[英]How does the compiler determine between a function using SFINAE and a standard function if both are viable?
Consider the following code:考虑以下代码:
#include <iostream>
#include <type_traits>
template <typename T>
class A
{
public:
// Allow func to be called if T is the const version of T2
// e.g., T is 'int const' and T2 is 'int'
template <typename T2,
typename = typename std::enable_if<
std::is_same<T, T2 const>::value>::type>
void func(A<T2> const &)
{
std::cout << "Conversion" << std::endl;
}
// Allow func to be called for same version of T
void func(A const &)
{
std::cout << "No conversion" << std::endl;
}
};
int main()
{
A<int const> a;
a.func(A<int const>{});
return 0;
}
This code, when compiled with GCC-8.3 compiles and produces the output No conversion
- it selected the version of func
that does not use std::enable_if
.此代码在使用 GCC-8.3 编译时编译并产生输出
No conversion
- 它选择了不使用std::enable_if
的func
版本。 However, if I comment out the second version of func
, it will still compile and now produce the output Conversion
.但是,如果我注释掉
func
的第二个版本,它仍然会编译并现在生成输出Conversion
。 In other words, both versions of func
within A
are usable for this method.换句话说,
A
中的两个版本的func
都可用于此方法。 Given that both overloads are viable, what specific rule is the compiler using to select func(A const &)
over the other version ( func(A<T2> const &)
)?鉴于两个重载都是可行的,编译器使用什么特定规则来选择
func(A const &)
而不是另一个版本( func(A<T2> const &)
)?
The rule is that if a non function template and a function template specialization have the same signature, then the non function template is chosen over the template specialization.规则是,如果非函数模板和函数模板特化具有相同的签名,则选择非函数模板而不是模板特化。 This can be found in [over.match.best]/2
这可以在[over.match.best]/2 中找到
Given these definitions, a viable function F1 is defined to be a better function than another viable function F2 if for all arguments i, ICSi(F1) is not a worse conversion sequence than ICSi(F2), and then
鉴于这些定义,如果对于所有参数 i,ICSi(F1) 不是比 ICSi(F2) 更差的转换序列,那么一个可行函数 F1 被定义为比另一个可行函数 F2 更好的函数,然后
[...]
[...]
- F1 is not a function template specialization and F2 is a function template specialization [...]
F1 不是函数模板特化,F2 是函数模板特化 [...]
You can read about overload resolution here .您可以 在此处阅读有关重载解析的 信息。 In particular
特别是
If any candidate is a function template, its specializations are generated using template argument deduction, and such specializations are treated just like non-template functions except where specified otherwise in the tie-breaker rules.
如果任何候选函数是函数模板,则使用模板参数推导生成其特化,并且此类特化被视为非模板函数,除非在决胜局规则中另有规定。
And then进而
Best viable function
最佳可行功能
For each pair of viable function F1 and F2, the implicit conversion sequences from the i-th argument to i-th parameter are ranked to determine which one is better [...]
对于每对可行函数 F1 和 F2,从第 i 个参数到第 i 个参数的隐式转换序列被排序以确定哪个更好 [...]
F1 is determined to be a better function than F2 if implicit conversions for all arguments of F1 are not worse than the implicit conversions for all arguments of F2, and [...]
如果 F1 的所有参数的隐式转换不比 F2 的所有参数的隐式转换差,则确定 F1 是比 F2 更好的函数,并且 [...]
4) [...] F1 is a non-template function while F2 is a template specialization
4) [...] F1 是非模板函数,而 F2 是模板特化
Basically the same rules apply in this simpler example:基本上相同的规则适用于这个更简单的例子:
#include<iostream>
template <typename T>
void foo(T i) { std::cout << "template" ; }
void foo(int i) { std::cout << "non template"; }
int main() {
foo(1); // non template
}
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