This works because there cannot be a gap in the pages in memory. Here is some sample code:// Any unreasonably large value will work say for example 0x100000 or 100,000h void ChangeSizeOfImage(DWORD NewSize) { __
In Windows CE 3.0 it is 256 MB with a 32 MB limit on each file, but in Windows CE .NET this value has been increased to a rather large 4 GB. In addition, there is a limit to the number of files that can be stored in RAM (4 million) and to the number of programs that can ...
LOAD_LIBRARY_REQUIRE_SIGNED_TARGET :NULL);//Ignored because ActualFlags can't have 0x1000 (if called from LoadLibraryExW), this value is used probably in calls from different functions.FlagUsed |= ((ActualFlags &0x1000) ?0x100:0x0);//Ignored because ActualFlags can't be negative (if cal...
2.2.4.378 serarr:ArrayOfKeyValueOfTupleOflongDnsResourceRecordTypem1ahUJFxIpamExceptionVfr71_PXs 2.2.4.379 serarr:ArrayOflong 2.2.4.380 serarr:ArrayOfstring 2.2.4.381 serarr:ArrayOfunsignedByte 2.2.4.382 serarr:ArrayOfunsignedShort 2.2.4.383 ServerDataFormatter 2.2.4.384 ServerInfo 2.2.4.385 Ser...
u2: an unsigned 16-bit integer in big-endian byte order u4: an unsigned 32-bit integer in big-endian byte order table: an array of variable-length items of some type. The number of items in the table is identified by a preceding count number, but the size in bytes of the table can...
foriinrange(2,n+1): input[i] =input[i-1]+input[i-2] input[i] =c_uint(input[i]).value returnhex(input[n]) 个人遇到的最大问题是python中的变量,如何转换为 c 语言类型。 使用ctype 类,文档中给出了各种数据类型的关键字: 本题需要将结果转换为 unsigned long,所以选择 c_uint 即可。注意...
Defining the IOCTL Value and Buffer In this approach, we’ve decided to simply use the standard of beginning our IOCTL functions at 0x100, and picking the FILE_DEVICE_UNKNOWN device type instead of defining our own. The input data structure contains the actual pointers to the input and output...
Broadcom is one of the major vendors of wireless devices worldwide. Since these chips are so widespread they constitute a high value target to attackers and any vulnerability found in them should be considered to pose high risk. In this blog post I provi
Possibly the second byte is unused, or is padding. Another possibility, though, is that the first two bytes together represent a single 16-bit value, stored in little-endian order.What is little-endian? When you store a numerical value that is larger than a single byte, you have to ...
This leaves an unused 32-bit value that was wasted for alignment purposes on earlier versions of Windows. In Windows 8.1, this is now used to store the TypeInfo field, which is the Object Type Index in the Object Type Index Table (nt!ObTypeIndexTable). Dereferencing this index quickly rev...