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林與安德烈Alexandrescu的和彼得魯杜米特Marginean打scoped guard object禁用「未使用變量」爲ScopedGuard
當你與-Wall -Werror編譯你會得到「未使用變量」的錯誤。
ScopeGuard scope_guard = MakeGuard(&foo);
這只是
const ScopeGuardImplBase& scope_guard = ScopeGuardImpl0<void(*)()>(&foo);
即時通訊使用宏來得到在一月底的一些動作:下面的代碼是從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);
}
的問題是使用取應付:
#define SCOPE_GUARD ScopedGuard scope_guard = MakeGuard
這種方式,用戶可以叫
SCOPE_GUARD(&foo, param) ...
這個宏使其難以禁止未使用的警告。
有人可以幫助我更好地理解這一點,也許提供一個解決方案,而不使用-Wno-unused-variable?
的可能的複製[?如何使用範圍後衛時避免警告(http://stackoverflow.com/questions/35587076/how-to-avoid-預警時,使用範圍的後衛) –