C Questions
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
2
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.
3
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()
4
{
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;
5
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
6
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
7
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:
8
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:
9
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);
10
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
11
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()
{
41printf("%p",main);
}8Answer:
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()
{
12
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()
13
{
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++;
}
}
14
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
15
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
}
16
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;
17
str1=s;
printf("%d",++*p +
++*str1-32);
}
Answer:
M
Explanation:
p is pointing to character '\n'.str1
is pointing to character 'a' ++*p
meAnswer:"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
meAnswer:"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
{
18
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:
19
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( )
{
20
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:
21
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
22
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.
23
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.
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