我想知道这些代码行背后的数学原理。处理指针。有人能用if语句中的+运算符和&位运算符向我演示一下数学吗?我就是不太明白。
// check signatures -- must be a PE
pDosHeader = (PIMAGE_DOS_HEADER)hMap;
if(pDosHeader->e_magic != IMAGE_DOS_SIGNATURE) goto cleanup;
pNtHeaders = (PIMAGE_NT_HEADERS)((DWORD)hMap + pDosHeader->e_lfanew);
if(pNtHeaders->Signature != IMAGE_NT_SIGNATURE) goto cleanup;
// Not dll
if (pNtHeaders->FileHeader.Characteristics & IMAGE_FILE_DLL
&& pNtHeaders->FileHeader.Characteristics & IMAGE_FILE_EXECUTABLE_IMAGE) goto cleanup;
// get last section's header...
pSectionHeader = (PIMAGE_SECTION_HEADER)((DWORD)hMap + pDosHeader->e_lfanew + sizeof(IMAGE_NT_HEADERS));
pSection = pSectionHeader;
pSection += (pNtHeaders->FileHeader.NumberOfSections - 1);发布于 2012-11-25 05:31:02
+运算符只是一个算术加号。当与指针一起使用时,它现在变得更清楚为什么C和C++中的指针是类型化的-它不仅向地址添加字节,而且添加它所指向的任何类型的大小。
例如,如果我们有:
struct stuff x;
struct stuff *p = &x; /* p now points at x */
p = p + 1;
/* the address at p has been incremented by the sizeof(struct stuff),
and is pointing at the next struct stuff in memeory */用作二元运算符的&是按位AND,它携带在两个操作数中设置的位。例如:
unsigned int b = 99; /* 99 is binary 01100011 */
unsigned int w = b & 6; /* 6 is binary 00000110 */
/* w is now 2 2 is binary 00000010 */在您的示例代码中,它看起来是用来测试是否在结构成员中设置了位掩码IMAGE_FILE_DLL和IMAGE_FILE_EXECUTABLE_IMAGE。
https://stackoverflow.com/questions/13545666
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