MethodTable内存空间分配中加法运算算法解析

　　在分析MethodTable具体分配内存实现的时候，看到了计算MethodTable的大小，然后分配空间的算法。其中有个加法运算实现的非常赞，特地截取出来。

         所有的MethodTable的分配，都是通过methodtable中的一个static方法AllocagteNewMT来实现的，该方法定义如下：

MethodTable * MethodTable::AllocagteNewMT(EEClass *pClass,

                                         DWORD dwVtableSlots,

                                         DWORD dwGCSize,

                                         DWORD dwNumInterfaces,

                                         DWORD numGenericArgs,

                                         DWORD dwNumDicts,

                                         DWORD cbDict,

                                         ClassLoader *pClassLoader,

                                         BaseDomain *pDomain,

                                         BOOL isInterface,

                                         BOOL fHasGenericsStaticsInfo,

                                         BOOL fNeedsRemotableMethodInfo,

                                         BOOL fNeedsRemotingVtsInfo,

                                         BOOL fHasThreadOrContextStatics

        , AllocMemTracker *pamTracker

)

 

         下面是该方法中计算大小的一段，采用模板来忽略类型带来的影响：

   　　 DWORD cbTotalSize = 0;

    　　DWORD cbDicts = 0;

    　　if (!ClrSafeInt<DWORD>::multiply(dwNumDicts, sizeof(TypeHandle*), cbDicts) ||

     　　   !ClrSafeInt<DWORD>::addition((DWORD)size, cbDicts, cbTotalSize) ||

       　　 !ClrSafeInt<DWORD>::addition(cbTotalSize, dwGCSize, cbTotalSize))

  　　      ThrowHR(COR_E_OVERFLOW);

        

         然后转到addition((DWORD)size, cbDicts, cbTotalSize)的实现，加法的实现如下，加入了对各种情况的严格考虑：

    // Returns true if safe, false on overflow

    static inline bool addition(T lhs, T rhs, T &result)

{

                   //check for T first.

        if(IsSigned())

        {

            //test for +/- combo

            if(!IsMixedSign(lhs, rhs))

            {

                //either two negatives, or 2 positives, not mixed symbols

                if(rhs < 0)

                {

                    //two negatives

                    if(lhs < (T)(MinInt() - rhs)) //remember rhs < 0

                    {

                        return false;

                    }

                    //ok

                }

                else

                {

                    //two positives

                    if((T)(MaxInt() - lhs) < rhs)

                    {

                        return false;

                    }

                    //OK

                }

            }

            //else overflow not possible

            result = lhs + rhs;

            return true;

        }

        else //unsigned, and two symbols is mixed

        {

            if((T)(MaxInt() - lhs) < rhs)

            {

                return false;               

            }

            result = lhs + rhs;

            return true;

        }

}

 

其中，涉及到中间调用的几个方法如下：

static bool IsSigned()

{

return( (T)-1 < 0 );

}

 

//Check if lhs and rhs is mixed Sign symbols

static bool IsMixedSign(T lhs, T rhs)

{

return ((lhs ^ rhs) < 0);

}

 

 

//both of the following should optimize away

static T MinInt()

{

if(IsSigned())

                   {

return (T)((T)1 << (BitCount()-1));

                   }

    else

                   {

return ((T)0);

                   }

}

static T MaxInt()

{

if(IsSigned())

                   {

return (T)~((T)1 << (BitCount()-1));

}

//else

return (T)(~(T)0);

}

 

检查的挺详细的，实现的也挺不错。

 

lbq1221110@Cnblogs.  11.5 ; first post at sscli.cnblogs.com