??????????????????? ---------------by nasiry?
?????????????????????????????????????轉載請說明出處,并通知我
首先看看eboot的source文件，該文件約束了eboot下文件的編譯方式和參與編譯的文件等等信息

SYNCHRONIZE_DRAIN=1

TARGETNAME=EBOOT
TARGETTYPE=PROGRAM
WINCECPU=1RELEASETYPE=PLATFORM
WINCETARGETFILES=$(_RELEASEDIR)\$(TARGETNAME).nb0
EXEENTRY=StartUp
EXEBASE=0x80000000

INCLUDES=..\inc;$(_PUBLICROOT)\common\oak\drivers\block\msflashfmd\inc
#INCLUDES=$(_TARGETPLATROOT)\eboot\inc;$(INCLUDES)
INCLUDES=$(_TARGETPLATROOT)\Drivers\NandFlsh\FMD;$(INCLUDES)
ADEFINES=-pd "ALLOCATE_TABLE SETS \"FALSE\"" $(ADEFINES)

LDEFINES=-subsystem:native /DEBUG /DEBUGTYPE:CV /FIXED:NO
CDEFINES= $(CDEFINES) -DPPSH_PROTOCOL_NOTIMEOUT -DCOREDLL -DPLAT_ONBOARDEDBG=1 -DBOOT_LOADER=1 -DNOSYSCALL=1

TARGETLIBS=\
.......

Source文件僅僅提供了PROGRAM/LIBRARY/DYNLINK三個選項參數(見oak\MISC\makefile.def line 324),分別對應.exe,.lib,.dll文件。.exe和.dll文件都是符合windows程序要求的帶頭文件，而.lib僅僅是可重定位的函數集。這三者都不是我們通常用作loader的headless程序。所謂headless，不包含文件執行頭，按照最后所在的地址信息進行鏈接，不可重定位為特點，這與通常的單片機程序是相似的。而上面所能指定的三種編譯目標模式是用于winCE執行所用，其中包含了一個所謂類似PE頭的信息部。這樣產生的程序顯然就不能滿足運行的要求。下面是編譯后產生的程序用實例來看看：

