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Sample Technical Questions

C Sample Question

22. main()
    {
    char *p="hai friends",*p1;
    p1=p;
    while(*p!='\0') ++*p++;
    printf("%s   %s",p,p1);
    }

    Answer:

    ibj!gsjfoet

    Explanation:

    ++*p++ will be parse in the given order

    • *p that is value at the location currently pointed by p will be taken
    • ++*p the retrieved value will be incremented
    • when ; is encountered the location will be incremented that is p++ will be executed

    
    Hence, in the while loop initial value pointed by p is 'h', which is changed to 'i' by executing ++*p and pointer moves to point, 'a' which is similarly changed to 'b' and so on. Similarly blank space is converted to '!'. Thus, we obtain value in p becomes “ibj!gsjfoet” and since p reaches '\0' and p1 points to p thus p1doesnot print anything.

23. #include‹stdio.h›
    #define a 10
    main()
    {
    #define a 50
    printf("%d",a);
    }
    Answer:
    50
    Explanation:
    The preprocessor directives can be redefined anywhere in the program. So the most recently assigned value will be taken.
    
    #define clrscr() 100
    main()
    {
    clrscr();
    printf("%d\n",clrscr());
    }
    Answer:
    100
    Explanation:
    Preprocessor executes as a seperate pass before the execution of the compiler. So textual replacement of clrscr() to 100 occurs.The input  program to compiler looks like this :
    		
    main()
    		{
    		     100;
    		     printf("%d\n",100);
    		}
    Note:
    100; is an executable statement but with no action. So it doesn't give any problem
    
    main()
    {
    printf("%p",main);
    }
    Answer:
    Some address will be printed.
    Explanation:
    Function names are just addresses (just like array names are addresses).
    main() is also a function. So the address of function main will be printed. %p in printf specifies that the argument is an address. They are printed as hexadecimal numbers.
    

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