101) 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.
102) 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).
103) 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.
104) 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.
104) 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.
105) #include<conio.h>
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.
106) 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 !
107) #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
108) 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.
109) main()
{
int i=0;
while(+(+i--)!=0)
i-=i++;
printf("%d",i);
}
Answer:
-1
Explanation:
Unary
+ is the only dummy operator in C. So it has no effect on the expression and
now the while loop is, while(i--!=0)
which is false and so breaks out of while loop. The value -1 is printed due to
the post-decrement operator.
113) main()
{
float f=5,g=10;
enum{i=10,j=20,k=50};
printf("%d\n",++k);
printf("%f\n",f<<2);
printf("%lf\n",f%g);
printf("%lf\n",fmod(f,g));
}
Answer:
Line
no 5: Error: Lvalue required
Line
no 6: Cannot apply leftshift to float
Line
no 7: Cannot apply mod to float
Explanation:
Enumeration
constants cannot be modified, so you cannot apply ++.
Bit-wise
operators and % operators cannot be applied on float values.
fmod()
is to find the modulus values for floats as % operator is for ints.
110) main()
{
int i=10;
void pascal f(int,int,int);
f(i++,i++,i++);
printf(" %d",i);
}
void
pascal f(integer :i,integer:j,integer :k)
{
write(i,j,k);
}
Answer:
Compiler
error: unknown type integer
Compiler
error: undeclared function write
Explanation:
Pascal
keyword doesn’t mean that pascal code can be used. It means that the function
follows Pascal argument passing mechanism in calling the functions.
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