問題描述
我有以下代碼:
#include 模板 void f (T) { std::cout <<f(T)"<<std::endl;}模板 void f (bool) { std::cout <<"f(bool)" <<std::endl;}內部主(){f(真);//#1 打印 f(T)f<bool>(真);//#2 打印 f(bool)} #1 行調用 f(T)，而 #2 行調用 f(bool).
為什么會這樣?以及選擇重載模板函數的規則是什么?
更新
我知道在第一次調用中，編譯器在嘗試調用第二個函數時無法推斷出T，所以選擇了第一個.
在第二次調用中，第二個函數被認為在 gcc 上更匹配，而第一個是在 VS2013 下選擇的.誰在這里做正確的事?順便說一句，我仍然對過程的完整描述感興趣.
未特化的函數模板也稱為底層基模板. 基模板可以是特化的.查看在不同情況下調用哪些方法的重載規則非常簡單，至少在高層次上是這樣:
非模板函數是一等公民.與參數類型以及任何函數模板匹配的普通舊非模板函數將被選擇，而不是其他一樣好的函數模板.
如果沒有至少與二等公民一樣好的一等公民可供選擇，那么接下來將咨詢二等公民的函數基礎模板.根據一組相當神秘規則:
如果很明顯有一個最專業的"函數基礎模板，那么就會使用那個模板.如果該基模板恰好專用于正在使用的類型，則將使用該特化，否則將使用使用正確類型實例化的基模板.
Else(如您的情況)如果最專業"的函數基模板存在聯系，則調用是不明確的，因為編譯器無法決定哪個更匹配.程序員必須做一些事情來限定調用并說明需要哪個.
否則，如果沒有可以匹配的函數基礎模板，調用就會出錯，程序員將不得不修復代碼.
如果您想自定義函數基礎模板并希望該自定義參與重載解析(或者，始終在完全匹配的情況下使用)，請將其設為普通的舊函數，不是專業.而且，如果您確實提供了重載，請避免同時提供專業化.
以上摘自the Herb Sutter，在突出顯示的項目符號中，您可以看到問題的根源
編輯
如果你在 Visual Studio 2012 中嘗試(不要這樣做)上面的代碼，你會得到 <塊引用>
致命錯誤 LNK1179:無效或損壞的文件:重復的 COMDAT '??$f@_N@@YAX_N@Z'
正如此處所解釋的那樣，是因為
<塊引用>你做了一些 C++ 無效的詭計"，它通過了編譯器，但你現在有一個無效的 *.obj，它阻塞了鏈接器.
下面這行是罪魁禍首
f(true);//#1 打印 f(T)所以答案中解釋的歧義沒有保證解決
I have the following code:
#include <iostream>
template <typename T>
void f (T) { std::cout << "f(T)" << std::endl; }
template <typename T>
void f (bool) { std::cout << "f(bool)" << std::endl; }
int main ( )
{
f(true); // #1 prints f(T)
f<bool>(true); // #2 prints f(bool)
}
The #1 line calls f(T), while #2 line calls f(bool).
Why does this happen? And what are the rules for selecting an overloaded template function?
UPDATE
I understood that in the first call compiler is just unable to deduce T while trying to call the second function, so the first is chosen.
In the second call second function is considered a better match on gcc, while the first is chosen under VS2013. Who does the right thing here? By the way, I am still interested in the full description of the process.
The unspecialized function templates are also called the underlying base templates. Base templates can be specialized. The overloading rules to see which ones get called in different situations, are pretty simple, at least at a high level:
Nontemplate functions are first-class citizens. A plain old nontemplate function that matches the parameter types as well as any function template will be selected over an otherwise-just-as-good function template.
If there are no first-class citizens to choose from that are at least as good, then function base templates as the second-class citizens get consulted next. Which function base template gets selected depends on which matches best and is the "most specialized" (important note: this use of "specialized" oddly enough has nothing to do with template specializations; it's just an unfortunate colloquialism) according to a set of fairly arcane rules:
If it's clear that there's one "most specialized" function base template, that one gets used. If that base template happens to be specialized for the types being used, the specialization will get used, otherwise the base template instantiated with the correct types will be used.
Else (as in your case) if there's a tie for the "most specialized" function base template, the call is ambiguous because the compiler can't decide which is a better match. The programmer will have to do something to qualify the call and say which one is wanted.
Else if there's no function base template that can be made to match, the call is bad and the programmer will have to fix the code.
If you want to customize a function base template and want that customization to participate in overload resolution (or, to always be used in the case of exact match), make it a plain old function, not a specialization. And, if you do provide overloads, avoid also providing specializations.
The above is an extract from this post by the Herb Sutter and in the highlighted bullet you can see the source of your problem
EDIT
If you try (don't do it) the above code with Visual Studio 2012, you get
fatal error LNK1179: invalid or corrupt file: duplicate COMDAT '??$f@_N@@YAX_N@Z'
which, as explained here, is because
You did some "trickery" that is invalid C++, and it passed the compiler, but you now have an invalid *.obj, and it chokes the linker.
and the following line is to blame
f(true); // #1 prints f(T)
so the ambiguity explained in the answer has no guarandeed resolution
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