Citicorp Technical Questions with Answers 21st August 2012

Posted on :13-04-2016

 

Q1. 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; }
}

ANS: 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.

 

Q2. What is the output of the following program?

main()
{
char *p;
printf("%d %d ", sizeof(*p), sizeof(p));
}

ANS:  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.

 

Q3. What is the output of the following program?

main()
{
char s[ ] = "man";
int i;
for(i=0; s[ i ]; i++)
printf (" %c%c%c%c", s[ i ], *(s+i), *(i+s), i[s]);
}

ANS:
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].

 

Q4. What is the output of the following program?

#include
main()
{
char s[]={'a', 'b', 'c', ' ', 'c', ''};
char *p,*str, *str1;
p=&s[3];
str=p;
str1=s;
printf("%d", ++*p + ++* str1- 32);
}

ANS: 77 

Explanation:

p is pointing to character ' '. str1 is pointing to character 'a' ++*p. "p is pointing to ' ' and that is incremented by one." the ASCII value of ' ' is 10, which is then incremented to 11. The value of ++*p is 11. ++*str1, str1 is pointing to 'a' that is incremented by 1 and it becomes 'b'. ASCII value of 'b' is 98.


Now performing (11 + 98 – 32), we get 77("M");
So we get the output 77 :: "M" (Ascii is 77).

 

Q5. What is the output of the following program?

void main()
{
int const * p = 5;
printf("%d", ++(*p));
}

ANS: 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".

 

Q6. 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");
}

ANS: 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 (== , >, <, <=, >=,!= ) . 

 

Q7. What is the output of the following program?

main()
{
int i=10;
i=!i>14;
Printf ("i=%d", i);
}

ANS: i=0

Explanation:

In the expression !i>14 , NOT (!) operator has more precedence than ‘ >’ symbol. ! is a unary logical operator. !i (!10) is 0 (not of true is false). 0>14 is false (zero).

 

Q8. What is the output of the following program?

main()
{
static int var = 5;
printf("%d ", var--);
if(var)
main();
}

ANS: 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. 

 

Q9. What is the output of the following program?

main()
{
extern int i;
i=20;
printf("%d", i);
}

ANS: 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.

 

Q10. What is the output of the following program?

main()
{
int i=3;
switch(i)
{
default:printf("zero");
case 1: printf("one");
break;
case 2:printf("two");
break;
case 3: printf("three");
break;

}

ANS: three

Explanation:

The default case can be placed anywhere inside the loop. It is executed only when all other cases doesn't match.

 

Q11. What is the output of the following program?

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);
}

ANS: 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.

 

Q12. What is the output of the following program?

#define int char
main()
{
int i=65;
printf("sizeof(i) = %d", sizeof(i));
}

ANS: sizeof (i) = 1

Explanation:

Since the #define replaces the string int by the macro char

 

Q13. What is the output of the following program?

main()
{
char string[] = "Hello World";
display(string);
}
void display (char *string)
{
printf("%s", string);
}

ANS: Compiler Error : Type mismatch in redeclaration of function display 

Explanation:

In third line, when the function display is encountered, the compiler doesn't know anything about the function display. It assumes the arguments and return types to be integers, (which is the default type). When it sees the actual function display, the arguments and type contradicts with what it has assumed previously. Hence a compile time error occurs.

 

Q14. What is the output of the following program?

main()
{
printf("%x", -1<<4);
}

ANS: 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.

 

Q15. What is the output of the following program?

main()
{
int c=- -2;
printf("c=%d", c);
}

ANS: c=2;

Explanation:

Here unary minus (or negation) operator is used twice. Same maths rules applies, ie. minus * minus= plus.
Note: However you cannot give like --2. Because -- operator can only be applied to variables as a decrement operator (eg., i--). 2 is a constant and not a variable.

 




Citicorp Overseas Software Ltd Related Companies


Placement Papers for All Companies




FreshersLive - No.1 Job site in India. Here you can find latest 2023 government as well as private job recruitment notifications for different posts vacancies in India. Get top company jobs for both fresher and experienced. Job Seekers can get useful interview tips, resume services & interview Question and answer. Practice online test free which is helpful for interview preparation. Register with us to get latest employment news/rojgar samachar notifications. Also get latest free govt and other sarkari naukri job alerts daily through E-mail...
DMCA.com Protection Status