Question

我对C语言并不熟悉并在Linux中进行编译，但我有一些问题需要帮助。

我有这行代码使用已安装的命令及其参数来加入域。（见图）。运行gcc join.c进行编译后，它创建了一个文件a.out。到目前为止还不错，但当我运行vim a.out来查看该文件的内容时，我看到mypassword可以通过简单的文本编辑器轻松查看。（见第二张图片）

在编译我的C代码时，有什么办法可以避免这种情况吗？

#include <stdio.h>
#include <unistd.h>

int main ()
{
        printf("Running 'net join' with the following parameters: \n");
        char *domain="mydomain";
        char *user="domainjoinuser";
        char *pass="mypassword";
        char *vastool="/opt/quest/bin/vastool";
        char *ou="OU=test,DC=mtdomain,DC=local";
        char unjoin[512];

        sprintf(unjoin,"/opt/quest/in/vastool -u %s -w '%s' unjoin -f",user,pass);

        printf("Domain: %s\n",domain);
        printf("User: %s\n",user);
        printf("-----------------\n");

        printf("Unjoin.............\n");
        system(unjoin);

        printf("Join................\n");
        execl("/opt/quest/bin/vastool", "vastool", "-u", user, "-w", pass, "join", "-c", "ou", "-f", domain, (char*)0);

}

source code viewed in vim

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00000400  00 00 48 89 c7 b8 00 00  00 00 e8 71 02 00 00 48  |..H........q...H|
00000410  8b 45 f0 48 89 c6 48 8d  05 8c 13 00 00 48 89 c7  |.E.H..H......H..|
00000420  b8 00 00 00 00 e8 56 02  00 00 48 8d 05 82 13 00  |......V...H.....|
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00000530  4c 29 e5 41 89 ff 49 89  f6 48 c1 fd 03 49 89 d5  |L).A..I..H...I..|
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000006d0  00 00 00 00 00 00 00 00  00 00 00 00 00 00 00 00  |................|
000006e0  00 00 00 00 52 75 6e 6e  69 6e 67 20 27 6e 65 74  |....Running 'net|
000006f0  20 6a 6f 69 6e 27 20 77  69 74 68 20 74 68 65 20  | join' with the |
00000700  66 6f 6c 6c 6f 77 69 6e  67 20 70 61 72 61 6d 65  |following parame|
00000710  74 65 72 73 3a 20 0a 00  6d 79 64 6f 6d 61 69 6e  |ters: ..mydomain|
00000720  00 64 6f 6d 61 69 6e 6a  6f 69 6e 75 73 65 72 00  |.domainjoinuser.|
00000730  6d 79 70 61 73 73 77 6f  72 64 00 2f 6f 70 74 2f  |mypassword./opt/|
00000740  71 75 65 73 74 2f 62 69  6e 2f 76 61 73 74 6f 6f  |quest/bin/vastoo|
00000750  6c 00 4f 55 3d 74 65 73  74 2c 44 43 3d 6d 74 64  |l.OU=test,DC=mtd|
00000760  6f 6d 61 69 6e 2c 44 43  3d 6c 6f 63 61 6c 00 2f  |omain,DC=local./|
00000770  6f 70 74 2f 71 75 65 73  74 2f 69 6e 2f 76 61 73  |opt/quest/in/vas|
00000780  74 6f 6f 6c 20 2d 75 20  25 73 20 2d 77 20 27 25  |tool -u %s -w '%|
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000007a0  61 69 6e 3a 20 25 73 0a  00 55 73 65 72 3a 20 25  |ain: %s..User: %|
000007b0  73 0a 00 2d 2d 2d 2d 2d  2d 2d 2d 2d 2d 2d 2d 2d  |s..-------------|
000007c0  2d 2d 2d 2d 0a 00 55 6e  6a 6f 69 6e 2e 2e 2e 2e  |----..Unjoin....|
000007d0  2e 2e 2e 2e 2e 2e 2e 2e  2e 0a 00 4a 6f 69 6e 2e  |...........Join.|
000007e0  2e 2e 2e 2e 2e 2e 2e 2e  2e 2e 2e 2e 2e 2e 2e 0a  |................|
000007f0  00 2f 6f 70 74 2f 71 75  65 73 74 2f 62 69 6e 2f  |./opt/quest/bin/|
00000800  76 61 73 74 6f 6f 6c 00  76 61 73 74 6f 6f 6c 00  |vastool.vastool.|
00000810  2d 75 00 2d 77 00 6a 6f  69 6e 00 2d 63 00 6f 75  |-u.-w.join.-c.ou|
00000820  00 2d 66 00 01 00 00 00  00 00 00 00 44 00 00 00  |.-f.........D...|
00000830  00 00 00 00 04 00 00 00  00 00 00 00 4c 82 04 08  |............L...|
...

binary file viewed in vim

Answer 1

编译我的C代码时，有什么办法可以避免这种情况吗？

您唯一能做的就是不将密码或任何敏感信息硬编码到C程序中：即使您以某种方式加密此敏感信息，也需要在以下方式提供解密密钥：运行程序的用户运行时，或者信息可以由充满动力的人解密。您也可以提示用户输入密码。

请注意，任何其他形式的隐藏都只是一种混淆 - 这是试图访问您的秘密信息的用户的一个小障碍。它可能会阻止一些“脚本小子”，但它将属于第一个知识渊博的用户。

Answer 2

作为dasblinkenlight pointed out，无论你做什么都不会阻止某人获取你的密码。例如，他可以在你的程序上运行strace来确定哪些参数被传递给vastool，或者他可以使用调试器。如果您为了安全而编写程序，请始终牢记Kerkhoff's principle：

即使系统的所有内容都是公共知识，（加密）系统也应该是安全的。

但是，如果您喜欢一些额外的~~烟幕安全性~~模糊处理，您可能需要查看memfrob(3)函数：

void *memfrob(void *s, size_t n);
memfrob （）函数   通过异或来加密存储区 s 的第一个 n 字节   每个字符的编号为42.可以通过使用来反转该效果   加密内存区域上的 memfrob （）。

请注意，此功能不是XOR的正确加密例程   常量是固定的，仅适用于隐藏字符串。

Answer 3

一种方法是使用函数（rot13（），reverse（），chartobinary（）f.e.）并使用编码的字符串调用函数。

如何在编译的C代码中隐藏文本？

3 个答案: