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Chris · Guest

Just a quick compile time, template, optimized out question for anyone
that has time to answer this.

Let's say I define some template class like such:

typedef char One;
typedef struct { char a[2]; } Two;

template<typename B>
struct CaseOneCheck
{
private:
template<typename C> static One DepTest(typename C::CaseOne *);
template<typename C> static Two DepTest(...);

public:

enum { Dependent = sizeof(CaseOneCheck<B>::DepTest<B>(0)) ==
1 };

typedef double Type;

inline static bool Check(Type amount)
{
return(amount > 0);
}
};

Where I define:

struct WidgetOne
{
typedef bool CaseOne;
};

struct WidgetTwo
{
typedef bool CaseTwo;
};

Then, a method:

template<typename Arg,
template<typename> class CheckPolicy>
bool ShouldEnable(typename CheckPolicy<Arg>::Type type)
{
bool returnCode = true;

if(CheckPolicy<Arg>::Dependent)
{
returnCode = CheckPolicy<Arg>::Check(type);
}

return returnCode;
}

Will, since I am assuming the if(CheckPolicy<Arg>::Dependent) is
filled in to true or false at compile time, then cause the compiler to
optimize out the returnCode = CheckPolicy<Arg>::Check(type); line if
the if is false at the "second" compile stage? Or will it always be
there? The compiler I am using is g++ version 3.3.3.

That is, depending on the dependence, the code will be

ShouldEnable()
{
bool returnCode = true;
return returnCode;
}

When I use:

bool test = ShouldEnable<WidgetTwo, CheckOne>(5.5);

?

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Mathias Gaunard · Guest

On May 19, 2:04 am, Chris <foureightye...@yahoo.com> wrote:

Guest · Posted: Tue May 22, 2007 12:10 am Post subject: Re: Optimized out?

On May 19, 2:04 am, Chris <foureightye...@yahoo.com> wrote: