Thursday 15 August 2013

BasicLevelCPractice_Functions_ SET1


Note :    All the programs are tested under Turbo C/C++ compilers. 

All the programs are tested on DOS environment, machine is an x86 system and compiled using Turbo C/C++ compiler

The program output may depend on the information based on this assumptions (for example sizeof(int) == 2 may be assumed).

Predict the output or error(s) for the following:

1.      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.

2.      main()

{

clrscr();

}

clrscr();

           

Answer:

No output/error

Explanation:

The first clrscr() occurs inside a function. So it becomes a function call. In the second clrscr(); is a function declaration (because it is not inside any function).

3.      main()

{

int i;

printf("%d",scanf("%d",&i));  // value 10 is given as input here

}

Answer:

1

Explanation:

Scanf returns number of items successfully read and not 1/0.  Here 10 is given as input which should have been scanned successfully. So number of items read is 1.
4.      main()

{

printf("%d", out);

}

int out=100;

Answer:

Compiler error: undefined symbol out in function main.

Explanation:

The rule is that a variable is available for use from the point of declaration. Even though a is a global variable, it is not available for main. Hence an error.
5.      main()

{

 show();

}

void show()

{

 printf("I'm the greatest");

}

Answer:

Compier error: Type mismatch in redeclaration of show.

Explanation:

When the compiler sees the function show it doesn't know anything about it. So the default return type (ie, int) is assumed. But when compiler sees the actual definition of show mismatch occurs since it is declared as void. Hence the error.

The solutions are as follows:

1. declare void show() in main() .

2. define show() before main().

3. declare extern void show() before the use of show().
6.      main()

            {

            char name[10],s[12];

            scanf(" \"%[^\"]\"",s);

            }

            How scanf will execute?

Answer:

First it checks for the leading white space and discards it.Then it matches with a quotation mark and then it  reads all character upto another quotation mark.
7.      main()

            {

            main();

            }

Answer:

 Runtime error : Stack overflow.

Explanation:

main function calls itself again and again. Each time the function is called its return address is stored in the call stack. Since there is no condition to terminate the function call, the call stack overflows at runtime. So it terminates the program and results in an error.
8.      main()

            {

            int k=1;

            printf("%d==1 is ""%s",k,k==1?"TRUE":"FALSE");

            }

Answer:

1==1 is TRUE

Explanation:

When two strings are placed together (or separated by white-space) they are concatenated (this is called as "stringization" operation). So the string is as if it is given as "%d==1 is %s". The conditional operator( ?: ) evaluates to "TRUE".
9.      main()

            {

            int y;

            scanf("%d",&y); // input given is 2000

            if( (y%4==0 && y%100 != 0) || y%100 == 0 )

                 printf("%d is a leap year");

            else

                 printf("%d is not a leap year");

            }

Answer:

2000 is a leap year

Explanation:

An ordinary program to check if leap year or not.
10.  main()

{

 int i=_l_abc(10);

             printf("%d\n",--i);

}

int _l_abc(int i)

{

 return(i++);

}

Answer:

9

Explanation:

return(i++) it will first return i and then increments. i.e. 10 will be returned.

 

Wednesday 14 August 2013

BasicLevelCPractice_Pointer_ SET1


Note :    All the programs are tested under Turbo C/C++ compilers. 

All the programs are tested on DOS environment, machine is an x86 system and compiled using Turbo C/C++ compiler

The program output may depend on the information based on this assumptions (for example sizeof(int) == 2 may be assumed).

Predict the output or error(s) for the following:

1.      void main()

{

            int  const * p=5;

            printf("%d",++(*p));

}

Answer:

                        Compiler error: Cannot modify a constant value.

Explanation:   

p is a pointer to a "constant integer". But we tried to change the value of the "constant integer".

2.      void main()

{

 char far *farther,*farthest;

 

 printf("%d..%d",sizeof(farther),sizeof(farthest));

  

 }

Answer:

4..2 

Explanation:

            the second pointer is of char type and not a far pointer

3.      void main()

{

 char *p;

 p="Hello";

 printf("%c\n",*&*p);

}

Answer:

H

Explanation:

* is a dereference operator & is a reference  operator. They can be    applied any number of times provided it is meaningful. Here  p points to  the first character in the string "Hello". *p dereferences it and so its value is H. Again  & references it to an address and * dereferences it to the value H.

4.      void main()

{

 static int  a[ ]   = {0,1,2,3,4};

 int  *p[ ] = {a,a+1,a+2,a+3,a+4};

 int  **ptr =  p;

 ptr++;

 printf(“\n %d  %d  %d”, ptr-p, *ptr-a, **ptr);

 *ptr++;

 printf(“\n %d  %d  %d”, ptr-p, *ptr-a, **ptr);

 *++ptr;

 printf(“\n %d  %d  %d”, ptr-p, *ptr-a, **ptr);

 ++*ptr;

         printf(“\n %d  %d  %d”, ptr-p, *ptr-a, **ptr);

}

Answer:

                111

                222

                333

                344

Explanation:

Let us consider the array and the two pointers with some address

a     

0
1
2
3
4

   100      102      104      106      108

                                                                             p

100
102
104
106
108

                                                   1000    1002    1004    1006    1008

             ptr    

1000

2000

After execution of the instruction ptr++ value in ptr becomes 1002, if scaling factor for integer is 2 bytes. Now ptr – p is value in ptr – starting location of array p, (1002 – 1000) / (scaling factor) = 1,  *ptr – a = value at address pointed by ptr – starting value of array a, 1002 has a value 102  so the value is (102 – 100)/(scaling factor) = 1,  **ptr is the value stored in the location pointed by  the pointer of ptr = value pointed by value pointed by 1002 = value pointed by 102 = 1. Hence the output of the firs printf is  1, 1, 1.

