一个比较奇怪的问题 memcpy拷贝

senic

使用memcpy前后 目标内容不一致.......  很奇怪的问题   大家有什么解决的思路吗？sBuff  >>> sendbuff ,bufflen 是合法


        void* p = memcpy(sBuff, git->gi[iPos].sendbuff, git->gi[iPos].bufflen);
        printf("%d --- %d \n", p, sBuff);  //用来检测前后指针是否发生变化
        //bcopy(git->gi[iPos].sendbuff, sBuff, git->gi[iPos].bufflen);
        for (i = 0; i < git->gi[iPos].bufflen; i++)
                printf("%02x ", sBuff);  // memcpy dest 打印
        printf("\n-----------------------------\n");
        for (i = 0; i < git->gi[iPos].bufflen; i++)
                printf("%02x ", git->gi[iPos].sendbuff); // memcpy src 打印       
        iRet = git->gi[iPos].bufflen;  //返回上层函数缓冲长度
       


打印的结果如下
1096524316 --- 1096524316
00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 36 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 01 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 02 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 03 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 04 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 05 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 06 00 00 00
-----------------------------
00 00 00 00 36 00 00 00 00 00 00 00 00 00 00 00 00 00 01 00 00 00 00 00 00 00 00 00 02 00 00 00 00 00 00 00 00 00 03 00 00 00 00 00 00 00 00 00 04 00 00 00 00 00 00 00 00 00 05 00 00 00 00 00 00 00 00 00 06 00 00 00 00 00 00 00 00 00 07 00 00 00 00 00 00 00 00 00 08 00 00 00 00 00 00 00 00 00 09 00 00 00 00 00 00 00 00 00 0a 00 00 00 00 00 00 00 00 00 0b 00 00 00 00 00 00 00 00 00 0c 00 00 00 00 00 00 00 00 00 0d 00 00 00 00 00 00 00 00 00 0e 00 00 00 00 00 00 00 00 00 0f 00 00 00 00 00 00 00 00 00 10 00 00 00 00 00 00 00 00 00 11 00 00 00 00 00 00 00 00 00 12 00 00 00 00 00 00 00 00 00 13 00 00 00 00 00 00 00 00 00 14 00 00 00 00 00 00 00 00 00 15 00 00 00 00 00 00 00 00 00 16 00 00 00 00 00 00 00 00 00 17 00 00 00 00 00 00 00 00 00 18 00 00 00 00 00 00 00 00 00 19 00 00 00 00 00 00 00 00 00 1a 00 00 00 00 00 00 00 00 00 1b 00 00 00 00 00 00 00 00 00 1c 00 00 00 00 00 00 00 00 00 1d 00 00 00 00 00 00 00 00 00 1e 00 00 00 00 00 00 00 00 00 1f 00 00 00 00 00 00 00 00 00 20 00 00 00 00 00 00 00 00 00 21 00 00 00 00 00 00 00 00 00 22 00 00 00 00 00 00 00 00 00 23 00 00 00 00 00 00 00 00 00 24 00 00 00 00 00 00 00 00 00 25 00 00 00 00 00 00 00 00 00 26 00 00 00 00 00 00 00 00 00 27 00 00 00 00 00 00 00 00 00 28 00 00 00 00 00 00 00 00 00 29 00 00 00 00 00 00 00 00 00 2a 00 00 00 00 00 00 00 00 00 2b 00 00 00 00 00 00 00 00 00 2c 00 00 00 00 00 00 00 00 00 2d 00 00 00 00 00 00 00 00 00 2e 00 00 00 00 00 00 00 00 00 2f 00 00 00 00 00 00 00 00 00 30 00 00 00 00 00 00 00 00 00 31 00 00 00 00 00 00 00 00 00 32 00 00 00 00 00 00 00 00 00 33 00 00 00 00 00 00 00 00 00 34 00 00 00 00 00 00 00 00 00 35 00 00 00 00 00 00 00 00 00

senic

解决了..... 很土的办法   

将memcpy  改成

        for (i = 0; i < git->gi[iPos].bufflen; i++)
                sBuff[i] = git->gi[iPos].sendbuff[i];

