#22. 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!!

71) struct aaa{
struct aaa *prev;
int i;
struct aaa *next;
};
main()
{
struct aaa abc,def,ghi,jkl;
int x=100;
abc.i=0;abc.prev=&jkl;
abc.next=&def;
def.i=1;def.prev=&abc;def.next=&ghi;
ghi.i=2;ghi.prev=&def;
ghi.next=&jkl;
jkl.i=3;jkl.prev=&ghi;jkl.next=&abc;
x=abc.next->next->prev->next->i;
printf("%d",x);
}

72) struct point
{
int x;
int y;
};

struct point origin,*pp;
main()
{
pp=&origin;
printf("origin is(%d%d)\n",(*pp).x,(*pp).y);
printf("origin is (%d%d)\n",pp->x,pp->y);
}

73) main()
{
int i=_l_abc(10);
printf("%d\n",--i);
}
int _l_abc(int i)
{
return(i++);
}

74) main()
{
char *p;
int *q;
long *r;
p=q=r=0;

p++;
}

75) main()
{
char c=' ',x,convert(z);
getc(c);
if((c>='a') && (c<='z'))
x=convert(c);
printf("%c",x);
}
convert(z)
{
return z-32;
}

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

71.
Answer:
2
Explanation:
above all statements form a double circular linked list;
abc.next->next->prev->next->i
this one points to "ghi" node the value of at particular node is 2.

72.
Answer:
origin is(0,0)
origin is(0,0)
Explanation:
pp is a pointer to structure. we can access the elements of the structure
either with arrow mark or with indirection operator.
Note:
Since structure point is globally declared x & y are initialized as zeroes

73.
Answer:
9
Explanation:
return(i++) it will first return i and then increments. i.e. 10 will be returned.

74.
Answer:
0001...0002...0004
Explanation:
++ operator when applied to pointers increments address according to their
corresponding data-types.

75.
Answer:
Compiler error
Explanation:

declaration of convert and format of getc() are wrong.

========================================================
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#21. 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!!

66) #include<stdio.h>
main()
{
const int i=4;
float j;
j = ++i;
printf("%d %f", i,++j);
}

67) #include<stdio.h>
main()
{
int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} };
int *p,*q;
p=&a[2][2][2];
*q=***a;
printf("%d..%d",*p,*q);
}

68) #include<stdio.h>
main()
{
register i=5;
char j[]= "hello";
printf("%s %d",j,i);
}

69) main()
{
int i=5,j=6,z;
printf("%d",i+++j);
}

70) #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);
}

========================================================
66.
Answer:
Compiler error
Explanation:
i is a constant. you cannot change the value of constant

67.
Answer:
garbagevalue..1
Explanation:
p=&a[2][2][2] you declare only two 2D arrays. but you are trying to access
the third 2D(which you are not declared) it will print garbage values. *q=***a
starting address of a is assigned integer pointer. now q is pointing to starting
address of a.if you print *q meAnswer:it will print first element of 3D array.

68.
Answer:
hello 5
Explanation:
if you declare i as register compiler will treat it as ordinary integer and it will
take integer value. i value may be stored either in register or in memory.

69.
Answer:
11
Explanation:
the expression i+++j is treated as (i++ + j)

70.
Answer:
Compiler Error
Explanation:
Initialization should not be done for structure members inside the structure
declaration

========================================================
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#20. 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!!

61) 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");
}

62) #define max 5
#define int arr1[max]
main()
{
typedef char arr2[max];
arr1 list={0,1,2,3,4};
arr2 name="name";
printf("%d %s",list[0],name);
}

63) int i=10;
main()
{
extern int i;
{
int i=20;
{
const volatile unsigned i=30;
printf("%d",i);
}
printf("%d",i);
}
printf("%d",i);
}

64) main()
{
int *j;
{
int i=10;
j=&i;
}
printf("%d",*j);
}

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


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

61.
Answer:
2000 is a leap year
Explanation:
An ordinary program to check if leap year or not.

62.
Answer:
Compiler error (in the line arr1 list = {0,1,2,3,4})
Explanation:
arr2 is declared of type array of size 5 of characters. So it can be used to
declare the variable name of the type arr2. But it is not the case of arr1.
Hence an error.
Rule of Thumb:
#defines are used for textual replacement whereas typedefs are used for
declaring new types.

63.
Answer:
30,20,10
Explanation:
'{' introduces new block and thus new scope. In the innermost block i is
declared as,
const volatile unsigned
which is a valid declaration. i is assumed of type int. So printf prints 30. In the
next block, i has value 20 and so printf prints 20. In the outermost block, i is
declared as extern, so no storage space is allocated for it. After compilation
is over the linker resolves it to global variable i (since it is the only variable
visible there). So it prints i's value as 10.

64.
Answer:
10
Explanation:
The variable i is a block level variable and the visibility is inside that block
only. But the lifetime of i is lifetime of the function so it lives upto the exit of
main function. Since the i is still allocated space, *j prints the value stored in i
since j points i.

65.
Answer:
i = -1, -i = 1
Explanation:
-i is executed and this execution doesn't affect the value of i. In printf first you
just print the value of i. After that the value of the expression -i = -(-1) is
printed.

To be continued...

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

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#19. 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!!

56) main()
{
char *cptr,c;
void *vptr,v;
c=10; v=0;
cptr=&c; vptr=&v;
printf("%c%v",c,v);
}

57) main()
{
char *str1="abcd";
char str2[]="abcd";
printf("%d %d %d",sizeof(str1),sizeof(str2),sizeof("abcd"));
}

58)main()
{
char not;
not=!2;

printf("%d",not);
}

59) #define FALSE -1
#define TRUE 1
#define NULL 0
main() {
if(NULL)
puts("NULL");
else if(FALSE)
puts("TRUE");
else
puts("FALSE");
}


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

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

56.
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.

57.
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.

58.
Answer:
0
Explanation:
! is a logical operator. In C the value 0 is considered to be the boolean value
FALSE, and any non-zero value is considered to be the boolean value
TRUE. Here 2 is a non-zero value so TRUE. !TRUE is FALSE (0) so it prints
0.

59.
Answer:
TRUE
Explanation:
The input program to the compiler after processing by the preprocessor is,
main(){
if(0)
puts("NULL");
else if(-1)
puts("TRUE");
else
puts("FALSE");
}
Preprocessor doesn't replace the values given inside the double quotes. The
check by if condition is boolean value false so it goes to else. In second if -1
is boolean value true hence "TRUE" is printed.

60.
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".

To be continued...

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

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#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...

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

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#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...

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

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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...

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

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SQL DBA