我是std::thread
C ++ 11的新手。尝试从https://en.cppreference.com/w/cpp/thread/thread/thread的示例中弄些玩具,我试图查看是否可以std::thread
使用带有非空参数列表的类成员函数调用运算符生成一个如下面的代码:
// main.cpp
#include <iostream>
#include <iostream>
#include <thread>
class Foo {
public:
void operator()( int& i ) {
std::cout << i << std::endl;
}
};
int main( int argc, char* argv[] ) {
Foo f;
int i = 42;
std::thread t1( f, i );
t1.join();
return 0;
}
错误消息是神秘的:
$ g++ --version && g++ ./main.cpp -lpthread && ./a.out
g++ (Debian 6.3.0-18+deb9u1) 6.3.0 20170516
Copyright (C) 2016 Free Software Foundation, Inc.
This is free software; see the source for copying conditions. There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
In file included from /usr/include/c++/6/thread:39:0,
from ./main.cpp:5:
/usr/include/c++/6/functional: In instantiation of ‘struct std::_Bind_simple<Foo(int)>’:
/usr/include/c++/6/thread:138:26: required from ‘std::thread::thread(_Callable&&, _Args&& ...) [with _Callable = Foo&; _Args = {int&}]’
./main.cpp:19:24: required from here
/usr/include/c++/6/functional:1365:61: error: no type named ‘type’ in ‘class std::result_of<Foo(int)>’
typedef typename result_of<_Callable(_Args...)>::type result_type;
^~~~~~~~~~~
/usr/include/c++/6/functional:1386:9: error: no type named ‘type’ in ‘class std::result_of<Foo(int)>’
_M_invoke(_Index_tuple<_Indices...>)
相反,空的参数列表调用运算符可以正常工作:
// main.cpp
#include <iostream>
#include <iostream>
#include <thread>
class Foo {
public:
void operator()() {
std::cout << 42 << std::endl;
}
};
int main( int argc, char* argv[] ) {
Foo f;
int i = 42;
std::thread t1( f );
t1.join();
return 0;
}
$ g++ --version && g++ ./main.cpp -lpthread && ./a.out
g++ (Debian 6.3.0-18+deb9u1) 6.3.0 20170516
Copyright (C) 2016 Free Software Foundation, Inc.
This is free software; see the source for copying conditions. There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
42
我的第一次尝试是否完全可行-我是否只有语法错误?有没有一种方法可以使用对象及其非空参数列表调用运算符生成std :: thread?
我认为这个问题与使用成员函数启动线程不同,因为这个问题专门涉及使用成员对象调用运算符生成线程,我知道这可以使用lambda来完成。
std::thread
除非您明确指出,否则不要让您通过引用,因为这是生命周期问题的简单来源。用于std::ref
明确表示您i
通过引用传递:
std::thread t1( f, std::ref(i) );
或者,按值传递。在通过引用将某些内容传递给线程之前,请认真考虑并确保有必要。您传递的变量必须在线程内超过其使用期限。
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