英文:
Passing designated initializers for a template class to a template function
问题
CTAD在最后一种情况下失败是因为编译器无法推断出模板类型B。有什么办法可以让它成功吗？
https://godbolt.org/z/7Es4Y11Yf
英文:
My goal is to create an API interface that looks like that this:
struct myAB{ int a,b; };
void function(myAB ab) {}
...
function({.a = 1, .b = 3});
The above works just fine. But if I want struct AB to have a templated type, CTAD fails.
template <class B>
struct myAB2{ int a; B b; };
template<typename B> myAB2(int, B) -> myAB2<B>;
template<typename B>
void function2(myAB2<B> ab) {}
...
myAB2 ab = {.a = 1, .b = 3}; //works just fine with CTAD
function2(ab); //fine as expected
function2(myAB2{.a = 1, .b = 3}); //works just fine with CTAD
function2({.a = 1, .b = 3}); //fails to compile, can't deduce type 'B'
Why does CTAD fail in the last case? Is there anything I can do to get it to succeed?
答案1
得分: 3
一个花括号初始化列表，包括其中包含的指定初始化器，没有类型。你只能从中推断出std::initializer_list或数组类型，但在这里并没有发生。
由于{.a = 1, .b = 3}不是一个表达式，也没有类型，所以从中无法推断出B的任何信息，因此无法调用该函数。
function2<int>({.a = 1, .b = 3})会起作用（因为myAB2<int> ab可以用指定的初始化器进行初始化，不需要进行更多的推断）。
function2(myAB2{.a = 1, .b = 3})也会起作用（因为它进行了CTAD以获得类型为myAB2<int>的prvalue，从中可以推断出B = int）。
英文:
A braced-init-list, including one that contains designated initializers, doesn't have a type. You can only ever deduce std::initializer_list or array types from them, which isn't happening here.
Since {.a = 1, .b = 3} is not an expression and doesn't have a type, function2({.a = 1, .b = 3}) cannot deduce anything for B from it, so the function can't be called.
function2<int>({.a = 1, .b = 3}) would work (because the myAB2<int> ab can be initialized with the designated initializers, no more deduction necessary).
So would function2(myAB2{.a = 1, .b = 3}) (because that does CTAD to get a prvalue of type myAB2<int>, which B = int can be deduced from).
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