36 00 00 00 00 00 00 00 00 00 00 00 00 00 01 00 00 00 00 00 00 00 00 00 02 00 00 00 00 00 00 00 00 00 03 00 00 00 00 00 00 00 00 00 04 00 00 00 00 00 00 00 00 00 05 00 00 00 00 00 00 00 00 00 06 00 00 00 00 00 00 00 00 00 07 00 00 00 00 00 00 00 00 00 08 00 00 00 00 00 00 00 00 00 09 00 00 00 00 00 00 00 00 00 0a 00 00 00 00 00 00 00 00 00 0b 00 00 00 00 00 00 00 00 00 0c 00 00 00 00 00 00 00 00 00 0d 00 00 00 00 00 00 00 00 00 0e 00 00 00 00 00 00 00 00 00 0f 00 00 00 00 00 00 00 00 00 10 00 00 00 00 00 00 00 00 00 11 00 00 00 00 00 00 00 00 00 12 00 00 00 00 00 00 00 00 00 13 00 00 00 00 00 00 00 00 00 14 00 00 00 00 00 00 00 00 00 15 00 00 00 00 00 00 00 00 00 16 00 00 00 00 00 00 00 00 00 17 00 00 00 00 00 00 00 00 00 18 00 00 00 00 00 00 00 00 00 19 00 00 00 00 00 00 00 00 00 1a 00 00 00 00 00 00 00 00 00 1b 00 00 00 00 00 00 00 00 00 1c 00 00 00 00 00 00 00 00 00 1d 00 00 00 00 00 00 00 00 00 1e 00 00 00 00 00 00 00 00 00 1f 00 00 00 00 00 00 00 00 00 20 00 00 00 00 00 00 00 00 00 21 00 00 00 00 00 00 00 00 00 22 00 00 00 00 00 00 00 00 00 23 00 00 00 00 00 00 00 00 00 24 00 00 00 00 00 00 00 00 00 25 00 00 00 00 00 00 00 00 00 26 00 00 00 00 00 00 00 00 00 27 00 00 00 00 00 00 00 00 00 28 00 00 00 00 00 00 00 00 00 29 00 00 00 00 00 00 00 00 00 2a 00 00 00 00 00 00 00 00 00 2b 00 00 00 00 00 00 00 00 00 2c 00 00 00 00 00 00 00 00 00 2d 00 00 00 00 00 00 00 00 00 2e 00 00 00 00 00 00 00 00 00 2f 00 00 00 00 00 00 00 00 00 30 00 00 00 00 00 00 00 00 00 31 00 00 00 00 00 00 00 00 00 32 00 00 00 00 00 00 00 00 00 33 00 00 00 00 00 00 00 00 00 34 00 00 00 00 00 00 00 00 00 35 00 00 00 00 00 00 00 00 00
-----------------------------
36 00 00 00 00 00 00 00 00 00 00 00 00 00 01 00 00 00 00 00 00 00 00 00 02 00 00 00 00 00 00 00 00 00 03 00 00 00 00 00 00 00 00 00 04 00 00 00 00 00 00 00 00 00 05 00 00 00 00 00 00 00 00 00 06 00 00 00 00 00 00 00 00 00 07 00 00 00 00 00 00 00 00 00 08 00 00 00 00 00 00 00 00 00 09 00 00 00 00 00 00 00 00 00 0a 00 00 00 00 00 00 00 00 00 0b 00 00 00 00 00 00 00 00 00 0c 00 00 00 00 00 00 00 00 00 0d 00 00 00 00 00 00 00 00 00 0e 00 00 00 00 00 00 00 00 00 0f 00 00 00 00 00 00 00 00 00 10 00 00 00 00 00 00 00 00 00 11 00 00 00 00 00 00 00 00 00 12 00 00 00 00 00 00 00 00 00 13 00 00 00 00 00 00 00 00 00 14 00 00 00 00 00 00 00 00 00 15 00 00 00 00 00 00 00 00 00 16 00 00 00 00 00 00 00 00 00 17 00 00 00 00 00 00 00 00 00 18 00 00 00 00 00 00 00 00 00 19 00 00 00 00 00 00 00 00 00 1a 00 00 00 00 00 00 00 00 00 1b 00 00 00 00 00 00 00 00 00 1c 00 00 00 00 00 00 00 00 00 1d 00 00 00 00 00 00 00 00 00 1e 00 00 00 00 00 00 00 00 00 1f 00 00 00 00 00 00 00 00 00 20 00 00 00 00 00 00 00 00 00 21 00 00 00 00 00 00 00 00 00 22 00 00 00 00 00 00 00 00 00 23 00 00 00 00 00 00 00 00 00 24 00 00 00 00 00 00 00 00 00 25 00 00 00 00 00 00 00 00 00 26 00 00 00 00 00 00 00 00 00 27 00 00 00 00 00 00 00 00 00 28 00 00 00 00 00 00 00 00 00 29 00 00 00 00 00 00 00 00 00 2a 00 00 00 00 00 00 00 00 00 2b 00 00 00 00 00 00 00 00 00 2c 00 00 00 00 00 00 00 00 00 2d 00 00 00 00 00 00 00 00 00 2e 00 00 00 00 00 00 00 00 00 2f 00 00 00 00 00 00 00 00 00 30 00 00 00 00 00 00 00 00 00 31 00 00 00 00 00 00 00 00 00 32 00 00 00 00 00 00 00 00 00 33 00 00 00 00 00 00 00 00 00 34 00 00 00 00 00 00 00 00 00 35 00 00 00 00 00 00 00 00 00

senic

哪位仁兄能解释下 什么情况下会出现这种情形么？   多核环境？

hellioncu

sBuff, git->gi[iPos].sendbuff两者内存有重叠？

senic

sBuff, git->gi.sendbuff两者内存有重叠？
hellioncu 发表于 2011-06-28 11:51

    应该不是这个问题   sbuff是从函数外传入的一个字符数组缓冲  每次使用前都初始化过
                      git->gi.sendbuff  是放在共享内存中的数据， 是打包好的下发数据包

KBTiller

没写函数原型(#include )函数调用有时会有一些奇怪现象
不过，楼主不至于没写吧

也可以试试memmove函数
如果memmove好用说明有重叠