Clicker Questions - 06-More-Pointers
Your machine has four byte pointers and is little endian.
memcpy(dest, src, size) copies size bytes from src to dest.
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int main() {
unsigned int k[4];
unsigned char *cp;
unsigned int *ip;
int i;
cp = (unsigned char *) k;
ip = k+2;
for (i = 0; i < 16; i++) {
cp[i] = i*16 + 15-i;
}
printf("0x%x\n", cp[3]); /* Output line 1. */
printf("0x%x\n", cp[7]); /* Output line 2. */
printf("0x%x\n", k[0]); /* Output line 3. */
printf("0x%x\n", *ip); /* Output line 4. */
memcpy(cp+2, cp+10, 4);
printf("0x%x\n", cp[3]); /* Output line 5. */
printf("0x%x\n", cp[4]); /* Output line 6. */
printf("0x%x\n", k[0]); /* Output line 7. */
printf("0x%x\n", k[1]); /* Output line 8. */
cp += 12;
ip = (unsigned int *) cp;
i = ip - k;
printf("%d\n", i); /* Output line 9. */
i = cp - (unsigned char *) k;
printf("%d\n", i); /* Output line 10. */
return 0;
}
Suppose k = 0x1000.
Help
Use this drawing:
Question 1: What is line 1 of the output?
Answer
From the first few lines, there are only 16 bytes in question -- k[0] through k[3]. Let's label them before they are set:
3 2 1 0
k[0] cp[0] through cp[3]
k[1] cp[4] through cp[7]
ip[0] k[2] cp[8] through cp[11]
ip[1] k[3] cp[12] through cp[15]
The for loop sets the bytes. Each cp[i] will be of the form 0xwy, where w is equal to i, and y is equal to 15-i. Here is what the 16 bytes look like:
3 2 1 0
---- ---- ---- ----
k[0] 0x3c 0x2d 0x1e 0x0f cp[0] through cp[3]
k[1] 0x78 0x69 0x5a 0x4b cp[4] through cp[7]
ip[0] k[2] 0xb4 0xa5 0x96 0x87 cp[8] through cp[11]
ip[1] k[3] 0xf0 0xe1 0xd2 0xc3 cp[12] through cp[15]
So, the answer is 0x3c
Question 2: What is line 2 of the output?
Answer
Here is what the 16 bytes look like:
3 2 1 0
---- ---- ---- ----
k[0] 0x3c 0x2d 0x1e 0x0f cp[0] through cp[3]
k[1] 0x78 0x69 0x5a 0x4b cp[4] through cp[7]
ip[0] k[2] 0xb4 0xa5 0x96 0x87 cp[8] through cp[11]
ip[1] k[3] 0xf0 0xe1 0xd2 0xc3 cp[12] through cp[15]
So, the answer is 0x78
Question 3: What is line 3 of the output?
Answer
Here is what the 16 bytes look like:
3 2 1 0
---- ---- ---- ----
k[0] 0x3c 0x2d 0x1e 0x0f cp[0] through cp[3]
k[1] 0x78 0x69 0x5a 0x4b cp[4] through cp[7]
ip[0] k[2] 0xb4 0xa5 0x96 0x87 cp[8] through cp[11]
ip[1] k[3] 0xf0 0xe1 0xd2 0xc3 cp[12] through cp[15]
So, the answer is 0x3c2d1e0f
Question 4: What is line 4 of the output?
Answer
Here is what the 16 bytes look like:
3 2 1 0
---- ---- ---- ----
k[0] 0x3c 0x2d 0x1e 0x0f cp[0] through cp[3]
k[1] 0x78 0x69 0x5a 0x4b cp[4] through cp[7]
ip[0] k[2] 0xb4 0xa5 0x96 0x87 cp[8] through cp[11]
ip[1] k[3] 0xf0 0xe1 0xd2 0xc3 cp[12] through cp[15]
So, the answer is 0xb4a5987
Question 5: What is line 5 of the output?
Answer
Thememcpy() statement will move the four bytes starting at index cp+10 to the four bytes starting at index cp+2. When it's done, here are the 16 bytes of memory:
3 2 1 0
---- ---- ---- ----
k[0] 0xb4 0xa5 0x1e 0x0f cp[0] through cp[3]
k[1] 0x78 0x69 0xd2 0xc3 cp[4] through cp[7]
ip[0] k[2] 0xb4 0xa5 0x96 0x87 cp[8] through cp[11]
ip[1] k[3] 0xf0 0xe1 0xd2 0xc3 cp[12] through cp[15]
So, the answer is 0xb4
Question 6: What is line 6 of the output?
Answer
Here is what the 16 bytes look like:
3 2 1 0
---- ---- ---- ----
k[0] 0xb4 0xa5 0x1e 0x0f cp[0] through cp[3]
k[1] 0x78 0x69 0xd2 0xc3 cp[4] through cp[7]
ip[0] k[2] 0xb4 0xa5 0x96 0x87 cp[8] through cp[11]
ip[1] k[3] 0xf0 0xe1 0xd2 0xc3 cp[12] through cp[15]
So, the answer is 0xc3
Question 7: What is line 7 of the output?
Answer
Here is what the 16 bytes look like:
3 2 1 0
---- ---- ---- ----
k[0] 0xb4 0xa5 0x1e 0x0f cp[0] through cp[3]
k[1] 0x78 0x69 0xd2 0xc3 cp[4] through cp[7]
ip[0] k[2] 0xb4 0xa5 0x96 0x87 cp[8] through cp[11]
ip[1] k[3] 0xf0 0xe1 0xd2 0xc3 cp[12] through cp[15]
So, the answer is 0xb4a51e0f
Question 8: What is line 8 of the output?
Answer
Here is what the 16 bytes look like:
3 2 1 0
---- ---- ---- ----
k[0] 0xb4 0xa5 0x1e 0x0f cp[0] through cp[3]
k[1] 0x78 0x69 0xd2 0xc3 cp[4] through cp[7]
ip[0] k[2] 0xb4 0xa5 0x96 0x87 cp[8] through cp[11]
ip[1] k[3] 0xf0 0xe1 0xd2 0xc3 cp[12] through cp[15]
So, the answer is 0x7869d2c3
Question 9: What is line 9 of the output?
Answer
cp and ip are now set to be (k+3). So the answer to question 9 is 3.
Question 10: What is line 10 of the output?
Answer
When we do the pointer arithmetic by bytes rather than ints, the difference betweencp and k is 12, so the answer to question to is 12.
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