英文:
how to deduce function template argument from function return type?
问题
我明白float可以被转换为double甚至int；是否可能根据所请求的返回类型自动实例化get<T>？如果可以，如何实现？
英文:
Example:
template<typename T>
T get() {
return T{};
}
void test() {
float f = get();//requires template argument; e.g. get<float>();
}
I understand that float can be converted to double or even int; is it possible to have get<T> instanciated automatically based on the requested return type? If so how?
答案1
得分: 8
不，仅在转换运算符模板的情况下才会从返回类型进行模板参数推断：
struct A {
template<typename T>
operator T() {
//...
}
};
//...
// 调用`operator T()`，其中`T == float`，将`A` 临时对象转换为`float`
float f = A{};
这也可以用于使get返回A对象，以便float f = get();语法也能正常工作。但是，是否使用此机制如您所愿是一个问题。存在许多注意事项，很容易变得难以理解。例如，在auto f = get();中会发生什么？在调用g(get())中如果函数g有多个重载会发生什么？等等。
相反，将类型说明符移到模板参数上，您就不必重复自己：
auto f = get<float>();
英文:
No, template argument deduction from the return type works only for conversion operator templates:
struct A {
template<typename T>
operator T() {
//...
}
};
//...
// calls `operator T()` with `T == float` to convert the `A` temporary to `float`
float f = A{};
This can also be used to have get return the A object so that float f = get(); syntax will work as well.
However, it is questionable whether using this mechanism as you intent is a good idea. There are quite a few caveats and can easily become difficult to follow. For example what happens in auto f = get();? What happens if there are multiple overloads of a function g in a call g(get())? Etc.
Instead move your type specifier to the template argument and you wont have to repeat yourself:
auto f = get<float>();
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