00000000h: 4D 5A 90 00 03 00 00 00 04 00 00 00 FF FF 00 00 ; MZ?..........
00000010h: B8 00 00 00 00 00 00 00 40 00 00 00 00 00 00 00 ; ?......@.......
00000020h: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ; ................
00000030h: 00 00 00 00 00 00 00 00 00 00 00 00 D8 00 00 00 ; ............?..
00000040h: 0E 1F BA 0E 00 B4 09 CD 21 B8 01 4C CD 21 54 68 ; ..?.???L?Th
00000050h: 69 73 20 70 72 6F 67 72 61 6D 20 63 61 6E 6E 6F ; is program canno
00000060h: 74 20 62 65 20 72 75 6E 20 69 6E 20 44 4F 53 20 ; t be run in DOS
00000070h: 6D 6F 64 65 2E 0D 0D 0A 24 00 00 00 00 00 00 00 ; mode....$.......
00000080h: 6A 57 E0 F7 2E 36 8E A4 2E 36 8E A4 2E 36 8E A4 ; jW圜.6帳.6帳.6帳
00000090h: A1 13 BF A4 2C 36 8E A4 A1 13 BE A4 22 36 8E A4 ; ?郡,6帳?兢"6帳
000000a0h: 01 10 BE A4 29 36 8E A4 01 10 BF A4 2F 36 8E A4 ; ..兢)6帳..郡/6帳
000000b0h: 2E 36 8E A4 3F 36 8E A4 D6 3D D8 A4 2F 36 8E A4 ; .6帳?6帳?丐/6帳
000000c0h: 52 69 63 68 2E 36 8E A4 00 00 00 00 00 00 00 00 ; Rich.6帳........
000000d0h: 00 00 00 00 00 00 00 00 50 45 00 00 C2 01 04 00 ; ........PE..?..
000000e0h: B2 42 8F 41 00 00 00 00 00 00 00 00 E0 00 2E 01 ; 睟廇........?..
000000f0h: 0B 01 06 18 00 38 01 00 00 28 00 00 00 00 00 00 ; .....8...(......
00000100h: 00 10 00 00 00 10 00 00 00 50 01 00 00 00 00 80 ; .........P.....€
00000110h: 00 10 00 00 00 02 00 00 04 00 00 00 00 00 00 00 ; ................
00000120h: 04 00 14 00 00 00 00 00 00 90 01 00 00 04 00 00 ; .........?.....
00000130h: 28 2F 02 00 09 00 00 00 00 00 01 00 00 10 00 00 ; (/..............
00000140h: 00 00 10 00 00 10 00 00 00 00 00 00 10 00 00 00 ; ................
00000150h: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ; ................
00000160h: 00 00 00 00 00 00 00 00 00 70 01 00 C8 05 00 00 ; .........p..?..
00000170h: 00 00 00 00 00 00 00 00 00 80 01 00 04 07 00 00 ; .........€......
00000180h: D0 14 00 00 1C 00 00 00 00 00 00 00 00 00 00 00 ; ?..............
00000190h: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ; ................
000001a0h: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ; ................
000001b0h: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ; ................
000001c0h: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ; ................
000001d0h: 2E 74 65 78 74 00 00 00 9C 36 01 00 00 10 00 00 ; .text...?......
000001e0h: 00 38 01 00 00 04 00 00 00 00 00 00 00 00 00 00 ; .8..............
000001f0h: 00 00 00 00 20 00 00 60 2E 64 61 74 61 00 00 00 ; .... ..`.data...
00000200h: C8 18 00 00 00 50 01 00 00 02 00 00 00 3C 01 00 ; ?...P.......<..
00000210h: 00 00 00 00 00 00 00 00 00 00 00 00 40 00 00 C0 ; ............@..?
00000220h: 2E 70 64 61 74 61 00 00 C8 05 00 00 00 70 01 00 ; .pdata..?...p..
00000230h: 00 06 00 00 00 3E 01 00 00 00 00 00 00 00 00 00 ; .....>..........
00000240h: 00 00 00 00 40 00 00 40 2E 72 65 6C 6F 63 00 00 ; ....@..@.reloc..
00000250h: F0 07 00 00 00 80 01 00 00 08 00 00 00 44 01 00 ; ?...€.......D..
00000260h: 00 00 00 00 00 00 00 00 00 00 00 00 40 00 00 42 ; ............@..B
00000270h: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ; ................
00000280h: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ; ................
00000290h: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ; ................
000002a0h: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ; ................
000002b0h: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ; ................
000002c0h: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ; ................
000002d0h: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ; ................
000002e0h: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ; ................
000002f0h: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ; ................
00000300h: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ; ................
00000310h: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ; ................
00000320h: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ; ................
00000330h: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ; ................
00000340h: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ; ................
00000350h: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ; ................
00000360h: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ; ................
00000370h: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ; ................
00000380h: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ; ................
00000390h: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ; ................
000003a0h: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ; ................
000003b0h: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ; ................
000003c0h: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ; ................
000003d0h: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ; ................
000003e0h: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ; ................
000003f0h: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ; ................
00000400h: 2A 00 00 EA FD FF FF EA FC FF FF EA FB FF FF EA ; *..挲挈犒?
00000410h: FA FF FF EA F9 FF FF EA F8 FF FF EA 74 00 9F E5 ; ?犍犋阾.熷
00000420h: 55 1C A0 E3 00 10 80 E5 6C 00 9F E5 20 10 A0 E3 ; U.犮..€錶.熷 .犮
00000430h: 00 10 80 E5 64 50 9F E5 00 60 95 E5 01 65 86 E3 ; ..€錮P熷.`曞.e嗐
00000440h: 5C 30 9F E5 00 40 93 E5 06 48 84 E3 07 46 C4 E3 ; \0熷.@撳.H勩.F你
00000450h: 06 46 84 E3 4C 20 9F E5 4C 10 9F E5 10 00 A0 E3 ; .F勩L 熷L.熷..犮
從上面dump出來的內容可以看到，程序的前0x400區域都是由windowsCE的編譯器產生的頭文件，而這部分內容本身不是我們的代碼指定的內容也不具備可執行的能力，所以實際的執行內容是在0x400位置開始的，但是是不是考慮直接拷貝0x400后的區域去掉頭的位置就可以直接作為我們需要的執行入口點呢？答案是否定的，這中間還有一個偏移量的問題，由于偏移不同的原因，絕對跳轉的代碼在程序中不會得到正確執行。所以這個程序需要使用romimage命令處理后才能成為可用的image.

