C Language Placement Question with Solution

1)

main()

{

static int a[3][3]={1,2,3,4,5,6,7,8,9};

int i,j;

static *p[ ]={a,a+1,a+2};

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

{

for(j=0;j<3;j++)

printf("%d\t%d\t%d\t%d\n",*(*(p+i)+j),

*(*(j+p)+i),*(*(i+p)+j),*(*(p+j)+i));

}

}

Answer:

1       1       1       1

2       4       2       4

3       7       3       7

4       2       4       2

5       5       5       5

6       8       6       8

7       3       7       3

8       6       8       6

9       9       9       9

Explanation:

*(*(p+i)+j) is equivalent to p[i][j].

2)

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.

3)

   main()

    {

int i = 257;

int *iPtr = &i;

printf("%d %d", *((char*)iPtr), *((char*)iPtr+1) );

    }

Answer:

1 1 

Explanation:

The integer value 257 is stored in the memory as, 00000001 00000001, so the individual bytes are taken by casting it to char * and get printed.

4)

main()

{

int i = 258;

int *iPtr = &i;

printf("%d %d", *((char*)iPtr), *((char*)iPtr+1) );

}

Answer:

2 1 

Explanation:

The integer value 257 can be represented in binary as, 00000001 00000001. Remember that the INTEL machines are ‘small-endian’ machines. Small-endian means that the lower order bytes are stored in the higher memory addresses and the higher order bytes are stored in lower addresses. The integer value 258 is stored in memory as: 00000001 00000010.   

5)

main()

{

int i=300;

char *ptr = &i;

*++ptr=2;

printf("%d",i);

}

Answer:

556

Explanation:-

The integer value 300  in binary notation is: 00000001 00101100. It is  stored in memory (small-endian) as: 00101100 00000001. Result of the expression *++ptr = 2 makes the memory representation as: 00101100 00000010. So the integer corresponding to it  is  00000010 00101100 => 556.

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