#18. Concepts in C -- only through Examples & Explanations

This is a continuation of last post. 

Before seeing the answer try to predict it & then see the explanation.

All the Best!!

51) What are the files which are automatically opened when a C file is executed?

52) what will be the position of the file marker?
a: fseek(ptr,0,SEEK_SET);
b: fseek(ptr,0,SEEK_CUR);

53) main()
{
char name[10],s[12];
scanf(" \"%[^\"]\"",s);
}
How scanf will execute?

54) What is the problem with the following code segment?
while ((fgets(receiving array,50,file_ptr)) != EOF);

55) main()
{
main();
}

========================================================
51.
Answer:
stdin, stdout, stderr (standard input,standard output,standard error).

52.
Answer :
a: The SEEK_SET sets the file position marker to the starting of the file.
b: The SEEK_CUR sets the file position marker to the current position
of the file.

53.
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.

54.
Answer & Explanation:
fgets returns a pointer. So the correct end of file check is checking for !=
NULL.

55.
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.

To be continued...

========================================================

Thanks for coming till this point!!
VSG
SQL DBA

#17. Concepts in C -- only through Examples & Explanations

This is a continuation of last post. 

Before seeing the answer try to predict it & then see the explanation.

All the Best!!

46) 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);
}

47) 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);
}

48) main()
{
int i, n;
char *x = “girl”;
n = strlen(x);
*x = x[n];
for(i=0; i<n; ++i)
{
printf(“%s\n”,x);
x++;
}
}

49) int i,j;
for(i=0;i<=10;i++)
{
j+=5;
assert(i<5);
}

50) main()
{
int i=-1;
+i;
printf("i = %d, +i = %d \n",i,+i);
}

========================================================
46.
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’.

47.
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’.

48.
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.

49.
Answer:
Runtime error: Abnormal program termination.
assert failed (i<5), <file name>,<line number>
Explanation:
asserts are used during debugging to make sure that certain conditions are
satisfied. If assertion fails, the program will terminate reporting the same.
After debugging use,
#undef NDEBUG
and this will disable all the assertions from the source code. Assertion
is a good debugging tool to make use of.

50.
Answer:
i = -1, +i = -1
Explanation:
Unary + is the only dummy operator in C. Where-ever it comes you can just
ignore it just because it has no effect in the expressions (hence the name dummy operator).

To be continued...

========================================================

Thanks for coming till this point!!
VSG
SQL DBA

#16. Concepts in C -- only through Examples & Explanations

This is a continuation of last post. 

Before seeing the answer try to predict it & then see the explanation.

All the Best!!

The Questions 44 & 45 are based on pointers former is Integer & the later one is Character pointer.

41) main()
{
show();
}
void show()
{
printf("I'm the greatest");
}

42) main( )
{
int a[2][3][2] = {{{2,4},{7,8},{3,4}},{{2,2},{2,3},{3,4}}};
printf(“%u %u %u %d \n”,a,*a,**a,***a);
printf(“%u %u %u %d \n”,a+1,*a+1,**a+1,***a+1);
}

43) main( )
{
int a[ ] = {10,20,30,40,50},j,*p;
for(j=0; j<5; j++)
{
printf(“%d” ,*a);
a++;
}
p = a;
for(j=0; j<5; j++)
{
printf(“%d ” ,*p);
p++;
}
}

**44) 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);
}

45) 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));
}

========================================================

41.
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().

42.
Answer:
100, 100, 100, 2
114, 104, 102, 3
Explanation:
The given array is a 3-D one. It can also be viewed as a 1-D array.
2 4 7 8 3 4 2 2 2 3 3 4
100 102 104 106 108 110 112 114 116 118 120 122
thus, for the first printf statement a, *a, **a give address of first element .
since the indirection ***a gives the value. Hence, the first line of the output.
for the second printf a+1 increases in the third dimension thus points to value
at 114, *a+1 increments in second dimension thus points to 104, **a +1
increments the first dimension thus points to 102 and ***a+1 first gets the
value at first location and then increments it by 1. Hence, the output.

43.
Answer:
Compiler error: lvalue required.
Explanation:
Error is in line with statement a++. The operand must be an lvalue and may
be of any of scalar type for the any operator, array name only when
subscripted is an lvalue. Simply array name is a non-modifiable lvalue.

44.
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.

45.
Answer:
MTV
MTVIRTUAL TVIRTUAL VIRTUAL
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.

To be continued...

========================================================

Thanks for coming till this point!!
VSG
SQL DBA

#15. Concepts in C -- only through Examples & Explanations

This is a continuation of last post. 

Before seeing the answer try to predict it & then see the explanation.