romimage使用的是boot.bib的內容來對上面的程序進行處理的。我們來看看這部分的內容。

MEMORY
?? DRV_GLB? 8c020000? 00001000? RESERVED
??? BIN_FS?? 8c021000? 00005000? RESERVED
??? RAM????? 8c026000? 00006000? RAM
??? STACK??? 8c02c000? 00004000? RESERVED
??? EBOOT??? 8c038000? 00020000? RAMIMAGE
;??? EBOOT??? 8c038000? 00040000? RAMIMAGE

???? FLSCACHE 8D000000? 01000000? RESERVED

CONFIG
???????? COMPRESSION=OFF
?????????PROFILE=OFF
?????????KERNELFIXUPS=ON
?????????SRE=ON
?????????ROMSTART=8c038000
?????????ROMWIDTH=32

??????? ROMSIZE=16000

MODULES
nk.exe $(_TARGETPLATROOT)\target\$(_TGTCPU)\$(WINCEDEBUG)\EBOOT.exe EBOOT

?

簡單說一次上面的內容的意義：

1.MEMORY部：
這部分指定了一些空間分配的情況，eboot的位置就是上面指定的，romimage通過該信息重新定位上面eboot.exe的偏移，以及內存使用的靜態分布信息.
2.Config部
該部分描述生成image的規則，包括壓縮與否，sre文件是否產生，起始地址,image的存放介質位寬，單片介質大小等等。
3.Modules部
該部分描述的是參與該image生成的元文件列表。

通過上面的信息，按道理romimage就會將0x400后的部分重新計算偏移后放到0x8c038000的位置。也就是我們文件的0x0位置。事實上呢？

我們來看看是什么樣的情況。

00000000h: FE 03 00 EA 00 00 00 00 00 00 00 00 00 00 00 00 ; ?.?...........
00000010h: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ; ................
00000020h: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ; ................
00000030h: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ; ................
00000040h: 45 43 45 43 88 C7 04 8C 00 00 00 00 00 00 00 00 ; ECEC埱.?.......
00000050h: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ; ................
00000060h: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ; ................
.................. zero .................................................
00000fc0h: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ; ................
00000fd0h: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ; ................
00000fe0h: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ; ................
00000ff0h: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ; ................
00001000h: 2A 00 00 EA FD FF FF EA FC FF FF EA FB FF FF EA ; *..挲挈犒?
00001010h: FA FF FF EA F9 FF FF EA F8 FF FF EA 74 00 9F E5 ; ?犍犋阾.熷
00001020h: 55 1C A0 E3 00 10 80 E5 6C 00 9F E5 20 10 A0 E3 ; U.犮..€錶.熷 .犮
00001030h: 00 10 80 E5 64 50 9F E5 00 60 95 E5 01 65 86 E3 ; ..€錮P熷.`曞.e嗐
00001040h: 5C 30 9F E5 00 40 93 E5 06 48 84 E3 07 46 C4 E3 ; \0熷.@撳.H勩.F你
00001050h: 06 46 84 E3 4C 20 9F E5 4C 10 9F E5 10 00 A0 E3 ; .F勩L 熷L.熷..犮
00001060h: 06 00 00 EA 00 00 00 00 00 00 00 00 00 00 00 00 ; ...?...........
00001070h: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ; ................
00001080h: 00 60 85 E5 01 00 50 E2 FD FF FF 1A 00 40 83 E5 ; .`呭..P愷..@冨

結果是程序最前端的4Kb空出來了！！！windowsCE的編譯器是否保留這部分作為chain的空間還無法得知，但是這樣一來所有的中斷向量在這里都無法使能了。除reset的中斷向量是可以間接生效的外，其他的所有的中斷向量都被寫成了0x0如果產生一個任意的異常/中斷的話，就只會連續的產程一系列數據異常，無法復位也無法執行。這就是為什么在windowsCE的啟動代碼中都不直接使用0x0區域的rom作為中斷向量的安裝手段。
?
后1：
這個現象造成的原因是eboot和應用程序使用相同的編譯器，該編譯器會將開始的4KB作為PE頭的放置，所以該部分要空出來。

總結

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