是否可以在lambda中捕获可变数量的参数? [英] Is it possible to capture a variable number of parameters in a lambda?
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问题描述
请考虑以下一组示例。
Consider the following set of examples.
- 函数
takeOnlyVoidFunction
使用零参数的函数,只是执行它。 - 函数
takeVariableArguments
采用可变数量的参数,并使用参数执行函数。 - 函数
captureVariableArgs
尝试将第二个函数转换为第一个函数可接受的lambda形式,但它不编译。
- The function
takeOnlyVoidFunction
takes a function with zero arguments and simply executes it. - The function
takeVariableArguments
takes a variable number of arguments and executes the function using the arguments. - The function
captureVariableArgs
attempts to convert the second function into a lambda form that is acceptable by the first function, but it does not compile.
如何使函数 captureVariableArgs
将具有可变数量参数的函数转换为没有参数的闭包的正确行为?
How can I make the function captureVariableArgs
compile and exhibit the correct behavior of converting a function with a variable number of arguments into a closure with no arguments?
#include <stdio.h>
#include <functional>
void takeOnlyVoidFunction(std::function<void()> task) {
task();
}
template<typename _Callable, typename... _Args>
void takeVariableArguments(_Callable&& __f, _Args&&... __args) {
__f(__args...);
}
// How can I make this function compile?
template<typename _Callable, typename... _Args>
void captureVariableArgs(_Callable&& __f, _Args&&... __args) {
takeOnlyVoidFunction([=]() { __f(__args...);});
}
void normalFunction(int a, int b) {
printf("I am a normal function which takes params (%d,%d)\n", a, b);
}
int main() {
int a = 7;
int b = 8;
takeVariableArguments(normalFunction, a, b);
takeOnlyVoidFunction([=](){ normalFunction(a,b);});
captureVariableArgs(normalFunction, a, b);
}
我正在运行 gcc 4.9.2
。这是我看到的编译器错误。
I'm running gcc 4.9.2
. Here is the compiler error I see.
g++ -std=c++11 Test.cc -o Test
Test.cc: In instantiation of ‘captureVariableArgs(_Callable&&, _Args&& ...)::<lambda()> [with _Callable = void (&)(int, int); _Args = {int&, int&}]’:
Test.cc:16:38: required from ‘struct captureVariableArgs(_Callable&&, _Args&& ...) [with _Callable = void (&)(int, int); _Args = {int&, int&}]::<lambda()>’
Test.cc:16:50: required from ‘void captureVariableArgs(_Callable&&, _Args&& ...) [with _Callable = void (&)(int, int); _Args = {int&, int&}]’
Test.cc:28:45: required from here
Test.cc:16:34: error: variable ‘__f’ has function type
takeOnlyVoidFunction([=]() { __f(__args...);});
^
Test.cc:16:34: error: variable ‘__f’ has function type
Test.cc: In instantiation of ‘struct captureVariableArgs(_Callable&&, _Args&& ...) [with _Callable = void (&)(int, int); _Args = {int&, int&}]::<lambda()>’:
Test.cc:16:50: required from ‘void captureVariableArgs(_Callable&&, _Args&& ...) [with _Callable = void (&)(int, int); _Args = {int&, int&}]’
Test.cc:28:45: required from here
Test.cc:16:34: error: field ‘captureVariableArgs(_Callable&&, _Args&& ...) [with _Callable = void (&)(int, int); _Args = {int&, int&}]::<lambda()>::<__f capture>’ invalidly declared function type
In file included from Test.cc:2:0:
/usr/include/c++/4.9/functional:2418:7: error: ‘std::function<_Res(_ArgTypes ...)>::function(_Functor) [with _Functor = captureVariableArgs(_Callable&&, _Args&& ...) [with _Callable = void (&)(int, int); _Args = {int&, int&}]::<lambda()>; <template-parameter-2-2> = void; _Res = void; _ArgTypes = {}]’, declared using local type ‘captureVariableArgs(_Callable&&, _Args&& ...) [with _Callable = void (&)(int, int); _Args = {int&, int&}]::<lambda()>’, is used but never defined [-fpermissive]
function<_Res(_ArgTypes...)>::
^
更新 :一个更小的示例演示这个问题。
Update: A more minimal example demonstrating this problem.
#include <stdio.h>
// How can I make this function compile?
template<typename _Callable>
void captureVariableArgs(_Callable&& __f) {
takeOnlyVoidFunction( [=]{ __f(); } );
}
void normalFunction() {
printf("I am a normal function\n");
}
int main(){
captureVariableArgs(normalFunction);
}
推荐答案
,而不是使用lambda,您可以使用 std :: bind
:
As another potential workaround for GCC, instead of using a lambda, you could use std::bind
:
template <typename F, typename... Args>
auto captureVariable(F&& f, Args&&... args)
{
return std::bind(std::forward<F>(f), std::forward<Args>(args)...);
}
这适用于GCC 4.9.3下的我。
This works for me under GCC 4.9.3.
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