All the Best!!

36) #include<stdio.h>
main()
{
struct xx
{
int x=3;
char name[]="hello";
};
struct xx *s=malloc(sizeof(struct xx));
printf("%d",s->x);
printf("%s",s->name);
}

37) #include<stdio.h>
main()
{
struct xx

{
int x;
struct yy
{
char s;
struct xx *p;
};
struct yy *q;
};
}

38) main()
{
extern int i;
i=20;
printf("%d",sizeof(i));
}

44) main()
{
printf("%d", out);
}
int out=100;

40) main()
{
extern out;
printf("%d", out);
}
int out=100;

========================================================
36.
Answer:
Compiler Error
Explanation:
Initialization should not be done for structure members inside the structure
declaration

37.

Answer:
Compiler Error
Explanation:
in the end of nested structure yy a member have to be declared.

38.

Answer:
Linker error: undefined symbol '_i'.
Explanation:
extern declaration specifies that the variable i is defined somewhere else.
The compiler passes the external variable to be resolved by the linker. So
compiler doesn't find an error. During linking the linker searches for the
definition of i. Since it is not found the linker flags an error.

39.

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.

40.

Answer:
100
Explanation:

This is the correct way of writing the previous program.

To be continued...

========================================================

Thanks for coming till this point!!
VSG
SQL DBA

#14. Concepts in C -- only through Examples & Explanations

This is a continuation of last post. 

Before seeing the answer try to predict it & then see the explanation.

All the Best!!

31) #include<stdio.h>
main()
{
int i=1,j=2;
switch(i)
{
case 1: printf("GOOD");
break;
case j: printf("BAD");
break;
}
}

32) main()
{
int i;
printf("%d",scanf("%d",&i)); // value 10 is given as input here
}

33) main()
{
clrscr();
}
clrscr();

34) main()
{
int i=0;
for(;i++;printf("%d",i)) ;
printf("%d",i);
}

35) #include<stdio.h>
main()
{
char s[]={'a','b','c','\n','c','\0'};
char *p,*str,*str1;
p=&s[3];
str=p;
str1=s;
printf("%C",++*p + ++*str1-32);
}

======================================================
31.
Answer:
Compiler Error: Constant expression required in function main.
Explanation:
The case statement can have only constant expressions (this implies that we
cannot use variable names directly so an error).
Note:
Enumerated types can be used in case statements.

32.
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.

33.
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).

34.
Answer:
1
Explanation:
before entering into the for loop the checking condition is "evaluated". Here it
evaluates to 0 (false) and comes out of the loop, and i is incremented (note
the semicolon after the for loop).

35.
Answer:
M
Explanation:
p is pointing to character '\n'.str1 is pointing to character 'a' ++*p
 Answer:"p is pointing to '\n' and that is incremented by one." the ASCII
value of '\n' is 10. then it is incremented to 11. the value of ++*p is 11. ++*str1
 Answer:"str1 is pointing to 'a' that is incremented by 1 and it becomes 'b'.
ASCII value of 'b' is 98. both 11 and 98 is added and result is subtracted from
32.
i.e. (11+98-32)=77("M");

To be continued...

========================================================

Thanks for coming till this point!!
VSG
SQL DBA

#12. Script to check detached_attached database

Why?
There are multiple ways on which you could maintain your databases numbers along with the details of newly created databases & those which got deleted from your Sql Server , this can be easily identified from SQL logs. But to my scenario, i don't want to do that often rather designed a script which can be run on a weekly basis that will log the details i need. So let's start.

Scenarios:
My idea is to use XP_cmdshell. to check databases available from sys.master_files & load their names into text files & then compare on a weekly basis. & it will log the results into another text file with info.We are going to handle 3 scenarios within this .
 a. No databases got deleted or created.
 b. Databases got newly created.
 c. Databases got deleted.

Script :

/* script to identify detached or attached databases from a SQL server by VSG

 the detached_added_db.txt will give you difference in how database names were there in last week

& how they are in current week along with the databases got detached or deleted or added.