After execution of *ptr++ increments value of the value in ptr by scaling factor, so it becomes1004. Hence, the outputs for the second printf are ptr – p = 2, *ptr – a = 2, **ptr = 2.

After execution of *++ptr increments value of the value in ptr by scaling factor, so it becomes1004. Hence, the outputs for the third printf are ptr – p = 3, *ptr – a = 3, **ptr = 3.

After execution of ++*ptr value in ptr remains the same, the value pointed by the value is incremented by the scaling factor. So the value in array p at location 1006 changes from 106 10 108,. Hence, the outputs for the fourth printf are ptr – p = 1006 – 1000 = 3, *ptr – a = 108 – 100 = 4, **ptr = 4.

5.      void main()

{

 char  *q;

 int  j;

 for (j=0; j<3; j++) scanf(“%s” ,(q+j));

 for (j=0; j<3; j++) printf(“%c” ,*(q+j));

 for (j=0; j<3; j++) printf(“%s” ,(q+j));

}

Explanation:

Here we have only one pointer to type char and since we take input in the same pointer thus we keep writing over in the same location, each time shifting the pointer value by 1. Suppose the inputs are MOUSE,  TRACK and VIRTUAL. Then for the first input suppose the pointer starts at location 100 then the input one is stored as

M

O

U

S
E
\0

When the second input is given the pointer is incremented as j value becomes 1, so the input is filled in memory starting from 101.

M

T

R

A

C

K

\0


The third input  starts filling from the location 102

M

T

V

I

R

T

U

A

L

\0

This is the final value stored .

The first printf prints the values at the position q, q+1 and q+2  = M T V

The second printf prints three strings starting from locations q, q+1, q+2

 i.e  MTVIRTUAL, TVIRTUAL and VIRTUAL.  

6.      void main()

{

 void *vp;

 char ch = ‘g’, *cp = “goofy”;

 int j = 20;

 vp = &ch;

 printf(“%c”, *(char *)vp);

 vp = &j;

 printf(“%d”,*(int *)vp);

 vp = cp;

 printf(“%s”,(char *)vp + 3);

}

Answer:

                g20fy

Explanation:

Since a void pointer is used it can be type casted to any  other type pointer. vp = &ch  stores address of char ch and the next statement prints the value stored in vp after type casting it to the proper data type pointer. the output is ‘g’. Similarly  the output from second printf is ‘20’. The third printf statement type casts it to print the string from the 4th value hence the output is ‘fy’.

7.      void main()

{

 static char *s[ ]  = {“black”, “white”, “yellow”, “violet”};

 char **ptr[ ] = {s+3, s+2, s+1, s}, ***p;

 p = ptr;

 **++p;

 printf(“%s”,*--*++p + 3);

}

Answer:

                ck

Explanation:

In this problem we have an array of char pointers pointing to start of 4 strings. Then we have ptr which is a pointer to a pointer of type char and a variable p which is a pointer to a pointer to a pointer of type char. p hold the initial value of ptr, i.e. p = s+3. The next statement increment value in p by 1 , thus now value of p =  s+2. In the printf statement the expression is evaluated *++p causes gets value s+1 then the pre decrement is executed and we get s+1 – 1 = s . the indirection operator now gets the value from the array of s and adds 3 to the starting address. The string is printed starting from this position. Thus, the output is ‘ck’. 

8.      void main()

{

 int  i, n;

 char *x = “girl”;

 n = strlen(x);

 *x = x[n];

 for(i=0; i<n; ++i)

   {

printf(“%s\n”,x);

x++;

   }

 }

Answer:

(blank space)

irl

rl

l


Explanation:

Here a string (a pointer to char) is initialized with a value “girl”.  The strlen function returns the length of the string, thus n has a value 4. The next statement assigns value at the nth location (‘\0’) to the first location. Now the string becomes “\0irl” . Now the printf statement prints the string after each iteration it increments it starting position.  Loop starts from 0 to 4. The first time x[0] = ‘\0’ hence it prints nothing and pointer value is incremented. The second time it prints from x[1] i.e “irl” and the third time it prints “rl” and the last time it prints “l” and the loop terminates.

9.      void main()

            {

            char *cptr,c;

            void *vptr,v;

            c=10;  v=0;

            cptr=&c; vptr=&v;

            printf("%c%v",c,v);

            }

Answer:

Compiler error (at line number 4): size of v is Unknown.

Explanation:

You can create a variable of type void * but not of type void, since void is an empty type. In the second line you are creating variable vptr of type void * and v of type void hence an error.

10.  void main()

            {

            char *str1="abcd";

            char str2[]="abcd";

            printf("%d %d %d",sizeof(str1),sizeof(str2),sizeof("abcd"));

            }

Answer:

2 5 5

Explanation:

In first sizeof, str1 is a character pointer so it gives you the size of the pointer variable. In second sizeof the name str2 indicates the name of the array whose size is 5 (including the '\0' termination character). The third sizeof is similar to the second one.