Placement Questions-5

What is the output of the following Program?
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.

Placement Question-4

What is the output of the following program ?
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.

Placement Question-3

What is the output of the following program?
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.

Placement Question-2

What is the output of the following program ?
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 (== , >, <, <=, >=,!= ) .

Placement Question-1

What is the output of the following program ?
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".