Note : All the
programs are tested under Turbo C/C++ compilers.
Predict the output or
error(s) for the following:
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).
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 Cit 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).
Last Updated Date: March 22, 2012
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