Some of the C puzzle which will explore C fundamental strongly,
C Apti Questions and answer with explanation,
Note : All the programs are tested under Turbo
C/C++ compilers.
It is assumed that,
·
Programs run
under DOS environment,
·
The underlying
machine is an x86 system,
·
Program is
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. main()
{
char s[ ]="man";
int i;
for(i=0;s[ i ];i++)
printf("\n%c%c%c%c",s[ i
],*(s+i),*(i+s),i[s]);
}
Answer:
mmmm
aaaa
nnnn
Explanation:
s[i], *(i+s), *(s+i), i[s] are all different ways of
expressing the same idea. Generally
array name is the base address for that array. Here s is the base
address. i is the index number/displacement from the base address. So,
indirecting it with * is same as s[i]. i[s] may be surprising. But in the case of
C it is same as s[i].
3. main()
{
float me = 1.1;
double you = 1.1;
if(me==you)
printf("I love U");
else
printf("I hate U");
}
Answer:
I hate U
Explanation:
For floating point numbers (float, double, long
double) the values cannot be predicted exactly. Depending on the number
of bytes, the precession with of the value
represented varies. Float takes 4 bytes and long double takes 10 bytes.
So float stores 0.9 with less precision than long double.
Rule of Thumb:
Never compare or at-least be cautious when using
floating point numbers with relational operators (== , >, <,
<=, >=,!= ) .
4. main()
{
static int var = 5;
printf("%d ",var--);
if(var)
main();
}
Answer:
5 4 3 2 1
Explanation:
When static storage class is given, it is
initialized once. The change in the value of a static variable is
retained even between the function calls. Main is also treated like any other
ordinary function, which can be called recursively.
5. main()
{
int c[ ]={2.8,3.4,4,6.7,5};
int j,*p=c,*q=c;
for(j=0;j<5;j++) {
printf(" %d ",*c);
++q; }
for(j=0;j<5;j++){
printf(" %d ",*p);
++p; }
}
Answer:
2
2 2 2 2 2 3 4 6 5
Explanation:
Initially pointer c is assigned to both p and q.
In the first loop, since only q is incremented and not c , the value 2
will be printed 5 times. In second loop p itself is incremented. So the
values 2 3 4 6 5 will be printed.
6. main()
{
extern int i;
i=20;
printf("%d",i);
}
Answer:
Linker Error
: Undefined symbol '_i'
Explanation:
extern
storage class in the following declaration,
extern int i;
specifies to the compiler that the memory for i
is allocated in some other program and that address will be given to the
current program at the time of linking. But linker finds that no other variable
of name i is available in any other program with memory space allocated
for it. Hence a linker error has occurred .
7. main()
{
int i=-1,j=-1,k=0,l=2,m;
m=i++&&j++&&k++||l++;
printf("%d %d %d %d
%d",i,j,k,l,m);
}
Answer:
0
0 1 3 1
Explanation :
Logical operations always give a result of 1 or 0 .
And also the logical AND (&&) operator has higher priority over the
logical OR (||) operator. So the expression
‘i++ && j++ && k++’ is executed first. The result
of this expression is 0 (-1 &&
-1 && 0 = 0). Now the expression is 0 || 2 which evaluates to 1
(because OR operator always gives 1 except for ‘0 || 0’ combination- for which
it gives 0). So the value of m is 1. The values of other variables are also
incremented by 1.
8. main()
{
char *p;
printf("%d %d
",sizeof(*p),sizeof(p));
}
Answer:
1
2
Explanation:
The sizeof() operator gives the number of bytes taken
by its operand. P is a character pointer, which needs one byte for storing its
value (a character). Hence sizeof(*p) gives a value of 1. Since it needs two
bytes to store the address of the character pointer sizeof(p) gives 2.
9. main()
{
int i=3;
switch(i)
{
default:printf("zero");
case 1: printf("one");
break;
case 2:printf("two");
break;
case 3: printf("three");
break;
}
}
Answer :
three
Explanation :
The default case can be placed anywhere inside the
loop. It is executed only when all other cases doesn't match.
10.main()
{
printf("%x",-1<<4);
}
Answer:
fff0
Explanation :
-1 is internally represented as all 1's. When left
shifted four times the least significant 4 bits are filled with 0's.The %x
format specifier specifies that the integer value be printed as a hexadecimal
value.
