Is the following code legal?

    Is the following code legal?

struct a

    {

int x;

 struct a b;

    }

Answer:

         No

Explanation:

Is it not legal for a structure to contain a member that is of the same

type as in this case. Because this will cause the structure declaration to be recursive without end.

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Is there any difference between the two declarations,

    Is there any difference between the two declarations,

1. int foo(int *arr[]) and

2. int foo(int *arr[2])

Answer:

No

Explanation:

Functions can only pass pointers and not arrays. The numbers that are allowed inside the [] is just for more readability. So there is no difference between the two declarations.

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*a ^= *b, *b ^= *a, *a ^= *b;

    main()

{

         void swap();

    int x=10,y=8;    

         swap(&x,&y);

    printf("x=%d y=%d",x,y);

}

void swap(int *a, int *b)

{

   *a ^= *b,  *b ^= *a, *a ^= *b;

}  

Answer:

x=10 y=8

Explanation:

Using ^ like this is a way to swap two variables without using a temporary variable and that too in a single statement.

Inside main(), void swap(); means that swap is a function that may take any number of arguments (not no arguments) and returns nothing. So this doesn’t issue a compiler error by the call swap(&x,&y); that has two arguments.

This convention is historically due to pre-ANSI style (referred to as Kernighan and Ritchie style) style of function declaration. In that style, the swap function will be defined as follows,

void swap()

int *a, int *b

{

   *a ^= *b,  *b ^= *a, *a ^= *b;

}

where the arguments follow the (). So naturally the declaration for swap will look like, void swap() which means the swap can take any number of arguments.

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*f2+=*f2+=a+=2.5;

    main()

{

         int a=2,*f1,*f2;

    f1=f2=&a;

         *f2+=*f2+=a+=2.5;

    printf("\n%d %d %d",a,*f1,*f2);

}

Answer:

16 16 16

Explanation:

f1 and f2 both refer to the same memory location a. So changes through f1 and f2 ultimately affects only the value of a.

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extern variable

    main()

{  

         extern i;

    printf("%d\n",i);

         {

             int i=20;

         printf("%d\n",i);

         }

}

Answer:

Linker Error : Unresolved external symbol i

Explanation:

The identifier i is available in the inner block and so using extern has no use in resolving it.

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switch(float)

    main()

{

         float i=1.5;

    switch(i)

         {

         case 1: printf("1");

             case 2: printf("2");

             default : printf("0");

    }

}

Answer:

Compiler Error: switch expression not integral

Explanation:

         Switch statements can be applied only to integral types.

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register int a=2;

    main()

{

         register int a=2;

    printf("Address of a = %d",&a);

         printf("Value of a   = %d",a);

}

Answer:

Compier Error: '&' on register variable

Rule to Remember:

          & (address of ) operator cannot be applied on register variables.

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j = j || i++ && printf("YOU CAN");

    main()

{

         int i=4,j=7;

    j = j || i++ && printf("YOU CAN");

         printf("%d %d", i, j);

}

Answer:

4 1

Explanation:

The boolean expression needs to be evaluated only till the truth value of the expression is not known. j is not equal to zero itself means that the expression’s truth value is 1. Because it is followed by || and true || (anything) => true where (anything) will not be evaluated. So the remaining expression is not evaluated and so the value of i remains the same.

Similarly when && operator is involved in an expression, when any of the operands become false, the whole expression’s truth value becomes false and hence the remaining expression will not be evaluated.    

    false && (anything) => false where (anything) will not be evaluated.

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i=i&=j&&10;

    main()

{

         int i=5,j=10;

    i=i&=j&&10;

         printf("%d %d",i,j);

}

Answer:

1 10

Explanation:

The expression can be written as i=(i&=(j&&10)); The inner expression (j&&10) evaluates to 1 because j==10. i is 5. i = 5&1 is 1. Hence the result.

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Differentiate the above declarations

    1. const char *a;

2. char* const a;

3. char const *a;

-Differentiate the above declarations.

Answer:

1. 'const' applies to char * rather than 'a' ( pointer to a constant char )

    *a='F'       : illegal

             a="Hi"       : legal

2. 'const' applies to 'a'  rather than to the value of a (constant pointer to char )

    *a='F'       : legal

    a="Hi"       : illegal

3. Same as 1.

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j = i, j?(i,j)?i:j:j;

    main()

{

         int i=10,j=20;

    j = i, j?(i,j)?i:j:j;

         printf("%d %d",i,j);

}

Answer:

10 10

Explanation:

         The Ternary operator ( ? : ) is equivalent for if-then-else statement. So the question can be written as:

         if(i,j)

            {

if(i,j)

              j = i;

         else

             j = j;           

         }

       else

         j = j;      

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(i>2)?i++:i--;

    void main()

{

static int i;

while(i<=10)

(i>2)?i++:i--;

    printf(“%d”, i);

}

Answer:

         32767

Explanation:

Since i is static it is initialized to 0. Inside the while loop the conditional operator evaluates to false, executing i--. This continues till the integer value rotates to positive value (32767). The while condition becomes false and hence, comes out of the while loop, printing the i value.

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for (;i++=0;) printf(“%d”,i);

    main()

{

int i = 3;

for (;i++=0;) printf(“%d”,i);

}

Answer:

         Compiler Error: Lvalue required.

Explanation:

As we know that increment operators return rvalues and  hence it cannot appear on the left hand side of an assignment operation.

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while (strcmp(str1,str2)) (2)

    main()

{

char str1[] = {‘s’,’o’,’m’,’e’};

char str2[] = {‘s’,’o’,’m’,’e’,’\0’};

while (strcmp(str1,str2))

printf(“Strings are not equal\n”);

}

Answer:

“Strings are not equal”

“Strings are not equal”

….

Explanation:

If a string constant is initialized explicitly with characters, ‘\0’ is not appended automatically to the string. Since str1 doesn’t have null termination, it treats whatever the values that are in the following positions as part of the string until it randomly reaches a ‘\0’. So str1 and str2 are not the same, hence the result.

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while (strcmp(“some”,”some\0”))

    main()

{

while (strcmp(“some”,”some\0”))

printf(“Strings are not equal\n”);

    }

Answer:

No output

Explanation:

Ending the string constant with \0 explicitly makes no difference. So “some” and “some\0” are equivalent. So, strcmp returns 0 (false) hence breaking out of the while loop.

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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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Important Errors

    main()

{

    float f=5,g=10;

    enum{i=10,j=20,k=50};

    printf("%d\n",++k);

    printf("%f\n",f<<2);

    printf("%lf\n",f%g);

    printf("%lf\n",fmod(f,g));

}

Answer:

Line no 5: Error: Lvalue required

Line no 6: Cannot apply leftshift to float

Line no 7: Cannot apply mod to float

Explanation:

         Enumeration constants cannot be modified, so you cannot apply ++.

         Bit-wise operators and % operators cannot be applied on float values.

         fmod() is to find the modulus values for floats as % operator is for ints. 

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