自动占位符可用于推导非类型模板参数的函数结果吗? [英] Can auto placeholder be used to deduce function result in non-type template parameter?

问题描述

考虑简单的示例:

template <auto(*X)()>
struct Foo {
    decltype(X()) x;
};

int bar();

int main() {
    static_cast<void>(Foo<bar>{});
}

[gcc] 和

Both [gcc] and [clang] seem to accept the code. Is the code really c++17 compliant? If so is there some other rule that makes the following code ill formed?

template <class T, auto(*X)(T)>
struct Foo {
    decltype(X(0)) x;
};

int bar(int);

int main() {
    static_cast<void>(Foo<int, bar>{});
}

这仅使 [gcc] 不满意.

错误消息:

prog.cc: In function 'int main()':
prog.cc:9:35: error: unable to deduce 'auto (*)(T)' from 'bar'
     static_cast<void>(Foo<int, bar>{});
                                   ^
prog.cc:9:35: note:   mismatched types 'T' and 'int'

推荐答案

是的，可以使用auto ( [temp.param] /4.6 ， [dcl.type.auto.deduct] ).我相信gcc在您的第二个示例中是错误的:在执行扣除之前，已替换显式指定的int的T([temp.deduct]/2.3、/5和/6，由[dcl.type引用. auto.deduct]/2.3和/4).

Yes, auto may be used inside a compound type ([temp.param]/4.6, [dcl.type.auto.deduct]). I believe that gcc is in error in your second example: your explicitly specified T of int is substituted before performing deduction ([temp.deduct]/2.3, /5, and /6, referenced by [dcl.type.auto.deduct]/2.3 and /4).

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