11.main()
{
char string[]="Hello World";
display(string);
}
void display(char *string)
{
printf("%s",string);
}
Answer:
Compiler Error : Type mismatch in redeclaration of function display
Explanation
:
In third line, when the function display is
encountered, the compiler doesn't know anything about the function display. It
assumes the arguments and return types to be integers, (which is the default
type). When it sees the actual function display, the arguments and type
contradicts with what it has assumed previously. Hence a compile time error
occurs.
12.main()
{
int c=- -2;
printf("c=%d",c);
}
Answer:
c=2;
Explanation:
Here unary minus (or negation) operator is used twice.
Same maths rules applies, ie. minus *
minus= plus.
Note:
However you cannot give like --2. Because -- operator
can only be applied to variables as a decrement
operator (eg., i--). 2 is a constant and not a variable.
13.#define int char
main()
{
int i=65;
printf("sizeof(i)=%d",sizeof(i));
}
Answer:
sizeof(i)=1
Explanation:
Since the #define replaces the string int by the macro char
14.main()
{
int i=10;
i=!i>14;
printf("i=%d",i);
}
Answer:
i=0
Explanation:
In the expression !i>14 , NOT (!) operator
has more precedence than ‘ >’ symbol.
! is a unary logical operator. !i (!10) is 0 (not of true is
false). 0>14 is false (zero).
15.#include<stdio.h>
main()
{
char s[]={'a','b','c','\n','c','\0'};
char *p,*str,*str1;
p=&s[3];
str=p;
str1=s;
printf("%d",++*p + ++*str1-32);
}
Answer:
77
Explanation:
p is pointing to character '\n'. str1 is pointing to
character 'a' ++*p. "p is pointing to '\n' and that is incremented by
one." the ASCII value of '\n' is 10, which is then incremented to 11. The
value of ++*p is 11. ++*str1, str1 is pointing to 'a' that is incremented by 1
and it becomes 'b'. ASCII value of 'b' is 98.
Now performing
(11 + 98 - 32), we get 77("M");
So we get the
output 77 :: "M" (Ascii is 77).
16.#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);
}
Answer:
SomeGarbageValue---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, it will print first element of 3D array.
17.#include<stdio.h>
main()
{
struct xx
{
int x=3;
char
name[]="hello";
};
struct xx *s;
printf("%d",s->x);
printf("%s",s->name);
}
Answer:
Compiler Error
Explanation:
You should not initialize variables in declaration
18.#include<stdio.h>
main()
{
struct xx
{
int x;
struct yy
{
char s;
struct
xx *p;
};
struct yy *q;
};
}
Answer:
Compiler Error
Explanation:
The structure yy is nested within structure xx. Hence,
the elements are of yy are to be accessed through the instance of structure xx,
which needs an instance of yy to be known. If the instance is created after
defining the structure the compiler will not know about the instance relative
to xx. Hence for nested structure yy you have to declare member.
19.main()
{
printf("\nab");
printf("\bsi");
printf("\rha");
}
Answer:
hai
Explanation:
\n - newline
\b - backspace
\r - linefeed
20.main()
{
int i=5;
printf("%d%d%d%d%d%d",i++,i--,++i,--i,i);
}
Answer:
45545
Explanation:
The arguments in a function call are pushed into the
stack from left to right. The evaluation is by popping out from the stack. and
the evaluation is from right to left,
hence the result.
21.#define square(x) x*x
main()
{
int i;
i = 64/square(4);
printf("%d",i);
}
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.main()
{
char *p="hai friends",*p1;
p1=p;
while(*p!='\0') ++*p++;
printf("%s
%s",p,p1);
}
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.#include <stdio.h>
#define a 10
main()
{
#define a 50
printf("%d",a);
}
Answer:
50
Explanation:
The preprocessor directives can be redefined anywhere
in the program. So the most recently assigned value will be taken.
24.#define clrscr() 100
main()
{
clrscr();
printf("%d\n",clrscr());
}
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.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.
27) 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).
