Predict the output or error(s) for the
following:
1.
void main()
{
int const * p=5;
printf("%d",++(*p));
}
Answer:
Compiler
error: Cannot modify a constant value.
Explanation:
p is a pointer to a
"constant integer". But we tried to change the value of the
"constant integer".
2.
main()
{
char
s[ ]="man";
int
i;
for(i=0;s[
i ];i++)
printf("\n%c%c%c%c",s[
i ],*(s+i),*(i+s),i[s]);
}
Answer:
mmmm
aaaa
nnnn
Explanation:
s[i], *(i+s),
*(s+i), i[s] are all different ways of expressing the same idea. Generally array name is the base address for that
array. Here s is the base address. i is the index number/displacement from the
base address. So, indirecting it with * is same as s[i]. i[s] may be
surprising. But in the case of C it
is same as s[i].
3.
main()
{
float
me = 1.1;
double
you = 1.1;
if(me==you)
printf("I
love U");
else
printf("I
hate U");
}
Answer:
I hate
U
Explanation:
For floating point
numbers (float, double, long double) the values cannot be predicted exactly.
Depending on the number of bytes, the precession with of the value represented varies. Float takes 4 bytes and
long double takes 10 bytes. So float stores 0.9 with less precision than long
double.
Rule of Thumb:
Never compare or
at-least be cautious when using floating point numbers with relational
operators (== , >, <, <=, >=,!= ) .
4.
main()
{
static
int var = 5;
printf("%d
",var--);
if(var)
main();
}
Answer:
5 4 3 2
1
Explanation:
When static storage
class is given, it is initialized once. The change in the value of a static
variable is retained even between the function calls. Main
is also treated like any other ordinary function, which can be called
recursively.
5.
main()
{
int c[ ]={2.8,3.4,4,6.7,5};
int j,*p=c,*q=c;
for(j=0;j<5;j++) {
printf("
%d ",*c);
++q; }
for(j=0;j<5;j++){
printf("
%d ",*p);
++p; }
}
Answer:
2 2 2 2 2 2 3 4 6 5
Explanation:
Initially pointer c
is assigned to both p and q. In the first loop, since only q is incremented and
not c , the value 2 will be printed 5 times. In second loop p itself is
incremented. So the values 2 3 4 6 5 will be printed.
6.
main()
{
extern
int i;
i=20;
printf("%d",i);
}
Answer:
Linker
Error : Undefined symbol '_i'
Explanation:
extern storage class in the
following declaration,
extern int i;
specifies to the
compiler that the memory for i is allocated in some other program and that
address will be given to the current program at the time of linking. But linker
finds that no other variable of name i is available in any other program with
memory space allocated for it. Hence a linker error has occurred .
7.
main()
{
int
i=-1,j=-1,k=0,l=2,m;
m=i++&&j++&&k++||l++;
printf("%d
%d %d %d %d",i,j,k,l,m);
}
Answer:
0 0 1 3 1
Explanation :
Logical operations
always give a result of 1 or 0 . And also the logical AND (&&) operator
has higher priority over the logical OR (||) operator. So the expression ‘i++ && j++ && k++’ is
executed first. The result of this expression is 0 (-1 && -1 && 0 = 0). Now
the expression is 0 || 2 which evaluates to 1 (because OR operator always gives
1 except for ‘0 || 0’ combination- for which it gives 0). So the value of m is
1. The values of other variables are also incremented by 1.
8.
main()
{
char
*p;
printf("%d
%d ",sizeof(*p),sizeof(p));
}
Answer:
1 2
Explanation:
The sizeof()
operator gives the number of bytes taken by its operand. P is a character
pointer, which needs one byte for storing its value (a character). Hence
sizeof(*p) gives a value of 1. Since it needs two bytes to store the address of
the character pointer sizeof(p) gives 2.
9.
main()
{
int
i=3;
switch(i)
{
default:printf("zero");
case 1: printf("one");
break;
case 2:printf("two");
break;
case 3: printf("three");
break;
}
}
Answer :
three
Explanation :
The default case
can be placed anywhere inside the loop. It is executed only when all other
cases doesn't match.
10.
main()
{
printf("%x",-1<<4);
}
Answer:
fff0
Explanation :
-1 is
internally represented as all 1's. When left shifted four times the least
significant 4 bits are filled with 0's.The %x format specifier specifies that
the integer value be printed as a hexadecimal value.
No comments:
Post a Comment