64_t foo(int64_t) requires (!std::is_same_v<int64_t, long>);
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