28) enum colors {BLACK,BLUE,GREEN}
main()
{
printf("%d..%d..%d",BLACK,BLUE,GREEN);
return(1);
}
Answer:
0..1..2
Explanation:
enum assigns numbers starting from 0, if not
explicitly defined.
29) 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
30) main()
{
int
i=400,j=300;
printf("%d..%d");
}
Answer:
400..300
Explanation:
printf takes the values of the first two assignments
of the program. Any number of printf's may be given. All of them take only the
first two values. If more number of assignments given in the program,then
printf will take garbage values.
31) 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.
32)
main()
{
int i=1;
while
(i<=5)
{
printf("%d",i);
if
(i>2)
goto here;
i++;
}
}
fun()
{
here:
printf("PP");
}
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.
33) 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]);
}
Answer:
Compiler error: Lvalue required in function main
Explanation:
Array names are pointer constants. So it cannot be
modified.
34) void main()
{
int
i=5;
printf("%d",i++
+ ++i);
}
Answer:
Output Cannot be predicted exactly.
Explanation:
Side effects are involved in the evaluation of i
35) void main()
{
int
i=5;
printf("%d",i+++++i);
}
Answer:
Compiler Error
Explanation:
The expression i+++++i is parsed as i ++ ++ + i which
is an illegal combination of operators.
36) #include<stdio.h>
main()
{
int i=1,j=2;
switch(i)
{
case 1: printf("GOOD");
break;
case j: printf("BAD");
break;
}
}
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.
37) 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.
38) #define f(g,g2) g##g2
main()
{
int var12=100;
printf("%d",f(var,12));
}
Answer:
100
39) main()
{
int i=0;
for(;i++;printf("%d",i)) ;
printf("%d",i);
}
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).
40) #include<stdio.h>
main()
{
char
s[]={'a','b','c','\n','c','\0'};
char
*p,*str,*str1;
p=&s[3];
str=p;
str1=s;
printf("%d",++*p + ++*str1-32);
}
Answer:
M
Explanation:
p is pointing to character '\n'.str1 is pointing to
character 'a' ;++*p means :"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 means:"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");
41) #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);
}
Answer:
Compiler Error
Explanation:
Initialization should not be done for structure
members inside the structure declaration
42) #include<stdio.h>
main()
{
struct xx
{
int x;
struct yy
{
char s;
struct xx *p;
};
struct yy *q;
};
}
Answer:
Compiler Error
Explanation:
in the end of nested structure yy a member have to be
declared.
43) main()
{
extern int i;
i=20;
printf("%d",sizeof(i));
}
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.
44) 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.
45) main()
{
extern out;
printf("%d", out);
}
int out=100;
Answer:
100
Explanation:
This is the correct way of writing the previous
program.
46) 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().
47) 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);
}
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.
48) 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++;
}
}
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.
49) 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.
50) 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.
51) 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’.
52) 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’.
53) 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.
54) int i,j;
for(i=0;i<=10;i++)
{
j+=5;
assert(i<5);
}
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.
55) main()
{
int i=-1;
+i;
printf("i = %d, +i = %d
\n",i,+i);
}
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).
56) What are the files which are automatically
opened when a C file is executed?
Answer:
stdin, stdout, stderr (standard input,standard
output,standard error).
57)
what will be the position of the file marker?
a: fseek(ptr,0,SEEK_SET);
b: fseek(ptr,0,SEEK_CUR);
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.
58) 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.
59) What is the problem with the following code
segment?
while ((fgets(receiving
array,50,file_ptr)) != EOF)
;
Answer & Explanation:
fgets returns a pointer. So the correct end of file
check is checking for != NULL.
60) 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.
61) 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.
62) 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.
63) main()
{
char not;
not=!2;
printf("%d",not);
}
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.
64) #define FALSE -1
#define TRUE 1
#define NULL 0
main() {
if(NULL)
puts("NULL");
else if(FALSE)
puts("TRUE");
else
puts("FALSE");
}
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.
65) 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".
66) 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.
67) #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);
}
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.
68) int i=10;
main()
{
extern int i;
{
int i=20;
{
const volatile unsigned i=30;
printf("%d",i);
}
printf("%d",i);
}
printf("%d",i);
}
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.
69) main()
{
int *j;
{
int i=10;
j=&i;
}
printf("%d",*j);
}
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.
