如何调用 lambda 模板? [英] How to invoke a lambda template?
问题描述
我能够用 gcc 编译以下代码:
template自动功能(包... x){返回 (x + ...) ;}模板auto lamd = [](Pack ... x) {返回 (x + ...) ;};
我可以使用 func(1,2,3)
调用函数模板,但是在调用 lambda 时出现错误,使用 lamd(1,2,3)
或 lamd
.
第二个定义是变量模板.它没有将 lambda 的 operator()
定义为模板,而是采用 operator()
的参数类型的参数包.结果 operator()
是实例化变量的闭包类型的常规成员函数.此处无法进行模板参数推导.
因此,当您编写 lamd
时,该变量会获得带有 operator()(int)
的闭包类型,而不是可使用 3 个整数调用的类型.>
如前所述,您可以改用通用 lambda.
在 C++20 中,如果您需要命名和推导 lambda 的参数类型,您可以使用以下语法:
auto lamd = [](Pack...) {}
这将将运算符定义为模板,接受参数包,并为模板参数推导敞开大门.
I was able to compile the following code with gcc:
template<typename... Pack>
auto func(Pack... x) {
return (x + ...) ;
}
template<typename... Pack>
auto lamd = [](Pack... x) {
return (x + ...) ;
};
I can invoke the function template with func(1,2,3)
, but I get an error when invoking a the lambda, with lamd(1,2,3)
or lamd<int>(1,2,3)
.
The second definition is a variable template. It does not define the lambda's operator()
as a template, but rather takes the parameter pack for the argument types of operator()
. The resulting operator()
is a regular member function of the instantiated variable's closure type. There is no template argument deduction possible here.
So when you write lamd<int>
, the variable gets a closure type with a operator()(int)
, not something callable with 3 integers.
As mentioned already, you can use a generic lambda instead.
In C++20, if you'd need the lambda's argument types to be named and deduced, you can use the syntax:
auto lamd = []<typename... Pack>(Pack...) {}
This will define the operator as template, accepting a parameter pack, and leave the door open for template argument deduction.
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