Predict the output or
error(s) for the following:
24. void main()
{
void *v;
int
integer=2;
int
*i=&integer;
v=i;
printf("%d",(int*)*v);
}
Answer:
Compiler Error. We cannot apply indirection on type void*.
Explanation:
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()
{
int
i=i++,j=j++,k=k++;
printf(“%d%d%d”,i,j,k);
}
Answer:
Garbage values.
Explanation:
An identifier is available to use in program code from the point of its
declaration.
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()
{
static int
i=i++, j=j++, k=k++;
printf(“i = %d j = %d k = %d”, i, j, k);
}
Answer:
i = 1 j = 1 k = 1
Explanation:
Since static variables are initialized to zero by default.
27. void main()
{
while(1){
if(printf("%d",printf("%d")))
break;
else
continue;
}
}
Answer:
Garbage values
Explanation:
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.
28. main()
{
unsigned int
i=10;
while(i-->=0)
printf("%u ",i);
}
Answer:
10 9 8 7 6 5 4 3 2 1 0 65535 65534…..
Explanation:
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.
29. #include
main()
{
int
x,y=2,z,a;
if(x=y%2)
z=2;
a=2;
printf("%d %d
",z,x);
}
Answer:
Garbage-value 0
Explanation:
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.
30. main()
{
int a[10];
printf("%d",*a+1-*a+3);
}
Answer:
4
Explanation:
*a and -*a
cancels out. The result is as simple as 1 + 3 = 4 !
31. #define
prod(a,b) a*b
main()
{
int x=3,y=4;
printf("%d",prod(x+2,y-1));
}
Answer:
10
Explanation:
The macro
expands and evaluates to as:
x+2*y-1 =>
x+(2*y)-1 => 10
32. main()
{
unsigned int
i=65000;
while(i++!=0);
printf("%d",i);
}
Answer:
1
Explanation:
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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