Search Your Dream Job Here

C Programming Interview Question Paper part One

Note : All the programs are tested under Turbo C/C++ compilers.
It is assumed that,
_ Programs run under DOS environment,
_ The underlying machine is an x86 system,
_ Program is compiled using Turbo C/C++ compiler.
The program output may depend on the information based on this
assumptions (for example sizeof(int) == 2 may be assumed).


                                                 
1
2
3
4
5
6
7

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 Cit 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.

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

Answer:
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.


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


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

Answer:
sizeof(i)=1
Explanation:
Since the #define replaces the string int by the macro char


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

Answer:
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).


15. #include<stdio.h>
main()
{
char s[]={'a','b','c','\n','c','\0'};
char *p,*str,*str1;
p=&s[3];
str=p;
str1=s;
printf("%d",++*p + ++*str1-32);
}

Answer:
77
Explanation:
p is pointing to character '\n'. str1 is pointing to character 'a' ++*p. "p is pointing to '\n' and that is incremented by one."
the ASCII value of '\n' 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).


Last Updated Date: March 22, 2012

0 comments:

Post a Comment

Popular Posts in this Weak