为什么函数返回类型中不允许参数推导? [英] Why isn't argument deduction allowed in function return type?

问题描述

最明显的答案可能是-因为标准是这样.
很好，但是我想尽办法了解这种选择的原因.

The most obvious answer could be - because the standard says so.
That's fine, but I'm wrapping my head around it to understand the reasons behind this choice.

请考虑以下示例:

template<typename T>
struct S { S(T) {} };

S f() { return 0; }

int main() {
    auto s = f();
    (void)s;
}

它无法编译并显示以下错误:

It fails to compile with errors like:

错误:使用类模板'S'需要模板参数；函数返回类型中不允许参数推导

error: use of class template 'S' requires template arguments; argument deduction not allowed in function return type

很容易修复，这不是问题，类似这样的方法就可以了:

Quite easy to fix, it isn't a problem, something like this works just fine:

auto f() { return S{0}; }

但是，我想了解在函数返回类型中也允许类模板参数推导的缺点.
乍一看，这似乎是一个愚蠢的限制，但是我敢肯定，我在这里错过了一些重要的事情.

However, I'd like to understand what were the drawbacks of allowing class template arguments deduction also in function return types.
At a first glance, it looks just like a silly limitation, but I'm pretty sure I'm missing something important here.

推荐答案

此处没有任何法律限制:如果您指定返回类型(而不是auto或T，其中T是模板类型)，该返回类型必须有效.让我给您提供一个更简单，更好的示例:

There's nothing language-lawery here: If you specify a return type (and not auto or T where T is a template type), that return type has to be valid. let me give you even a simpler, better example:

std::vector function() {
    return std::vector<int>();
}

很明显，即使没有精美的模板，auto和类型推导，它也无法编译，因为std::vector不是类型，std::vector<int>是.

Obviously it fails to compile, even without fancy templates, auto and type deductions, because std::vector isn't a type, std::vector<int> is.

基本上，当您将S指定为返回类型时，

When you specify S as a return type you basically

• 防止编译器推断类型本身
• 指定无效的返回类型，因为S不是类型，S<int>是.

• Prevent the compiler from deducing the type itself
• Specify an invalid return type, as S isn't a type, S<int> is.

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