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对于MVA到PA转换过程,以及描述符所扮演的角色,请看下图:


来看代码:

unsigned long virtuladdr, physicaladdr;
unsigned Sale Id Id Cheap Card Buy Fake long *mmu_tlb_base = (unsigned long *)0x30000000;//对应上图TTB
virtuladdr = 0xA0000000;//对应上图MVA
physicaladdr = 0x56000000;//对应上图PA
*(mmu_tlb_base + (virtuladdr >> 20)) = (physicaladdr & 0xFFF00000) | \
                                            MMU_SECDESC;
                                            //*(mmu_tlb_base + (virtuladdr >> 20)) = 
                                            //*(TTB[31-14]+MVA[31-20]+00) = *(描述符地址) 
Sale Id Id Cheap Card Buy Fake //这里作用就是将描述符的值给定(bit[31-20]为PA的[31-20],bit[19-0]自己给定一些段描述符的操作),CPU系统自动根据此发生操作,获得实际PA地址。

我们根据上面可以得到:段描述符的地址=Translation base(高18位)-TTB bit[31-14]+Table index(低12位)-MVA bit[31-20]+00(低两位)不少人会以为*(mmu_tlb_base | virtuladdr>>18)才是正确的。

解释:

首先mmu_tlb_base是一个unsigned long *型,而virtuladdr是unsigned long型,位或操作会出错!

查阅资料可知,一级页表的地址TTB必须16KB对齐,且位[14:0]必为0!

所以我们用+Self Your Information Licence Skipton Drive Driving News About Important|是一样的。而我们的mmu_tlb_base是unsigned long*型的,

在32位中,mmu_tlb_base + 1 = mmu_tlb_base + 1*sizeof(unsigned long*)=mmu_tlb_base + 4

Sale Id Id Cheap Card Buy Fake 所以我们mmu_tlb_base + (virtuladdr >> 20) = mmu_tlb_base +( (virtuladdr >> 20))* 4 = mmu_tlb_base +( (virtuladdr >> 20))<<2 = mmu_tlb_base + (virtuladdr >>18

实验验证:

#include 

int main(Sale Id Id Cheap Card Buy Fake Sale Id Id Cheap Card Buy Fake int argc,char ** argv)
{
    unsigned long *a = (unsigned long *)0x30000000;
    unsigned long b = 0xA0000000;
    printf("a+(b>>20) = %#x.\n\n",a+(b>>20Maker Id Virtual License Louisiana Fake Drivers - Card));
    Sale Id Id Cheap Card Buy Fake printf("a | ((b>>20)<<2) = %#x",0x30000000 | ((b>>20)<<2));
    return 0;
} 

结果:



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