49. Predict the output or error(s) of the following c code

1
2
3
4
5
6
7
8
9
10
11
12
13
14

 main()
{
	static int  a[ ]   = {0,1,2,3,4};
	int  *p[ ] = {a,a+1,a+2,a+3,a+4};
	int  **ptr =  p;
	ptr++;
	printf(“\n %d  %d  %d”, ptr-p, *ptr-a, **ptr); 
	*ptr++;
	printf(“\n %d  %d  %d”, ptr-p, *ptr-a, **ptr); 
	*++ptr;
	printf(“\n %d  %d  %d”, ptr-p, *ptr-a, **ptr); 
	++*ptr;
	printf(“\n %d  %d  %d”, ptr-p, *ptr-a, **ptr); 
}

Answer:
111
222
333
344

Explanation:
Let us consider the array and the two pointers with some address
a
0 1 2 3 4
100 102 104 106 108
p
100 102 104 106 108
1000 1002 1004 1006 1008
ptr
1000
2000
After execution of the instruction ptr++ value in ptr becomes 1002, if scaling factor for integer is 2 bytes. Now ptr – p is value in ptr – starting location of array p, (1002 – 1000) / (scaling factor) = 1, *ptr – a = value at address pointed by ptr – starting value of array a, 1002 has a value 102 so the value is (102 – 100)/(scaling factor) = 1, **ptr is the value stored in the location pointed by the pointer of ptr = value pointed by value pointed by 1002 = value pointed by 102 = 1. Hence the output of the firs printf is 1, 1, 1.
After execution of *ptr++ increments value of the value in ptr by scaling factor, so it becomes1004. Hence, the outputs for the second printf are ptr – p = 2, *ptr – a = 2, **ptr = 2.
After execution of *++ptr increments value of the value in ptr by scaling factor, so it becomes1004. Hence, the outputs for the third printf are ptr – p = 3, *ptr – a = 3, **ptr = 3.
After execution of ++*ptr value in ptr remains the same, the value pointed by the value is incremented by the scaling factor. So the value in array p at location 1006 changes from 106 10 108,. Hence, the outputs for the fourth printf are ptr – p = 1006 – 1000 = 3, *ptr – a = 108 – 100 = 4, **ptr = 4.