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UTK Notes


Clicker Questions - 08-Pointers

In these questions, your machine is little endian. Here are the prototypes of C procedures:



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char *stpcpy(char *dest, const char *src);
char *strcat(char *dest, const char *src);
char *memcpy(void *dest, const void *src, int bytes);
size_t strlen(const char *s);

Question 1: If you are lucky enough to have a seqmentation violation on this code, which line will cause the segmentation violation?



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/* Line  1 */  int main()
/* Line  2 */  {
/* Line  3 */      char s[100];
/* Line  4 */      int *vec;
/* Line  5 */      void *v[100];
/* Line  6 */      void *x[100];
/* Line  7 */      int i;
/* Line  8 */      
/* Line  9 */      strcpy(s, "");
/* Line 10 */      for (i = 0; i < 50; i++) {
/* Line 11 */          strcat(s, "j");
/* Line 12 */          vec[i] = i;
/* Line 13 */          v[i] = (void *) i;
/* Line 14 */          x[i] = (void *) &i;
/* Line 15 */      }
/* Line 16 */      return 0;
/* Line 17 */  }
Answer Line 12 -- vec is an uninitialized variable. This is one of the reasons I urge you to set your pointers to NULL at the beginning of your procedures. That will help you find bugs like this, because you're guaranteed to generate a seg fault.

Question 2: What is the output of this program?



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int main() {
    char b[30];
    int i;

    strcpy(b, "ABCDEFGHIJKLMNOPQRSTUVWXYZ");
    i = ('C' << 16) | ('D' << 8) | 'E';
    memcpy(b+2, &i, 4);
    printf("%d\n", (int) strlen(b));
    return 0;
}
Answer Since your machine is little endian, the four bytes of i are going to be 'E', 'D', 'C', and 0, in that order. It's a three-character string, "EDC". That means that after the memcpy(), b becomes "ABEDC*GHIJKLMNOPQRSTUVWXYZ", where * is the null character. The answer is 5.

Question 3: What is the 1st line of output of the following program?
Question 4: What is the 2nd line of output of the following program?



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int main() {
    unsigned char p[8];
    unsigned int *x;

    x = (unsigned int *) p;
    p[0] = (0x7 << 4) | 0xc;
    p[1] = (0xa << 4) | 0x9;
    p[2] = (0x3 << 4) | 0x2;
    p[3] = (0xf << 4) | 0xe;
    x[1] = 0x15768;

    printf("0x%x\n", x[0]);
    printf("%x-%x-%x-%x\n", p[4], p[5], p[6], p[7]);
}
Answer Let's suppose that the value of p is 0x12000. Here's memory: 
Address     +3     +2     +1     +0
-------- | ---- | ---- | ---- | ---- |
0x1200   | 0xfe | 0x32 | 0xa9 | 0x7c |
0x1204   | 0x00 | 0x01 | 0x57 | 0x68 |
-------- | ---- | ---- | ---- | ---- |
The answer is 0xfe32a97c and 68-57-1-0

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