Linux C++11:如何给函数取别名?
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C++11: How to alias a function?
提问by Andrew Tomazos
If I have a class Foo in namespace bar:
如果我在命名空间栏中有一个 Foo 类:
namespace bar
{
class Foo { ... }
};
I can then:
然后我可以:
using Baz = bar::Foo;
and now it is just like I defined the class in my namespace with the name Baz.
现在就像我在我的命名空间中定义了名为 Baz 的类一样。
Is it possible to do the same for functions?
是否可以对函数做同样的事情?
namespace bar
{
void f();
}
And then:
进而:
using g = bar::f; // error: ‘f' in namespace ‘bar' does not name a type
What is the cleanest way to do this?
什么是最干净的方法来做到这一点?
The solution should also hold for template functions.
该解决方案也应适用于模板函数。
Definition:If some entity B is an aliasof A, than if any or all usages (not declarations or definitions of course) of A are replaced by B in the source code than the (stripped) generated code remains the same. For example typedef A B
is an alias. #define B A
is an alias (at least). T& B = A
is not an alias, B can effectively implemented as an indirect pointer, wheres an "unaliased" A can use "immediate semantics".
定义:如果某个实体 B 是A的别名,那么如果 A 的任何或所有用法(当然不是声明或定义)在源代码中被 B 替换,那么(剥离)生成的代码保持不变。例如typedef A B
是别名。 #define B A
是别名(至少)。 T& B = A
不是别名,B 可以有效地实现为间接指针,其中“无别名”A 可以使用“立即语义”。
采纳答案by Jeremiah Willcock
You can define a function alias (with some work) using perfect forwarding:
您可以使用完美转发定义函数别名(需要一些工作):
template <typename... Args>
auto g(Args&&... args) -> decltype(f(std::forward<Args>(args)...)) {
return f(std::forward<Args>(args)...);
}
This solution does apply even if f
is overloaded and/or a function template.
即使f
重载和/或函数模板,此解决方案也适用。
回答by Kerrek SB
Classes are types, so they can be aliased with typedef
and using
(in C++11).
类是types,因此它们可以使用typedef
and别名using
(在 C++11 中)。
Functions are much more like objects, so there's no mechanism to alias them. At best you could use function pointers or function references:
函数更像对象,所以没有机制来给它们取别名。充其量您可以使用函数指针或函数引用:
void (*g)() = &bar::f;
void (&h)() = bar::f;
g();
h();
In the same vein, there's no mechanism for aliasing variables(short of through pointers or references).
同样,没有别名变量的机制(缺少直通指针或引用)。
回答by Martin York
Absolutely:
绝对地:
#include <iostream>
namespace Bar
{
void test()
{
std::cout << "Test\n";
}
template<typename T>
void test2(T const& a)
{
std::cout << "Test: " << a << std::endl;
}
}
void (&alias)() = Bar::test;
void (&a2)(int const&) = Bar::test2<int>;
int main()
{
Bar::test();
alias();
a2(3);
}
Try:
尝试:
> g++ a.cpp
> ./a.out
Test
Test
Test: 3
>
A reference is an alias to an existing object.
I just created a reference to a function. The reference can be used in exactly the same way as the original object.
引用是现有对象的别名。
我刚刚创建了一个函数的引用。可以以与原始对象完全相同的方式使用引用。
回答by Tom
It's not standard C++, but most compilers provide a way of doing this. With GCC you can do this:
它不是标准的 C++,但大多数编译器都提供了一种方法。使用 GCC,您可以这样做:
void f () __attribute__ ((weak, alias ("__f")));
This creates the symbol f
as an alias for __f
. With VC++ you do the same thing this way:
这将创建符号f
作为 的别名__f
。使用 VC++,您可以通过以下方式做同样的事情:
#pragma comment(linker, "/export:f=__f")
回答by Jason Haslam
It is possible to introduce the function into a different scope without changing its name. So you can alias a function with a different qualified name:
可以在不更改其名称的情况下将函数引入不同的作用域。因此,您可以使用不同的限定名称为函数设置别名:
namespace bar {
void f();
}
namespace baz {
using bar::f;
}
void foo() {
baz::f();
}
回答by Romeno
You can use good old macros
你可以使用好的旧宏
namespace bar
{
void f();
}
#define f bar::f
int main()
{
f();
}
回答by Pawe? Bylica
The constexpr
function pointer can be used as a function alias.
的constexpr
函数指针可以用作函数别名。
namespace bar
{
int f();
}
constexpr auto g = bar::f;
It is highly likely (but not guaranteed by the language) that using g
uses bar::f
directly.
Specifically, this depends on compiler version and optimization level.
很可能(但语言不保证) using直接g
使用bar::f
。具体来说,这取决于编译器版本和优化级别。
In particular, this is the case for:
尤其是以下情况:
- GCC 4.7.1+, without optimization,
- Clang 3.1+, without optimization,
- MSVC 19.14+, with optimization.
- GCC 4.7.1+,无优化,
- Clang 3.1+,没有优化,
- MSVC 19.14+,优化。
See assembly generated by these compilers.
请参阅这些编译器生成的程序集。