char p[ ]="%d\n";

    main()

{

         char p[ ]="%d\n";

p[1] = 'c';

printf(p,65);

}

Answer:

A

Explanation:

Due to the assignment p[1] = ‘c’ the string becomes, “%c\n”. Since this string becomes the format string for printf and ASCII value of 65 is ‘A’, the same gets printed.

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printf(“%d”,i=++i ==6);

main()

{

int i=5;

printf(“%d”,i=++i ==6);

}

Answer:

1

Explanation:

The expression can be treated as i = (++i==6), because == is of higher precedence than = operator. In the inner expression, ++i is equal to 6 yielding true(1). Hence the result.

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Pointer Function

int aaa() {printf(“Hi”);}

int bbb(){printf(“hello”);}

iny ccc(){printf(“bye”);}

main()

{

int ( * ptr[3]) ();

ptr[0] = aaa;

ptr[1] = bbb;

ptr[2] =ccc;

ptr[2]();

}

Answer:

 bye

Explanation:

int (* ptr[3])() says that ptr is an array of pointers to functions that takes no arguments and returns the type int. By the assignment ptr[0] = aaa; it means that the first function pointer in the array is initialized with the address of the function aaa. Similarly, the other two array elements also get initialized with the addresses of the functions bbb and ccc. Since ptr[2] contains the address of the function ccc, the call to the function ptr[2]() is same as calling ccc(). So it results in printing  "bye".

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printf("%d",++i++);

    main()

    {

     int i=5;

     printf("%d",++i++);

}

Answer:

         Compiler error: Lvalue required in function main

Explanation:

         ++i yields an rvalue.  For postfix ++ to operate an lvalue is required.

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int swap(int *a,int *b)

 int swap(int *a,int *b)

{

 *a=*a+*b;*b=*a-*b;*a=*a-*b;

}

main()

{

         int x=10,y=20;

    swap(&x,&y);

         printf("x= %d y = %d\n",x,y);

}

Answer

    x = 20 y = 10

Explanation

This is one way of swapping two values. Simple checking will help understand this.

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if(~0 == (unsigned int)-1)

 void main()

         {

if(~0 == (unsigned int)-1)

printf(“You can answer this if you know how values are represented in memory”);

         }  

 Answer

You can answer this if you know how values are represented in memory

Explanation

~ (tilde operator or bit-wise negation operator) operates on 0 to produce all ones to fill the space for an integer. –1 is represented in unsigned value as all 1’s and so both are equal.

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typedef enum

What is the output for the program given below

     typedef enum errorType{warning, error, exception,}error;

     main()

    {

        error g1;

        g1=1;

        printf("%d",g1);

     }

Answer

         Compiler error: Multiple declaration for error

Explanation

The name error is used in the two meanings. One means that it is a enumerator constant with value 1. The another use is that it is a type name (due to typedef) for enum errorType. Given a situation the compiler cannot distinguish the meaning of error to know in what sense the error is used:

    error g1;

g1=error;

    // which error it refers in each case?

When the compiler can distinguish between usages then it will not issue error (in pure technical terms, names can only be overloaded in different namespaces).

Note: the extra comma in the declaration,

enum errorType{warning, error, exception,}

is not an error. An extra comma is valid and is provided just for programmer’s convenience.

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int (*x)[10];

 Is the following statement a declaration/definition. Find what does it mean?

int (*x)[10];

Answer

         Definition.

Explanation: x is a pointer to array of(size 10) integers. Apply clock-wise rule to find the meaning of this definition.

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for(;i>=0;i++) ; (3)

 main()

    {

       char i=0;

       for(;i>=0;i++) ;

       printf("%d\n",i);

       

 }

Answer:

         Behavior is implementation dependent.

Explanation:

The detail if the char is signed/unsigned by default is implementation dependent. If the implementation treats the char to be signed by default the program will print –128 and terminate. On the other hand if it considers char to be unsigned by default, it goes to infinite loop.

Rule:

You can write programs that have implementation dependent behavior. But dont write programs that depend on such behavior.

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for(;i>=0;i++) ; (2)

 main()

    {

       unsigned char i=0;

       for(;i>=0;i++) ;

       printf("%d\n",i);

    }

Answer

    infinite loop

Explanation

The difference between the previous question and this one is that the char is declared to be unsigned. So the i++ can never yield negative value and i>=0 never becomes false so that it can come out of the for loop.

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for(;i>=0;i++) ;

main()

    {

       signed char i=0;

       for(;i>=0;i++) ;

       printf("%d\n",i);

    }

Answer

         -128

Explanation

Notice the semicolon at the end of the for loop. THe initial value of the i is set to 0. The inner loop executes to increment the value from 0 to 127 (the positive range of char) and then it rotates to the negative value of -128. The condition in the for loop fails and so comes out of the for loop. It prints the current value of i that is -128.

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