Technical / logical interview questions in C language

1.)

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 .

2.)

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.

3.)

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.

4.)

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.

5.)

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.

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C programming Language Technical Questions And Answers

1.)

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.

2.)

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.

3.)

#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 

4.)

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). 

5.)

#include

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).

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