Clicker Questions - 09-Pointers
The following is the state of memory at address 0x61626324 through 0x6162637f. In each line, I show:
- The pointer address to the first of four bytes.
- The four bytes in decimal.
- The four bytes in hexadecimal.
- The four bytes in chars.
The machine is little endian, but I’ve printed the characters in big-endian (e.g. 0, 1,2, 3). The pointers are four bytes.
When you run the following procedure, the value of p is 0x61626324.
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typedef struct ms {
char a, b, c, d;
int e;
char *f;
struct ms *g;
} Mystruct;
void pm(Mystruct *p) {
Mystruct *q;
printf("%c\n", p->b);
printf("%c%c%c%c\n", p->f[0], p->f[1], p->f[2], p->f[3]);
printf("%d\n", p[1]->e);
q = p+4;
printf("0x%x\n", (unsigned int) q->f);
q = p->g;
printf("%c\n", (unsigned int) q->a);
q = q[1].g;
printf("%c\n", (unsigned int) q->a);
}
Hint #1
Your first question should be, "How many bytes are in a Mystruct?Hint #2
Your second question should be, "If I have a pointer to Mystruct, then where do I find the various fields, a, b, c, etc.?"Hint #3
Your third question should be, "If I have a pointer p, to a Mystruct, then what is the pointer value of p+x?"Hint #4 (Not Given)
The Mystruct struct has 16 bytes:- Bytes 0 through 3: the characters a through b.
- Bytes 4 through 7: the integer e.
- Bytes 8 through 11: the (char *) f.
- Bytes 12 through 15: the (Mystruct *)g
Question 1: What is the first line of output?
Answer
Since p is 0x61626324, then p->b is the second of the four characters that start at the line. It is the character 'c'.Question 2: What is the second line of output?
Answer
Since p is 0x61626324, then p->f is the pointer stored at 0x6162632c, whose value is 0x61626336. The four characters will be the ones that start at that address, so the output is 'b', 'a', 'k', and 'c'.Question 3: What is the third line of output?
Answer
Since p is 0x61626324 and the Mystruct is 16 bytes, p+1 will be 0x61626334. Therefore, p[1].e will be the four bytes at address 0x61626338: 1633837931.Question 4: What is the fourth line of output?
Answer
Since p is 0x61626324 and the Mystruct is 16 bytes, p+4 will be 0x61626364. Therefore, q->f will be the four bytes at address 0x6162636c: 0x6162633b.Question 5: What is the fiveth line of output?
Answer
Since p is 0x61626324, we know that p->q is the pointer at address 0x61626330, which has the value 0x6162635c. So, q is 0x6162635c. So, q 0x6162635c, and q->a is the first character there: 'v'.Question 6: What is the sixth line of output?
Answer
q is 0x6162635c, so q+1 is 0x6162636c.That means q[1].g is the four bytes 0x61626378 which happen to also equal 0x61626378. When we set q to 0x61626378, q->a is the first character there: 'x'.Program Output
c bakc 1633837931 0x6162633b v x
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