Note : All the programs are tested under Turbo
C/C++ compilers.
C/C++ compilers.
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
expressing the same
idea. Generally array name is the base address for
that array. Here s is the base
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].
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");
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C-Aptitude chaitanya9186.co.cc
else
printf("I hate U");
}
Answer:
I hate U
Explanation:
For floating point numbers (float, double, long
double) the values
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.
bytes and long double takes 10 bytes.
So float stores 0.9 with less precision than long
double.
double.
Rule of Thumb:
Never compare or at-least be cautious when
using floating point
using floating point
numbers with relational operators (== , >, <, <=,
>=,!= ) .
>=,!= ) .
4. main()
{s
tatic 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
between the function calls. Main is also
treated like any other ordinary function, which
can be called recursively.
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
the first loop, since
only q is incremented and not c , the value 2 will
be printed 5 times. In second loop p
be printed 5 times. In second loop p
itself is incremented. So the values 2 3 4 6 5 will be printed.
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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
allocated in some other program and
that address will be given to the current program
at the time of linking. But linker
at the time of linking. But linker
finds that no other variable of name i is available
in any other program with memory
in any other program with memory
space allocated for it. Hence a linker error has
occurred .
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 also the logical
AND (&&) operator has higher priority over the
logical OR (||) operator. So the
logical OR (||) operator. So the
expression ¡¥i++ && j++ && k++¡¦ is executed
first. The result of this expression is 0
first. The result of this expression is 0
(-1 && -1 && 0 = 0). Now the expression is 0 || 2
which evaluates to 1 (because OR
which evaluates to 1 (because OR
operator always gives 1 except for ¡¥0 || 0¡¦
combination- for which it gives 0). So the
combination- for which it gives 0). So the
value of m is 1. The values of other variables are
also incremented by 1.
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
taken by its operand. P
is a character pointer, which needs one byte for
storing its value (a character). Hence
storing its value (a character). Hence
sizeof(*p) gives a value of 1. Since it needs two
bytes to store the address of the
bytes to store the address of the
character pointer sizeof(p) gives 2.
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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
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
shifted four times the
least significant 4 bits are filled with 0's.The %x
format specifier specifies that the
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
encountered, the compiler
doesn't know anything about the function display.
It assumes the arguments and
It assumes the arguments and
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return types to be integers, (which is the default
type). When it sees the actual
type). When it sees the actual
function display, the arguments and type
contradicts with what it has assumed
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
twice. Same maths
rules applies, ie. minus * minus= plus.
Note:
However you cannot give like --2. Because --
operator can only be
operator can only be
applied to variables as a decrement operator (eg.,
i--). 2 is a constant and not a
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
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 ¡¥
more precedence than ¡¥
>¡¦ symbol. ! is a unary logical operator. !i (!10) is 0
(not of true is false). 0>14 is
(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
character 'a' ++*p. "p is
pointing to '\n' and that is incremented by one."
the ASCII value of '\n' is 10, which is
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
++*str1, str1 is pointing to 'a' that is
incremented by 1 and it becomes 'b'. ASCII value
of 'b' is 98.
of 'b' is 98.
Now performing (11 + 98 ¡V 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
you are trying to
access the third 2D(which you are not declared) it
will print garbage values. *q=***a
will print garbage values. *q=***a
starting address of a is assigned integer pointer.
Now q is pointing to starting address
Now q is pointing to starting address
of a. If you print *q, it will print first element of 3D
array.
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
Hence, the elements are
of yy are to be accessed through the instance of
structure xx, which needs an instance
structure xx, which needs an instance
of yy to be known. If the instance is created after
defining the structure the compiler
defining the structure the compiler
will not know about the instance relative to xx.
Hence for nested structure yy you
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
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.
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