Clicker Questions - 11-Pointers
Here is the state of addresses 0x7fa586401790 through 0x7fa58640182f. Our machine is little endian with 8-byte pointers.
Here is a procedure, pm():
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void pm(unsigned long *p) {
unsigned long **q;
unsigned long ***r;
unsigned long ****s;
q = (unsigned long **) p;
r = (unsigned long ***) p;
s = (unsigned long ****) p;
printf("0x%021x\n", (*p) & 0xff); /* Line 1 of output. */
printf("0x%021x\n", (**q) & 0xff); /* Line 2 of output. */
printf("0x%021x\n", (***r) & 0xff); /* Line 3 of output. */
printf("0x%021x\n", (****s) & 0xff); /* Line 4 of output. */
printf("0x%021x\n", (p[1]) & 0xff); /* Line 5 of output. */
printf("0x%021x\n", (q[1][1]) & 0xff); /* Line 6 of output. */
}
Suppose pm() is called with p equal to 0x7fa586401790:
Question 1: What is the first line of ouput?
Answer
p is 0x7fa586401790, so *p is the value at that address: 0x7fa5864017c8. The answer is 0xc8.Question 2: What is the second line of ouput?
Answer
*q is 0x7fa5864017c8, so **q is the value at that address: 0x7fa5864017c0. The answer is 0xc0.Question 3: What is the third line of ouput?
Answer
**r is 0x7fa5864017c0, so ***r is the value at that address: 0x7fa586401828. The answer is 0x28.Question 4: What is the fourth line of ouput?
Answer
***s is 0x7fa586401828, so ****s is the value at that address: 0x7fa5864017f8. The answer is 0xf8.Question 5: What is the fifth line of ouput?
Answer
p+1 is 0x7fa586401798, so p[1] is the value at that: 0x7fa5864017d8. The answer is 0xd8.Question 6: What is the sixth line of ouput?
Answer
q[1] is 0x7fa5864017d8, so q[1][1] is the value at that: 0x7fa5864017e0: 0x7fa58664017a8 The answer is 0xa8.PDF Download, 2022 Class Stats, 2023 Class Stats, 2024 Class Stats