大括号初始化列表和功能模板类型的推导顺序 [英] Braced-init-lists and function template type deduction order

问题描述

我对功能模板参数类型推导过程有疑问.

I have a question regarding the function template parameter type deduction procedure.

以这个例子为例:

#include <vector>
#include <sstream>
#include <string>
#include <iterator>
#include <fstream>

int main()
{
    std::ifstream file("path/to/file");
    std::vector<int> vec(std::istream_iterator<int>{file},{}); // <- This part
    return 0;
}

如果我理解正确，第二个参数将推导为默认构造函数称为std::istream_iterator的类型.

If I understand things correctly, the second parameter is deduced to be of type std::istream_iterator of which the default constructor is called.

适当的std::vector构造函数声明为:

The appropriate std::vector constructor is declared as:

template <class InputIterator>
         vector (InputIterator first, InputIterator last,
                 const allocator_type& alloc = allocator_type());

由于将第一参数类型推导为std::istream_iterator<int>，因此将第二参数也推导为std::istream_iterator<int>，因此可以应用统一的初始化语义.我不知道类型推导发生的顺序.我非常感谢您提供一些信息.

Since the first parameter type is deduced as std::istream_iterator<int> the second parameter is deduced as std::istream_iterator<int> too and so the uniform initialization semantics can be applied. What I have no idea about is at what order the type deduction happens. I would really appreciate some info on this.

提前谢谢！

推荐答案

让我们使用一个更简单的示例:

Let's use an even simpler example:

template<class T>
void foo(T, T);

foo(42, {});

函数调用有两个参数:

• 类型为int(整数文字)的prvalue表达式
• 一个括号初始化列表 {}

• a prvalue expression of type int (an integer literal)
• a braced-init-list {}

后者{}可以是 expression-list 的一部分，但它本身不是 expression .一个 expression-list 定义为一个 initializer-list . braced-init-lists 没有类型.

The latter, {}, can be part of an expression-list but it is not an expression itself. An expression-list is defined as an initializer-list. braced-init-lists do not have a type.

每个功能参数的单独模板类型推导[temp.deduct.type]/2. [temp.deduct.call]/1关于函数参数P的类型推导的状态:

Template type deduction is done for each function parameter individually [temp.deduct.type]/2. [temp.deduct.call]/1 states about type deduction for a function parameter P:

如果从P中删除引用和简历限定符，则给出 std::initializer_list< P' >表示某些 P'，且参数为 初始化列表，然后对每个元素执行推导 初始化列表列表，以 P'作为函数模板参数 类型，并将初始值设定项元素作为其参数. 否则， 初始化程序列表参数导致该参数被认为是 非推论上下文. [强调我]

If removing references and cv-qualifiers from P gives std::initializer_list<P'> for some P' and the argument is an initializer list, then deduction is performed instead for each element of the initializer list, taking P' as a function template parameter type and the initializer element as its argument. Otherwise, an initializer list argument causes the parameter to be considered a non-deduced context. [emphasis mine]

因此，在调用foo(42, {});中，不会从第二个自变量{}推导出T.但是，可以从第一个参数推导出T.

So in the call foo(42, {}); the T will not be deduced from the second argument {}. However, T can be deduced from the first argument.

通常，我们可以从多个函数参数推导出T.在这种情况下，推导的类型必须精确匹配[temp.deduct.type]/2.如果类型仅从一个函数参数推导而在其他地方使用(在非推论上下文中的另一个函数参数中，在返回类型中)，则没有问题.类型推导可能会失败，例如当无法从 any 函数参数推断出模板参数并且未明确设置模板参数时.

In general, we can deduce T from multiple function parameters. In that case, the deduced types have to match exactly [temp.deduct.type]/2. There is no problem if the type is only deduced from one function parameter but used elsewhere (in another function parameter that is in a non-deduced context, in the return type etc). Type deduction can fail e.g. when a template parameter cannot be deduced from any function parameter and is not set explicitly.

扣除后，T将被int替换，产生类似于以下内容的功能签名:

After deduction, T will be substituted by int, producing a function signature similar to:

void foo<int>(int, int);

可以使用两个参数42和{}调用此函数.后者将执行复制列表初始化，从而导致第二个参数的值初始化.

This function can be called with the two arguments 42 and {}. The latter will perform a copy-list-initialization leading to a value-initialization of the second parameter.

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