SSD6中Exercise1答案解析

 

#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);
    }

    return0;
}

1.The secret message:
    From: CTE

    To: You

    Excellent! You got everything!

2.The secret keys:

    9  777  -1  45

3.The function process_keys12 change the value of dummy. The expression (int *) (key1 + *key1) point to the address of dummy. *key2 is the new value of dummy. So the value of start is the first byte of dummy, the value of stride is the second byte of dummy.

4.The value of key1 is the distance between the address of dummy and the address of key1. As &dummy is 0x0012FF60, &key1 is 0x0012FF3C. So key1 is (0x0012FF60-0x0012FF3C)/4=9. Because the message is start with From:. So message[0] should be F. And message[0] is equal to *(((char *) data) + start + 1). The characters of the first 16 are cccccccccFFrromo. And the expression for (k = 1; k < stride; k++, j++, i++), so I know the value of start is 9, and the value of stride is 3. Then, the value of key2 is 0x00000309, which in decimal is 777.

 

5.The function process_keys34 change the return address of itself.

 

6.The expression *(((int *)&key3) + *key3) += *key4; is executed. The program jump over the expressions msg1= extract_message1(start, stride); and process_keys34(&key3, &key4);. It directly execute the expression msg2 = extract_message2(start, stride);.

 

7.|&key3|                   ----------------0x0012fe28

   |return address|      ----------------0x0012fe24

   So the expression ((int *)&key3) + *key3 should be the value of return address, then the value of key3 is -1. When process_keys34 is executed, the program should jump to msg2 = extract_message2(start, stride);. The address of msg2 = extract_message2(start, stride); is 0x004117E9. So when ((int *)&key3) + *key3 plus *key4 , its value should be 0x004117E9. Then we can calculate key4 is 45.