*/

EXEC master.dbo.sp_configure 'show advanced options', 1
RECONFIGURE

EXEC master.dbo.sp_configure 'xp_cmdshell', 1
RECONFIGURE

-- the following query should be executed only once before you schedule your job.
--please leave it as commented it only once you had executed.
--EXEC xp_cmdshell 'bcp "SELECT [name]FROM SYS.master_files" queryout "D:\current_week.txt" -T -c -t'

-- changing previous file to last_week.txt

EXEC xp_cmdshell'DEL D:\last_week.txt'

EXEC xp_cmdshell'RENAME D:\current_week.txt last_week.txt'

-- loading currnet data
EXEC xp_cmdshell 'bcp "SELECT [name] FROM SYS.master_files" queryout "D:\current_week.txt" -T -c -t'

-- script to compare
EXEC xp_cmdshell'CD D:'

EXEC xp_cmdshell'echo this job is running on >>D:\Detached_added_db.txt'

EXEC xp_cmdshell'DATE /T >>D:\Detached_added_db.txt'

EXEC xp_cmdshell'TIME /T/N>>D:\Detached_added_db.txt'

EXEC xp_cmdshell'FC /C /N D:\last_week.txt D:\current_week.txt >>D:\Detached_added_db.txt'

EXEC xp_cmdshell'echo if databases are detached it will be shown under last_week changes if databases are created it will show in current_week changes>>D:\Detached_added_db.txt'

EXEC xp_cmdshell'echo============================>>D:\Detached_added_db.txt'

Explanation:

 1. For using this script we need  to enable xp_cmdshell  in sp_configure to 1.
 2. we are using 3 text files here.All will be created by the script itself.
      i. last_week--> holds details of last weeks Db details
     i i.current_week--> holds details of current week Db details.
    i i i.Detached_added_db --> holds details of changes happened.
3. Note that we need to create current_week.txt only once using the script commented there & then you need to schedule this job.Also comment the same again.
4.the idea is to compare current_week with last_week files & report the difference into Detached_added_db

4. Ran the commented line & this will create current_week text file on D:\ location on my system.
5.Now I am running whole script. this will create the required Detached_added_db text file on D:\ location.
6.Since we haven't done any changes in databases, the detached_added text file shows following output. ( scenario 1)

7.Now adding one database to the server (scenario 2) and running the query again.the output looks like  

8. Now deleting newly created database and detaching ADVENTURE database (Scenario 3). Then running the query.New log will get appended to the same file. Now the output will be

9. Thus by scheduling this job we can keep a record of our databases created & deleted on a weekly basis & you could see & predict on which week your application or users created or deleted them.

========================================================

Thanks for coming till this point!!
VSG
SQL DBA

#13. Concepts in C -- only through Examples & Explanations

This is a continuation of  post #11. 

Before seeing the answer try to predict it & then see the explanation.

All the Best!!

26) main()
{
char *p;
p="Hello";
printf("%c\n",*&*p);
}

27) main()
{
int i=1;
while (i<=5)
{
printf("%d",i);
if (i>2)
goto here;
i++;
}

}
fun()
{
here:
printf("PP");
}

28) main()
{
static char names[5][20]={"pascal","ada","cobol","fortran","perl"};
int i;
char *t;
t=names[3];
names[3]=names[4];
names[4]=t;
for (i=0;i<=4;i++)
printf("%s",names[i]);
}

29) void main()
{
int i=5;
printf("%d",i++ + ++i);
}

30) void main()
{
int i=5;
printf("%d",i+++++i);
}

===================================================
26.
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.

27.
Answer:
Compiler error: Undefined label 'here' in function main
Explanation:
Labels have functions scope, in other words the scope of the labels is limited
to functions. The label 'here' is available in function fun() Hence it is not
visible in function main.

28.
Answer:
Compiler error: Lvalue required in function main
Explanation:
Array names are pointer constants. So it cannot be modified.

29.
Answer:
Output Cannot be predicted exactly.
Explanation:
Side effects are involved in the evaluation of i

30.
Answer:
Compiler Error
Explanation:
The expression i+++++i is parsed as i ++ ++ + i which is an illegal
combination of operators.

To be continued...

========================================================

Thanks for coming till this point!!
VSG
SQL DBA

#11. Concepts in C -- only through Examples & Explanations

This is a continuation of last post. 

Before seeing the answer try to predict it & then see the explanation.

All the Best!!

21. #define square(x) x*x
main()
{
int i;

i = 64/square(4);
printf("%d",i);
}

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

23. #include <stdio.h>
#define a 10
main()
{
#define a 50
printf("%d",a);
}

24. #define clrscr() 100
main()
{
clrscr();
printf("%d\n",clrscr());
}

25. main()
{
printf("%p",main);
}

==========================================================

21.
Answer:
64
Explanation:
the macro call square(4) will substituted by 4*4 so the expression becomes i
= 64/4*4 . Since / and * has equal priority the expression will be evaluated as (64/4)*4 i.e.
16*4 = 64

22.
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.
Answer:

50
Explanation:
The preprocessor directives can be redefined anywhere in the program. So
the most recently assigned value will be taken.

24.
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

25.
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.

========================================================

Thanks for coming till this point!!
VSG
SQL DBA