【发布时间】：2012-12-20 04:57:59
【问题描述】：
考虑下面的代码：
#include <stdio.h>
#include <stdlib.h>
#define FORCE_CAST(var, type) *(type*)&var
struct processor_status_register
{
unsigned int cwp:5;
unsigned int et:1;
unsigned int ps:1;
unsigned int s:1;
unsigned int pil:4;
unsigned int ef:1;
unsigned int ec:1;
unsigned int reserved:6;
unsigned int c:1;
unsigned int v:1;
unsigned int z:1;
unsigned int n:1;
unsigned int ver:4;
unsigned int impl:4;
}__attribute__ ((__packed__));
struct registers
{
unsigned long* registerSet;
unsigned long* globalRegisters;
unsigned long* cwptr;
unsigned long wim, tbr, y, pc, npc;
unsigned short registerWindows;
/* Though Intel x86 architecture allows un-aligned memory access, SPARC mandates memory accesses to be 8 byte aligned. Without __attribute__ ((aligned (8))) or a preceding dummy byte e.g. unsigned short dummyByte, the code below crashes with a dreaded Bus error and Core dump. For more details, follow the links below:
http://blog.jgc.org/2007/04/debugging-solaris-bus-error-caused-by.html
https://groups.google.com/forum/?fromgroups=#!topic/comp.unix.solaris/8SgFiMudGL4
*/
struct processor_status_register __attribute__ ((aligned (8))) psr;
}__attribute__ ((__packed__));
int getBit(unsigned long bitStream, int position)
{
int bit;
bit = (bitStream & (1 << position)) >> position;
return bit;
}
char* showBits(unsigned long bitStream, int startPosition, int endPosition)
{
// Allocate one extra byte for NULL character
char* bits = (char*)malloc(endPosition - startPosition + 2);
int bitIndex;
for(bitIndex = 0; bitIndex <= endPosition; bitIndex++)
bits[bitIndex] = (getBit(bitStream, endPosition - bitIndex)) ? '1' : '0';
bits[bitIndex] = '\0';
return bits;
}
int main()
{
struct registers sparcRegisters; short isLittleEndian;
// Check for Endianness
unsigned long checkEndian = 0x00000001;
if(*((char*)(&checkEndian)))
{printf("Little Endian\n"); isLittleEndian = 1;} // Little
Endian architecture detected
else
{printf("Big Endian\n"); isLittleEndian = 0;} // Big
Endian architecture detected
unsigned long registerValue = 0xF30010A7;
unsigned long swappedRegisterValue = isLittleEndian ? registerValue :
__builtin_bswap32(registerValue);
sparcRegisters.psr = FORCE_CAST(swappedRegisterValue, struct
processor_status_register);
registerValue = isLittleEndian ? FORCE_CAST (sparcRegisters.psr,
unsigned long) : __builtin_bswap32(FORCE_CAST (sparcRegisters.psr,
unsigned long));
printf("\nPSR=0x%0X, IMPL=%u, VER=%u, CWP=%u\n", registerValue,
sparcRegisters.psr.impl, sparcRegisters.psr.ver,
sparcRegisters.psr.cwp);
printf("PSR=%s\n",showBits(registerValue, 0, 31));
sparcRegisters.psr.cwp = 7;
sparcRegisters.psr.et = 1;
sparcRegisters.psr.ps = 0;
sparcRegisters.psr.s = 1;
sparcRegisters.psr.pil = 0;
sparcRegisters.psr.ef = 0;
sparcRegisters.psr.ec = 0;
sparcRegisters.psr.reserved = 0;
sparcRegisters.psr.c = 0;
sparcRegisters.psr.v = 0;
sparcRegisters.psr.z = 0;
sparcRegisters.psr.n = 0;
sparcRegisters.psr.ver = 3;
sparcRegisters.psr.impl = 0xF;
registerValue = isLittleEndian ? FORCE_CAST (sparcRegisters.psr,
unsigned long) : __builtin_bswap32(FORCE_CAST (sparcRegisters.psr,
unsigned long));
printf("\nPSR=0x%0X, IMPL=%u, VER=%u, CWP=%u\n", registerValue,
sparcRegisters.psr.impl, sparcRegisters.psr.ver,
sparcRegisters.psr.cwp);
printf("PSR=%s\n\n",showBits(registerValue, 0, 31));
return 0;
}
我在 SPARC 上的 Solaris 10 上使用 gcc-4.7.2 来编译以下内容 生成 Big-Endian 输出的代码：
Big Endian
PSR=0xF30010A7, IMPL=3, VER=15, CWP=20
PSR=11110011000000000001000010100111
PSR=0x3F00003D, IMPL=15, VER=3, CWP=7
PSR=00111111000000000000000000111101
我在 Intel-x86 上的 Ubuntu-10.04 上使用 gcc-4.4 来编译 以下代码生成 Little-Endian 输出：
Little Endian
PSR=0xF30010A7, IMPL=15, VER=3, CWP=7
PSR=11110011000000000001000010100111
PSR=0xF30000A7, IMPL=15, VER=3, CWP=7
PSR=11110011000000000000000010100111
虽然后一个符合预期，但谁能解释一下 大端对应？考虑 showBits() 方法是 正确，PSR=0x3F00003D 如何导致 IMPL=15、VER=3、CWP=7 价值观？位域是如何排列和解释的 Big-Endian 系统上的内存？
【问题讨论】：