Exercise P: Pointers

On each line, write the type and the numeric value of the expression on the left. For pointer types, write the numeric address (what you would get from printf("%p")). Assume an x86-like machine (with 32-bit addresses and integers and little endian storage). We did line 0 for you.

		Type	Numeric value
	char* p = (char) 0x1100; char q = (char*) 0x1110;
0.	p	`char *`	`0x1100`
1.	&p[1]	_________________	_________________
2.	&p[-1]	_________________	_________________
3.	&p[0]	_________________	_________________
4.	&p[1] - &p[0]	_________________	_________________
5.	&p[8]	_________________	_________________
6.	(p + 1) - p	_________________	_________________
7.	&p[16] - p	_________________	_________________
8.	q - p	_________________	_________________
9.	sizeof(p)	_________________	_________________
10.	sizeof(*p)	_________________	_________________
	int* ip = (int*) p;
11.	&ip[0]	_________________	_________________
12.	&ip[1]	_________________	_________________
13.	&ip[1] - &ip[0]	_________________	_________________
14.	(char*) &ip[1] - p	_________________	_________________
15.	sizeof(ip)	_________________	_________________
16.	sizeof(*ip)	_________________	_________________
17.	&ip[sizeof(int)]	_________________	_________________
18.	ip + sizeof(int)	_________________	_________________
19.	ip + 1	_________________	_________________
20.	p + sizeof(int)	_________________	_________________
	int* iq = (int*) q;
21.	iq - ip	_________________	_________________
22.	&iq[-1] - ip	_________________	_________________
	p[0] = p[1] = p[2] = p[3] = 0;
23.	*ip	_________________	_________________
	(char) ip = 1;
24.	*ip	_________________	_________________
	((char) ip + 1) = 1;
25.	p[1]	_________________	_________________
26.	*ip	_________________	_________________
	((char) ip) = 2;
27.	((char) ip)	_________________	_________________
28.	*ip	_________________	_________________

Solutions

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