[英]Template function argument deduction with an implicit conversion
I understand that template function argument deduction does not take implicit conversions into account. 我了解模板函数自变量推导不考虑隐式转换。
So this code doesn't compile: 因此,此代码无法编译:
#include <iostream>
template<class T>
struct A {};
struct B : public A<int> {};
struct C {
operator B() { return {}; }
};
template<class X>
void test(A<X> arg1, A<X> arg2) {
std::cout << "ok1";
}
int main() {
B b;
C c;
test(b, c); // error: no matching function for call to 'test'
}
What I don't understand is how adding an extra level of indirection with an identity typedef
somehow makes it work: 我不明白的是如何使用身份
typedef
添加额外的间接级别以某种方式使其工作:
#include <iostream>
template<class T>
struct A {};
struct B : public A<int> {};
struct C {
operator B() { return {}; }
};
template<typename U> struct magic { typedef U type; };
template<class T> using magic_t = typename magic<T>::type;
template<class X>
void test(A<X> arg1, A<X> arg2) {
std::cout << "ok1";
}
template<class X>
void test(A<X> arg3, magic_t<A<X>> arg4) {
std::cout << "ok2";
}
int main() {
B b;
C c;
test(b, c); // prints "ok2"
}
Live demo on Godbolt 在Godbolt上进行现场演示
How does magic_t<A<X>>
end up matching C
? magic_t<A<X>>
如何最终匹配C
?
The second parameter becomes non-deduced context and doesn't participate in template argument deduction. 第二个参数变为非推导上下文 ,并且不参与模板参数推导。
X
is then successfully deduced from the first argument. 然后从第一个参数成功推导出
X
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