Predict the output or
error(s) for the following:
24. void main()
Compiler Error. We cannot apply indirection on type void*.
Void pointer is a generic pointer type. No pointer arithmetic can be done on it.
Void pointers are normally used for,
1. Passing generic pointers to functions and
returning such pointers.
2. As a intermediate pointer type.
3. Used when the exact pointer type will be known
at a later point of time.
25. void main()
An identifier is available to use in program code from the point of its
So expressions such as i = i++ are valid statements. The i, j and k are
automatic variables and so they contain some garbage value. Garbage in is
garbage out (GIGO).
26. void main()
i=i++, j=j++, k=k++;
printf(“i = %d j = %d k = %d”, i, j, k);
i = 1 j = 1 k = 1
Since static variables are initialized to zero by default.
27. void main()
The inner printf executes first to print some garbage value. The printf returns
no of characters printed and this value also cannot be predicted. Still the
outer printf prints something and so returns a non-zero value. So it
encounters the break statement and comes out of the while statement.
10 9 8 7 6 5 4 3 2 1 0 65535 65534…..
Since i is an unsigned integer it can never become negative. So the expression
i-- >=0 will always be true, leading to an infinite loop.
The value of y%2 is 0. This value is assigned to x. The condition reduces to if
(x) or in other words if(0) and so z goes uninitialized.
Thumb Rule: Check all control paths to write bug free code.
*a and -*a
cancels out. The result is as simple as 1 + 3 = 4 !
expands and evaluates to as:
x+(2*y)-1 => 10
Note the semicolon after the while statement. When the value of i becomes 0 it
comes out of while loop. Due to post-increment on i the value of i while
printing is 1.
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