标签：char conf 优势 any str ssi output hit images
64位平台的介绍
- IA-64 is a 64-bit microprocessor architecture developed by Intel and Hewlett Packard companies together. It is implemented in Itanium and Itanium 2 microprocessors. To learn more about the architecture IA-64 see the following Wikipedia article "Itanium".
- Intel 64 (EM64T / AMD64 / x86-64 / x64) is an extension of x86 architecture with full backward compatibility. There are many variants of its name, and it causes some confusion, but all these names mean the same thing: x86-64, AA-64, Hammer Architecture, AMD64, Yamhill Technology, EM64T, IA-32e, Intel 64, x64. To learn how so many names appeared see the article in Wikipedia: "X86-64".
64位平台的优势
Win64 Program model
Like in Win32, the size of a page in Win64 is 4 Kbytes. The first 64 Kbytes of the address space are never displayed, so the lowest correct address is 0x10000. Unlike Win32, system DLL‘s take more than 4 Gbytes.
Compilers for Intel 64 have one peculiarity: they can use registers with great efficiency to pass parameters into functions, instead of using the stack. This allowed the Win64 architecture developers to get rid of notions such as a calling convention. In Win32, you may use various conventions: __stdcall, __cdecl, __fastcall, etc. In Win64, there is only one calling convention. Here is an example of how four arguments of integer type are passed through registers:
- RCX: the first argument
- RDX: the second argument
- R8: the third argument
- R9: the fourth argument
The arguments following the first four integers are passed through the stack. To pass float arguments XMM0-XMM3 registers are used as well as the stack.
The difference in the calling conventions makes it impossible to use both 64-bit and 32-bit code in one program. In other words, if an application has been compiled for the 64-bit mode, all the libraries (DLL) being used must also be 64-bit.
Passing parameters through registers is one of the innovations that make 64-bit programs faster than 32-bit ones.
64位平台应用程序的性能
After being recompiled for a 64-bit system a program can use huge amounts of memory and its speed will increase in 5-15%. 5-10% of speed gain is achieved due to architectural features of the 64-bit processor, for example, a larger number of registers. And another 1-5% performance gain is determined by the absence of the WoW64 layer that translates calls between 32-bit applications and the 64-bit operating system.
For example, Adobe company says that a new 64-bit "Photoshop CS4" is 12% faster than its 32-bit version".
64位平台下容易犯的错误
struct BitFieldStruct { unsigned short a:15; unsigned short b:13; }; BitFieldStruct obj; obj.a = 0x4000; size_t addr = obj.a << 17; //Sign Extension printf("addr 0x%Ix\n", addr); //Output on 32-bit system: 0x80000000 //Output on 64-bit system: 0xffffffff80000000 // obj.a is converted into signed int before shift.
int A = -2; unsigned B = 1; int array[5] = { 1, 2, 3, 4, 5 }; int *ptr = array + 3; ptr = ptr + (A + B); //Invalid pointer value on 64-bit platform printf("%i\n", *ptr); //Access violation on 64-bit platform
原因如下：
Let us follow the algorithm of calculating the expression "ptr + (A + B)":
- According to C++ rules, the variable A of the type int is converted to unsigned.
- A and B are summed and we get the value 0xFFFFFFFF of unsigned type.
- The expression "ptr + 0xFFFFFFFFu" is calculated.
The result of this process depends upon the size of the pointer on a particular architecture. If the addition takes place in the 32-bit program, the expression is equivalent to "ptr - 1" and the program successfully prints the value "3". In the 64-bit program, the value 0xFFFFFFFFu is fairly added to the pointer. As a result, the pointer gets far outside the array while we encounter some troubles when trying to get access to the item by this pointer.
Like in the first case, we recommend you to use only memsize-types in pointer arithmetic to avoid the situation described above. Here are one ways to correct the code:
ptr = ptr + (ptrdiff_t(A) + ptrdiff_t(B)); ptrdiff_t A = -2; size_t B = 1; ... ptr = ptr + (A + B);
指针类型的强制转换也容易导致错误
int array[4] = { 1, 2, 3, 4 }; enum ENumbers { ZERO, ONE, TWO, THREE, FOUR }; //safe cast (for MSVC) ENumbers *enumPtr = (ENumbers *)(array); cout << enumPtr[1] << " "; //unsafe cast size_t *sizetPtr = (size_t *)(array); cout << sizetPtr[1] << endl; //Output on 32-bit system: 2 2 //Output on 64-bit system: 2 17179869187=0x0000000400000003
原因：在64-bit系统中，size_t 占8bytes，因此，sizetPtr[1]指向的是内存各占4btyes的数字3和4拼接起来的内存块，也就是0x0000000400000003
Storage of integer values in double
size_t a = size_t(-1); double b = a; --a; --b; size_t c = b; // x86: a == c // x64: a != c
原因：
This code may be justified when it is executed on a 32-bit system because the type double has 52 significant bits and can store a 32-bit integer value without loss. But when you save a 64-bit integer number into double, the exact result will be lost.
结构体占用的内存字节数
struct MyStruct { bool m_bool; char *m_pointer; int m_int; }; // x86: sizeof(MyStruct) = 12 // x64: sizeof(MyStruct) = 24
struct MyStruct { char *m_pointer; int m_int; bool m_bool; }; // x86: sizeof(MyStruct) = 12 // x64: sizeof(MyStruct) = 16
【 字节对齐的窍门】：The process of optimizing a field arrangement may seem complicated. But there is a very simple and very effective method: you just need to arrange the fields in decreasing order of their sizes.
[参考文献]：
标签：char conf 优势 any str ssi output hit images
原文地址：http://www.cnblogs.com/cnpirate/p/7026779.html
