scoped_lock 如何避免发出“未使用的变量"?警告?

How does scoped_lock avoid emitting an quot;unused variablequot; warning?(scoped_lock 如何避免发出“未使用的变量?警告?)

问题描述

boost::mutex::scoped_lock 是一个方便的 RAII 包装器，用于锁定互斥锁.我对其他事情使用了类似的技术:一个 RAII 包装器，它要求数据接口从/重新连接到串行设备.

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.

但是，我想不通的是为什么在下面的代码中只有我的对象 mst —其实例化和销毁确实有副作用 —导致 g++ 发出未使用变量"警告错误，而 l 设法保持沉默.

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.

你知道吗?你能告诉我吗?

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)

推荐答案

请注意，自从写完其他答案后，问题已经发生了变化.

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

g++ 在当前表单中没有发出警告的原因可能是因为 mst 是一个引用，并且构造和破坏一个引用没有副作用.确实，这里的引用延长了临时对象的生命周期，这对其构造函数和析构函数都有影响，但显然 g++ 并没有意识到这会有所不同.

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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