70) main()
{
int i=-1;
-i;
printf("i = %d, -i = %d
\n",i,-i);
}
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.
71) #include<stdio.h>
main()
{
const int
i=4;
float j;
j = ++i;
printf("%d %f", i,++j);
}
Answer:
Compiler error
Explanation:
i is a constant. you cannot change the value of
constant
72) #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);
}
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.
73) #include<stdio.h>
main()
{
register
i=5;
char j[]=
"hello";
printf("%s %d",j,i);
}
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.
74) main()
{
int i=5,j=6,z;
printf("%d",i+++j);
}
Answer:
11
Explanation:
the expression i+++j is treated as (i++ + j)
76) 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);
}
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.
77) 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);
}
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
78) 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.
79) main()
{
char *p;
int *q;
long *r;
p=q=r=0;
p++;
q++;
r++;
printf("%p...%p...%p",p,q,r);
}
Answer:
0001...0002...0004
Explanation:
++ operator
when applied to pointers increments address according to their
corresponding data-types.
80) main()
{
char c='
',x,convert(z);
getc(c);
if((c>='a')
&& (c<='z'))
x=convert(c);
printf("%c",x);
}
convert(z)
{
return z-32;
}
Answer:
Compiler error
Explanation:
declaration of convert and format of getc() are wrong.
81) main(int argc, char **argv)
{
printf("enter the character");
getchar();
sum(argv[1],argv[2]);
}
sum(num1,num2)
int num1,num2;
{
return
num1+num2;
}
Answer:
Compiler error.
Explanation:
argv[1] & argv[2] are strings. They are passed to
the function sum without converting it to integer values.
82) # include <stdio.h>
int one_d[]={1,2,3};
main()
{
int *ptr;
ptr=one_d;
ptr+=3;
printf("%d",*ptr);
}
Answer:
garbage value
Explanation:
ptr pointer is pointing to out of the array range of
one_d.
83) # include<stdio.h>
aaa() {
printf("hi");
}
bbb(){
printf("hello");
}
ccc(){
printf("bye");
}
main()
{
int
(*ptr[3])();
ptr[0]=aaa;
ptr[1]=bbb;
ptr[2]=ccc;
ptr[2]();
}
Answer:
bye
Explanation:
ptr is array of pointers to functions of return type
int.ptr[0] is assigned to address of the function aaa. Similarly ptr[1] and
ptr[2] for bbb and ccc respectively. ptr[2]() is in effect of writing ccc(),
since ptr[2] points to ccc.
85) #include<stdio.h>
main()
{
FILE *ptr;
char i;
ptr=fopen("zzz.c","r");
while((i=fgetch(ptr))!=EOF)
printf("%c",i);
}
Answer:
contents of zzz.c followed by an infinite loop
Explanation:
The condition is checked against EOF, it should be
checked against NULL.
86) main()
{
int i =0;j=0;
if(i &&
j++)
printf("%d..%d",i++,j);
printf("%d..%d,i,j);
}
Answer:
0..0
Explanation:
The value of i is 0. Since this information is enough
to determine the truth value of the boolean expression. So the statement
following the if statement is not executed.
The values of i and j remain unchanged and get printed.
87) main()
{
int i;
i = abc();
printf("%d",i);
}
abc()
{
_AX = 1000;
}
Answer:
1000
Explanation:
Normally the return value from the function is through
the information from the accumulator. Here _AH is the pseudo global variable
denoting the accumulator. Hence, the value of the accumulator is set 1000 so
the function returns value 1000.
88) int i;
main(){
int t;
for (
t=4;scanf("%d",&i)-t;printf("%d\n",i))
printf("%d--",t--);
}
// If the inputs are 0,1,2,3 find the
o/p
Answer:
4--0
3--1
2--2
Explanation:
Let us assume some x= scanf("%d",&i)-t
the values during execution
will
be,
t i
x
4 0
-4
3 1
-2
2 2
0
89) main(){
int a= 0;int b
= 20;char x =1;char y =10;
if(a,b,x,y)
printf("hello");
}
Answer:
hello
Explanation:
The comma operator has associativity from left to
right. Only the rightmost value is returned and the other values are evaluated
and ignored. Thus the value of last variable y is returned to check in if.
