c语言问题，指针乱飞的那种
#include <stdio.h>
#include <stdlib.h>
int prologue [] = {
0x5920453A, 0x54756F0A, 0x6F6F470A, 0x21643A6F,
0x6E617920, 0x680A6474, 0x6F697661, 0x20646E69,
0x63636363, 0x63636363, 0x72464663, 0x6F6D6F72,
0x63636363, 0x63636363, 0x72464663, 0x6F6D6F72,
0x2C336573, 0x7420346E, 0x20216F74, 0x726F5966,
0x7565636F, 0x20206120, 0x6C616763, 0x74206C6F,
0x20206F74, 0x74786565, 0x65617276, 0x32727463,
0x594E2020, 0x206F776F, 0x79727574, 0x4563200A
};
int data [] = {
0x63636363, 0x63636363, 0x72464663, 0x6F6D6F72,
0x466D203A, 0x65693A72, 0x43646E20, 0x6F54540A,
0x5920453A, 0x54756F0A, 0x6F6F470A, 0x21643A6F,
0x594E2020, 0x206F776F, 0x79727574, 0x4563200A,
0x6F786F68, 0x6E696373, 0x6C206765, 0x796C656B,
0x2C336573, 0x7420346E, 0x20216F74, 0x726F5966,
0x7565636F, 0x20206120, 0x6C616763, 0x74206C6F,
0x20206F74, 0x74786565, 0x65617276, 0x32727463,
0x6E617920, 0x680A6474, 0x6F697661, 0x20646E69,
0x21687467, 0x63002065, 0x6C6C7861, 0x78742078,
0x6578206F, 0x72747878, 0x78636178, 0x00783174
};
int epilogue [] = {
0x594E2020, 0x206F776F, 0x79727574, 0x4563200A,
0x6E617920, 0x680A6474, 0x6F697661, 0x20646E69,
0x7565636F, 0x20206120, 0x6C616763, 0x74206C6F,
0x2C336573, 0x7420346E, 0x20216F74, 0x726F5966,
0x20206F74, 0x74786565, 0x65617276, 0x32727463
};
char message[100];
void usage_and_exit(char * program_name) {
fprintf(stderr, "USAGE: %s key1 key2 key3 key4\n", program_name);
exit(1);
}
void process_keys12 (int * key1, int * key2) {
*((int *) (key1 + *key1)) = *key2;
}
void process_keys34 (int * key3, int * key4) {
*(((int *)&key3) + *key3) += *key4;
}
char * extract_message1(int start, int stride) {
int i, j, k;
int done = 0;
for (i = 0, j = start + 1; ! done; j++) {
for (k = 1; k < stride; k++, j++, i++) {
if (*(((char *) data) + j) == '\0') {
done = 1;
break;
}
message[i] = *(((char *) data) + j);
}
}
message[i] = '\0';
return message;
}
char * extract_message2(int start, int stride) {
int i, j;
for (i = 0, j = start;
*(((char *) data) + j) != '\0';
i++, j += stride)
{
message[i] = *(((char *) data) + j);
}
message[i] = '\0';
return message;
}
int main (int argc, char *argv[])
{
int dummy = 1;
int start, stride;
int key1, key2, key3, key4;
char * msg1, * msg2;
key3 = key4 = 0;
if (argc < 3) {
usage_and_exit(argv[0]);
}
key1 = strtol(argv[1], NULL, 0);
key2 = strtol(argv[2], NULL, 0);
if (argc > 3) key3 = strtol(argv[3], NULL, 0);
if (argc > 4) key4 = strtol(argv[4], NULL, 0);
process_keys12(&key1, &key2);
start = (int)(*(((char *) &dummy)));
stride = (int)(*(((char *) &dummy) + 1));
if (key3 != 0 && key4 != 0) {
process_keys34(&key3, &key4);
}
msg1 = extract_message1(start, stride);
if (*msg1 == '\0') {
process_keys34(&key3, &key4);
msg2 = extract_message2(start, stride);
printf("%s\n", msg2);
}
else {
printf("%s\n", msg1);
}
return 0;
}
///////////////////////////////////////////////////////////
//题目要求输入4个参数使程序输出：
From: CTE
To: You
Excellent!You got everything!
第一个第二个参数我已经求出为：3，777，第三个可能是28。第四个不知道，
高手帮帮忙，
答案可以发到我邮箱里yangningemail@gmail.com
------解决方案--------------------
你那几个数组是什么东东？
[解决办法]
int dummy = 1; //这个其实不重要 看名字就知道
int start, stride;
int key1, key2, key3, key4;
//栈上的地址,一次dummy,start stride,key1,key2,key3,key4 后面一个比前面小4
char * msg1, * msg2;
key3 = key4 = 0;
//取传入的参数
if (argc < 3) {
usage_and_exit(argv[0]);
}
key1 = strtol(argv[1], NULL, 0);
key2 = strtol(argv[2], NULL, 0);
if (argc > 3) key3 = strtol(argv[3], NULL, 0);
if (argc > 4) key4 = strtol(argv[4], NULL, 0);
//这里通过地址运算修改了dummy的值
process_keys12(&key1, &key2);
//如果按传入3，777来解释
//取出低位0x03
start = (int)(*(((char *) &dummy)));
//取出高位0x09
stride = (int)(*(((char *) &dummy) + 1));
后面不知道怎么分析了好复杂
[解决办法]
没有分析出来，把那些数据转换成字符，贴出来，方便大家讨论
:E Y.ouT.Gooo:d! //prologue本行开始
yantd.havioind
cccccccccFFrromo
cccccccccFFrromo
se3,n4 tto! fYor
oceu a cgalol t
to eextvraectr2
NYowo tury. cE
cccccccccFFrromo //data本行开始
: mFr:ie ndC.TTo
:E Y.ouT.Gooo:d!
NYowo tury. cE
hoxoscineg lkely
se3,n4 tto! fYor
oceu a cgalol t
to eextvraectr2
yantd.havioind
gth!e .caxllx tx
o xexxtrxacxt1x.
NYowo tury. cE //epilogue本行开始
yantd.havioind
oceu a cgalol t
se3,n4 tto! fYor
to eextvraectr2
