为什么加的是0x80

1万 · 2020-10-14 22:33

用的TM4C129ENCPDT ,由hex文件可看出，每行数据长度是0x20个，正常情况不是应该下一行的地址在上一行的基础上+0x20么？

2万 · 2020-10-14 22:43

:2000000074A5AB956D6D6D000000006D6D006D1D6D6D6D6D6D036D6D6D6D6D6D6D6D6D6D76
:200080006D6D6D6D6D6D6D6D6D6D6D6D6D6D6D6D6DB36D6DE96D6D6D6D6DD96D6D6D6D6D92
:200100006D6D6D6D6D6D6D6D4B6D6D6D6D6D6D6D6DBD6D6D00006D6D6D6D006D6D6D6D6D58
:200180006D6D6D6D6D6D6D6D6D6D6D6D6D6D6D6D6D6D6D6D6D6D6D6D6D6D6D6D6D6D6D6DBF
:020200006D008F
:20020800080CC70CC50DC339A8010D08BD009C08200D08B6008E08D115000800008D00699D
:20028800000199000169C5FD080069D700FE000169AFE70001000169A1D9001800009D00B0
:200308008D9D0008008DB784010283819C8D9D049DF006040406AD029D9D40130413030013
:2003880009800506050145048D9D040004044304A006020204049D9B04920449049B040085
:2004080004D207410041418DBD0D00009D400404BB298D0D0010E50202241B0900D24040E7
:2004880040408D1C0501C5FD020090488205400D00AF00F81A00F1071600F107900218063E
:20050800000323C900000005D10B000078EC9840565A0000307403020304051C1C80B00CEE
:200588001C0E0E1617188D009DA682060C03994FC102C00200024002000E00300005A402D5
:20060800000281039104040D888803910404F10000163D9D0D30902D9DD3D1160D00168D18
:2006880002C0F008C2905102020302808801054016C148888803910400768102011B01AF12
:200708006C1A04C1601A079D1A071A686C011B73803F6C001367005F6E6E1F0D1501550D46
:200788000000140000AD041A01130000A6041A01026FADBD10066F0000116E011951C1028C
:2008080080010392480200180600031FA7C1309880000000AD0001010190510A2001912013
:200888000001010150E4020A250A250A5F71250080306C0081AB0000030082000A030C00D4
:200908008200000000000016038D0062000004180603038204818D9D069D02006B0B034787
:200988000000009D042D049D9CED9D406565D406920606030410060303820481030C990462
:200A080040040800FD82069A90EA00011200240020801CB106020611111288889203060652
:200A880003040606020611110B1106D3AD820111E100009D8D04019D809D408D1801E193B7
:200B080004008064004F0006080300001179110C03820602823D038211110E40110306067D
:200B880003A4111108D339820B401103060603FF00020682C1038211020C030E30021290AD
:200C08000C1103820B0A20100E16030053019D004040AD0391626200009D408D2B00D008DB
:200C8800CB6200004FAD1111D0D101D3490100400F0160620000F1019D06F040AD03CE628B
:200D08006200009D408DBD18017F115200B120A1624FADBD1801816020606200009D06F04B
:200D88004FAD039962110001220349800F6300408DBD180129201114905103400413F01390
:200E0800EA00B120CF6240AD1313D0D101D3490100400F01606200009D06F04FAD039B626B
:200E8800140001220349800F6300408DBD1801B3EB4FC1160803304F0069B9008D004F0DD9
:200F08000040000D00B1049D9D9DA929000014008D0000540700240300D003970291BD023F
:200F8800016B080F089D0008080091189F02CD0101D001D2D351630F0F00080C000001009B
:20100800AD01AC001406070000440013020B06F10D06F507004040F40001170380000118BB

1万 · 2020-10-14 22:46

哦，为什么要生成hex文件

2万 · 2020-10-14 22:49

TM4C129ENCPDT是32位的，所以占用0x20*4个字节

1万 · 2020-10-14 22:53

。。。。没有这个说法吧！以前用的32位都不是这样的，直接拿这个去烧写的话，在j-flash中显示，一看就不对！！

1万 · 2020-10-14 22:58

烧录啊，不过我没注意还生成了bin格式的文件，所以就不用这个了！！

1万 · 2020-10-14 23:01

用bin正常吗？

1万 · 2020-10-14 23:06

bin看起来正常，下载进去也正常

2万 · 2020-10-14 23:10

这是16位操作的啦。

2万 · 2020-10-14 23:15

烧录方式应该很多吧

1万 · 2020-10-14 23:20

16进制操作，0x20,为32（十进制）个字节数据，那就32*8位。ox0000到0x0080,为8个地址，每个地址32位。8*32.

1万 · 2020-10-14 23:24

唉，还是没有什么结果，算了，多谢大家啦

1万 · 2020-11-11 11:45

这个问题好深奥啊

1万 · 2020-11-11 11:46

那是0x20个4 所以就是0x80

1万 · 2020-11-11 11:48

哦哦原来是这样的啊

只看该作者 · 2020-11-11 11:51

32bit是4byte 不仅仅是0x20的问题

只看该作者 · 2020-11-11 11:51

我也明白了感谢各位的解答啊

为什么加的是0x80

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