Since it is a non zero value if becomes true so, "hello" will be
printed.
90) main(){
unsigned int i;
for(i=1;i>-2;i--)
printf("c
aptitude");
}
Explanation:
i is an unsigned integer. It is compared with a signed
value. Since the both types doesn't match, signed is promoted to unsigned
value. The unsigned equivalent of -2 is a huge value so condition becomes false
and control comes out of the loop.
91) In the following pgm add a stmt in the function fun such that the address of
'a' gets stored in 'j'.
main(){
int * j;
void fun(int
**);
fun(&j);
}
void fun(int
**k) {
int a =0;
/* add a stmt
here*/
}
Answer:
*k = &a
Explanation:
The
argument of the function is a pointer to a pointer.
92) What are the following notations of defining
functions known as?
i. int
abc(int a,float b)
{
/* some code */
}
ii. int
abc(a,b)
int a;
float b;
{
/*
some code*/
}
Answer:
i. ANSI C
notation
ii. Kernighan & Ritche notation
93) main()
{
char *p;
p="%d\n";
p++;
p++;
printf(p-2,300);
}
Answer:
300
Explanation:
The pointer points to % since it is incremented twice
and again decremented by 2, it points to '%d\n' and 300 is printed.
94) main(){
char a[100];
a[0]='a';a[1]]='b';a[2]='c';a[4]='d';
abc(a);
}
abc(char a[]){
a++;
printf("%c",*a);
a++;
printf("%c",*a);
}
Explanation:
The base address is modified only in function and as a
result a points to 'b' then after incrementing to 'c' so bc will be printed.
95) func(a,b)
int a,b;
{
return( a=
(a==b) );
}
main()
{
int process(),func();
printf("The value of process is %d !\n
",process(func,3,6));
}
process(pf,val1,val2)
int (*pf) ();
int val1,val2;
{
return((*pf) (val1,val2));
}
Answer:
The value if process is 0 !
Explanation:
The function 'process' has 3 parameters - 1, a pointer
to another function 2 and 3, integers.
When this function is invoked from main, the following substitutions for formal
parameters take place: func for pf, 3 for val1 and 6 for val2. This function
returns the result of the operation performed by the function 'func'. The
function func has two integer parameters. The formal parameters are substituted
as 3 for a and 6 for b. since 3 is not equal to 6, a==b returns 0. therefore
the function returns 0 which in turn is returned by the function 'process'.
96) void main()
{
static
int i=5;
if(--i){
main();
printf("%d
",i);
}
}
Answer:
0 0 0 0
Explanation:
The
variable "I" is declared as static, hence memory for I will be
allocated for only once, as it encounters the statement. The function main()
will be called recursively unless I becomes equal to 0, and since main() is
recursively called, so the value of static I ie., 0 will be printed every time
the control is returned.
97) void main()
{
int
k=ret(sizeof(float));
printf("\n
here value is %d",++k);
}
int ret(int ret)
{
ret +=
2.5;
return(ret);
}
Answer:
Here value is 7
Explanation:
The int ret(int ret), ie., the function name and the
argument name can be the same.
Firstly,
the function ret() is called in which the sizeof(float) ie., 4 is passed, after the first expression the value in ret
will be 6, as ret is integer hence the value stored in ret will have implicit
type conversion from float to int. The ret is returned in main() it is printed
after and preincrement.
98) void main()
{
char
a[]="12345\0";
int
i=strlen(a);
printf("here
in 3 %d\n",++i);
}
Answer:
here in 3 6
Explanation:
The
char array 'a' will hold the initialized string, whose length will be counted
from 0 till the null character. Hence the 'I' will hold the value equal to 5,
after the pre-increment in the printf statement, the 6 will be printed.
99) void main()
{
unsigned
giveit=-1;
int
gotit;
printf("%u
",++giveit);
printf("%u
\n",gotit=--giveit);
}
Answer:
0 65535
Explanation:
100) void main()
{
int i;
char
a[]="\0";
if(printf("%s\n",a))
printf("Ok
here \n");
else
printf("Forget
it\n");
}
Answer:
Ok here
Explanation:
Printf will return how many characters does it print.
Hence printing a null character returns 1 which makes the if statement true,
thus "Ok here" is printed.
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