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

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 main()
{
	char  *q;
	int  j;
	for (j=0; j<3; j++) scanf(“%s” ,(q+j));
	for (j=0; j<3; j++) printf(“%c” ,*(q+j));
	for (j=0; j<3; j++) printf(“%s” ,(q+j));
}

Explanation:
Here we have only one pointer to type char and since we take input in the same pointer thus we keep writing over in the same location, each time shifting the pointer value by 1. Suppose the inputs are MOUSE, TRACK and VIRTUAL. Then for the first input suppose the pointer starts at location 100 then the input one is stored as
M O U S E \0
When the second input is given the pointer is incremented as j value becomes 1, so the input is filled in memory starting from 101.
M T R A C K \0
The third input starts filling from the location 102
M T V I R T U A L \0
This is the final value stored .
The first printf prints the values at the position q, q+1 and q+2 = M T V
The second printf prints three strings starting from locations q, q+1, q+2
i.e MTVIRTUAL, TVIRTUAL and VIRTUAL.