从默认参数推断模板参数 [英] Infer template argument from default parameter

问题描述

考虑以下代码：

#include <functional>

template <typename T,typename COMP>
bool foo(T a,T b,COMP c = std::less<T>()) { 
    return c(a,b); 
}

bool bar(int a, int b){ return a<b;}

int main(){
    foo(1,2,bar);               // OK
    foo(1,2,std::less<int>());  // OK
    foo(1,2);                   // error
}

前两个调用很好，但似乎被禁止让编译器从默认参数推断 COMP 的类型：

The first two calls are fine, but it seems to be forbidden to let the compiler infer the type of COMP from the default parameter:

<source>:14:5: error: no matching function for call to 'foo'    
    foo(1,2);    
    ^~~    
<source>:4:6: note: candidate template ignored: couldn't infer template argument 'COMP'    
bool foo(T a,T b,COMP c = std::less<T>()) {    
     ^   
1 error generated.    
Compiler returned: 1

我错过了什么吗？我真的不明白为什么编译器无法推断模板参数'COMP'，而我却怀疑不允许这样做。

Am I missing something? I dont really understand why the compiler "couldn't infer template argument 'COMP'" and I rather suspect that it is not allowed to do so.

是否可以从默认参数推断出模板参数？如果不是，为什么？

推荐答案

我建议像在标准C ++容器中那样进行操作地图或集合。它可以推断类型，还可以使用默认参数：

I'd suggest doing it the way it is done in standard C++ containers like map or set. It allows to infer the type and also allows to use default argument:

template <typename T,typename COMP = std::less<T>>
bool foo(T a,T b, COMP c = COMP()) { /* As before. */ }

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