c++ 禁用ScopedGuard的“未使用的变量”

ybzsozfc  于 2023-08-09  发布在  其他
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我在和安德烈· Alexandria 雷斯库和彼得鲁·马吉尼安一起玩
当你用-Wall -Werror编译它时,你会得到“未使用的变量”错误。下面的代码取自LOKI

class ScopeGuardImplBase
{
    ScopeGuardImplBase& operator =(const ScopeGuardImplBase&);

protected:

    ~ScopeGuardImplBase()
    {}

    ScopeGuardImplBase(const ScopeGuardImplBase& other) throw() 
        : dismissed_(other.dismissed_)
    {
        other.Dismiss();
    }

    template <typename J>
    static void SafeExecute(J& j) throw() 
    {
        if (!j.dismissed_)
            try
            {
                j.Execute();
            }
            catch(...)
            {}
    }

    mutable bool dismissed_;

public:
    ScopeGuardImplBase() throw() : dismissed_(false) 
    {}

    void Dismiss() const throw() 
    {
        dismissed_ = true;
    }
};

////////////////////////////////////////////////////////////////
///
/// \typedef typedef const ScopeGuardImplBase& ScopeGuard
/// \ingroup ExceptionGroup
///
/// See Andrei's and Petru Marginean's CUJ article
/// http://www.cuj.com/documents/s=8000/cujcexp1812alexandr/alexandr.htm
///
/// Changes to the original code by Joshua Lehrer:
/// http://www.lehrerfamily.com/scopeguard.html
////////////////////////////////////////////////////////////////

typedef const ScopeGuardImplBase& ScopeGuard;

template <typename F>
class ScopeGuardImpl0 : public ScopeGuardImplBase
{
public:
    static ScopeGuardImpl0<F> MakeGuard(F fun)
    {
        return ScopeGuardImpl0<F>(fun);
    }

    ~ScopeGuardImpl0() throw() 
    {
        SafeExecute(*this);
    }

    void Execute() 
    {
        fun_();
    }

protected:
    ScopeGuardImpl0(F fun) : fun_(fun) 
    {}

    F fun_;
};

template <typename F> 
inline ScopeGuardImpl0<F> MakeGuard(F fun)
{
    return ScopeGuardImpl0<F>::MakeGuard(fun);
}

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问题在于使用:

ScopeGuard scope_guard = MakeGuard(&foo);


这只是

const ScopeGuardImplBase& scope_guard = ScopeGuardImpl0<void(*)()>(&foo);


我使用宏得到一些科普结束时的一个范围:

#define SCOPE_GUARD ScopedGuard scope_guard = MakeGuard


这样用户就可以直接调用

SCOPE_GUARD(&foo, param) ...


这个宏使得禁用未使用的警告很困难。
有人能帮助我更好地理解这一点,也许提供一个解决方案,而不使用-Wno-unused-variable?

vuktfyat

vuktfyat1#

你可以试试老方法:

(void)scope_guard;

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