经典C程序求解

[eileilei]

Problem 1
Read Ken Thompson's article ...

char s[] = {
'\t',
'0',
'\n',
'}',
';",
'\n',
'\n',
'/',
'*',
'\n',
(213 lines deleted)
0
};

/* The string s is a
* representation of the body
* of this program from '0'
* to the end.
*/

main(){
int i;
printf("char\ts{ } = {\n");
for(i=0; s[i]; i++)
printf("\t%d,\n",s[i]);
printf("%s",s);
}
Compile and run this program. You will have to figure out what goes in the 213 deleted lines. If you want, you can leave the comment out of the string s and out of the program. Explain the output. By itself, this program does not pose any sort of security threat. Explain in your own words what this program has to do with Thompson's Trojan Horse attack.

[eileilei]

printf("%s",s);什么意思？？？？？？？？

[brookmill]

找到原文了，这是Ken Thompson获得图灵奖的演讲的一部分。
这段代码是一个例子，它可以产生具有自我复制能力的程序。
不过……还是没看懂

Figure 1 shows a self-reproducing program in the C
programming language. (The purist will note that the
program is not precisely a self-reproducing program,
but will produce a self-reproducing program.) This entry
is much too large to win a prize, but it demonstrates
the technique and has two important properties that I
need to complete my story: 1) This program can be
easily written by another program. 2) This program can
contain an arbitrary amount of excess baggage that will
be reproduced along with the main algorithm. In the
example, even the comment is reproduced.

[brookmill]

省略的那213行，大概就是把后面那段注释和main函数的每个字符依次写下来，每个字符一行

[brookmill]

看看这个：Ken Thompson的经典c程序
http://topic.csdn.net/t/20020116/19/482123.html

[brookmill]

没看懂。
printf("%s",s); 执行这句的时候，s[1]的'0'为什么没打出来呢？

[tianya0609]

up

[brookmill]

找到那个'0'了，刚才看错行了。
不过还是没看明白木马是怎么回事

[hoomey]

mark