問題描述

我剛剛問了這個問題:std::numeric_limits 作為條件

我了解 std::enable_if 將有條件地定義方法的返回類型導致方法編譯失敗的用法.

template類型名稱 std::enable_if<std::numeric_limits<T>::is_integer, void>::type foo(const T &bar) { isInt(bar);}

我不明白的是第二個參數和對 std::enable_if 的看似毫無意義的賦值，當它被聲明為模板語句的一部分時，如 Rapptz answer.

template::value, int>::type = 0>void foo(const T& bar) { isInt();}

正如 40two 在評論中提到的，理解替換失敗不是錯誤是理解的先決條件std::enable_if.

std::enable_if 是一個專門的模板，定義為:

template結構 enable_if {};模板struct enable_if{ typedef T 類型；};

這里的關鍵在于typedef T type僅在bool Cond為true時才定義.

現在有了對 std::enable_if 的理解，很明顯 void foo(const T &bar) { isInt(bar);} 定義為:

template類型名稱 std::enable_if<std::numeric_limits<T>::is_integer, void>::type foo(const T &bar) { isInt(bar);}

如firda 的答案所述，= 0 是第二個模板的默認值范圍.template<typename T, typename std::enable_if<std::is_integral<T>::value, int>::type = 0> 中默認的原因是兩個選項都可以用 foo<調用int >( 1 );.如果 std::enable_if 模板參數沒有被默認，調用 foo 將需要兩個模板參數，而不僅僅是 int.

一般注意，通過明確輸入 typename std::enable_if::is_integer, void>::type 但 void 是 std::enable_if 的默認第二個參數，如果你有 c++14 enable_if_t 是一個定義的類型，應該使用.所以返回類型應該壓縮為:std::enable_if_t::is_integer>

給 visual- 的用戶的特別說明-工作室 visual-studio-2013:不支持默認模板參數，因此您只能在函數返回時使用 enable_if:std::numeric_limits 作為條件

I just asked this question: std::numeric_limits as a Condition

I understand the usage where std::enable_if will define the return type of a method conditionally causing the method to fail to compile.

template<typename T>
typename std::enable_if<std::numeric_limits<T>::is_integer, void>::type foo(const T &bar) { isInt(bar); }

What I don't understand is the second argument and the seemingly meaningless assignment to std::enable_if when it's declared as part of the template statement, as in Rapptz answer.

template<typename T, typename std::enable_if<std::is_integral<T>::value, int>::type = 0>
void foo(const T& bar) { isInt(); }

As is mentioned in comment by 40two, understanding of Substitution Failure Is Not An Error is a prerequisite for understanding std::enable_if.

std::enable_if is a specialized template defined as:

template<bool Cond, class T = void> struct enable_if {};
template<class T> struct enable_if<true, T> { typedef T type; };

The key here is in the fact that typedef T type is only defined when bool Cond is true.

Now armed with that understanding of std::enable_if it's clear that void foo(const T &bar) { isInt(bar); } is defined by:

template<typename T>
typename std::enable_if<std::numeric_limits<T>::is_integer, void>::type foo(const T &bar) { isInt(bar); }

As mentioned in firda's answer, the = 0 is a defaulting of the second template parameter. The reason for the defaulting in template<typename T, typename std::enable_if<std::is_integral<T>::value, int>::type = 0> is so that both options can be called with foo< int >( 1 );. If the std::enable_if template parameter was not defaulted, calling foo would require two template parameters, not just the int.

General note, this answer is made clearer by explicitly typing out typename std::enable_if<std::numeric_limits<T>::is_integer, void>::type but void is the default second parameter to std::enable_if, and if you have c++14 enable_if_t is a defined type and should be used. So the return type should condense to: std::enable_if_t<std::numeric_limits<T>::is_integer>

A special note for users of visual-studio prior to visual-studio-2013: Default template parameters aren't supported, so you'll only be able to use the enable_if on the function return: std::numeric_limits as a Condition

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