C Aptitude Questions with Answers Part 3

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.

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<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).