如何避免scoped_lock的发光的＆QUOT;未使用的变量＆QUOT;警告？ [英] How does scoped_lock avoid emitting an "unused variable" warning?

问题描述

的boost ::互斥:: scoped_lock的是围绕锁定一个互斥体一个方便的RAII包装。我使用了类似的技术，别的东西：一个RAII包装器询问一个数据接口，从分离/重新连接到串行设备

。

我想不出什么，虽然是为什么只有我的对象 MST ＆MDASH低于code;其实例化和破坏的确实有副作用的＆MDASH;而→管理​​，以保持沉默的原因 G ++ 发出一个未使用变量警告错误。

你知道吗？你能告诉我吗？

  [通用@定点〜] $猫TEST.CPP
＃包括LT＆;升压/ shared_ptr.hpp＆GT;
＃包括LT＆;升压/线程/ mutex.hpp＆GT;
＃包括LT＆;＆iostream的GT;结构MyScopedThing;
结构MyWorkerObject {
    一个无效（）{性病::法院LT＆;＆LT; 一个; }
    空穴B（）{性病::法院LT＆;＆LT; B; }    提高:: shared_ptr的＆LT; MyScopedThing＆GT; getScopedThing（）;
};结构MyScopedThing {
    MyScopedThing（MyWorkerObject＆安培; W）：W（W）{
        W.A（）;
    }
    〜MyScopedThing（）{
        W.B（）;
    }    MyWorkerObject＆安培;瓦;
};提高:: shared_ptr的＆LT; MyScopedThing＆GT; MyWorkerObject :: getScopedThing（）{
    返回的boost :: shared_ptr的＆LT; MyScopedThing＆GT;（新MyScopedThing（*本））;
}诠释主（）{
    提高::互斥米;
    提高::互斥:: scoped_lock的L（M）;    MyWorkerObject瓦;
    常量的boost :: shared_ptr的＆LT; MyScopedThing＆GT;＆安培; MST = w.getScopedThing（）;
}
[通用@定点〜] $ G ++ TEST.CPP -o测试-lboost_thread -Wall
TEST.CPP：在函数'廉政的main（）：
TEST.CPP：33：警告：未使用的变量'MST'[通用@定点〜] $ ./test
AB [通用@定点〜] $ G ++ -v 2  - ;＆放大器; 1 | grep的版本
gcc版本4.4.5 20110214（红帽4.4.5-6）（GCC）
 

解决方案

请注意，这个问题已经改变，因为其他答案写。

有可能的原因G ++不会在目前的形式警告是因为 MST 是一个参考，构建和销毁的引用没有副作用。这是真的，这里的参考延长临时的寿命，它在它的构造函数和析构函数的影响，但显然G ++并没有意识到有差别。

boost::mutex::scoped_lock is a handy RAII wrapper around locking a mutex. I use a similar technique for something else: a RAII wrapper around asking a data interface to detach from/re-attach to a serial device.

What I can't figure out, though, is why in the code below only my object mst — whose instantiation and destruction do have side effects — causes g++ to emit an "unused variable" warning error whereas l manages to remain silent.

Do you know? Can you tell me?

[generic@sentinel ~]$ cat test.cpp
#include <boost/shared_ptr.hpp>
#include <boost/thread/mutex.hpp>
#include <iostream>

struct MyScopedThing;
struct MyWorkerObject {
    void a() { std::cout << "a"; }
    void b() { std::cout << "b"; }

    boost::shared_ptr<MyScopedThing> getScopedThing();
};

struct MyScopedThing {
    MyScopedThing(MyWorkerObject& w) : w(w) {
        w.a();
    }
    ~MyScopedThing() {
        w.b();
    }

    MyWorkerObject& w;
};

boost::shared_ptr<MyScopedThing> MyWorkerObject::getScopedThing() {
    return boost::shared_ptr<MyScopedThing>(new MyScopedThing(*this));
}

int main() {
    boost::mutex m;
    boost::mutex::scoped_lock l(m);

    MyWorkerObject w;
    const boost::shared_ptr<MyScopedThing>& mst = w.getScopedThing();
}


[generic@sentinel ~]$ g++ test.cpp -o test -lboost_thread -Wall
test.cpp: In function ‘int main()’:
test.cpp:33: warning: unused variable ‘mst’

[generic@sentinel ~]$ ./test
ab[generic@sentinel ~]$ g++ -v 2>&1 | grep version
gcc version 4.4.5 20110214 (Red Hat 4.4.5-6) (GCC)

解决方案

Note that the question has changed since the other answers were written.

Likely the reason g++ doesn't warn in the current form is because mst is a reference, and constructing and destructing a reference has no side effects. It's true that here the reference is extending the lifetime of a temporary, which has effects in its constructor and destructor, but apparently g++ doesn't realise that makes a difference.

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